贪心part
1
http://ybt.ssoier.cn:8088/problem_show.php?pid=1223
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
int a[31];
int n;
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
while(scanf("%d",&n),n)
{
memset(a,0,sizeof(a));
int cnt=0;
int s=0,pos=1,sum=0;
for(int j=n;j;j>>=1) a[++cnt]=j&1;a[++cnt]=0;//一般要多一位防止进位
bool flag=false;
for(int j=1;j<=cnt;j++)
{
if(flag)
{
if(!a[j]) {sum+=1<<(j-1),sum-=1<<(s-1);break;}//位数包含第0位，当前位置是1<<j-1
else sum+=1<<s,s++,n-=(n&-n);
}
else
{
if(!a[j])continue ;
else s++,flag=1,sum+=1,n-=(n&-n);
}
}
sum+=n;
printf("%d\n",sum);
}
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1224
枚举左右顶点等于枚举矩阵
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
int n;
const int N=1e3;
int a[N][N],sum[N][N];
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
n=read();
rep(i,1,n+1)
rep(j,1,n+1) a[i][j]=read(),sum[i][j]+=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j];
int ans=-INF;
rep(i,1,n+1)
rep(j,1,n+1)
rep(k,i,n+1)
rep(l,j,n+1)
ans=max(ans,sum[k][l]-sum[i-1][l]-sum[k][j-1]+sum[i-1][j-1]);
printf("%d\n",ans);
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1225
大的先放，不足的取比例
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
const int N=1e4+52;
struct jewel
{
int v,w;
double val;
bool operator <(const jewel p) const
{
return val> p.val;
}
}a[N];
int k,w,s;
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
k=read();
while(k--)
{
w=read(),s=read();
rep(i,1,s+1) a[i].w=read(),a[i].v=read(),a[i].val=a[i].v*1.0/a[i].w;
sort(a+1,a+1+s);
double sum=0;
rep(i,1,s+1)
{
if(a[i].w<=w) w-=a[i].w,sum+=a[i].v;
else {sum+=a[i].val*w;break;}
}
printf("%.2lf\n",sum);
}
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1226
箱子和箱子匹配，尽可能的利用空间
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
int a,b,c,d,e,f;
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
while(scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f),a||b||c||d||e||f)
{
int res=0;
res+=f;
res+=e;a=max(0,a-6*e);
res+=d;
if(b-5*d<0) a=max(0,a-(5*d-b)*4);
b=max(0,b-5*d);
res+=(c+3)/4;
if(c%4!=0)
{
c%=4;
if(c==1) b=max(0,b-5),a=max(0,a-7);
if(c==2) b=max(0,b-3),a=max(0,a-6);
if(c==3) b=max(0,b-1),a=max(0,a-5);
}
if(b) res+=(b+8)/9;
if(b%9!=0) a=max(0,a-(36-b%9*4));
if(a) res+=(a+35)/36;
/*
res+=e+f+d+(c+3)/4;
x=5*d+g[c%4];
if(b>x) n+=(b-x+8)/9;
y=36*n-36*f-25*e-16*d-9*c-4*b;
if(a>y) n+=(a-y+35)/36;
cout<<n<<'\n';
*/
printf("%d\n",res);
}
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1233
非常严重的错误，每次选取最小，INF要在内部更新，否则的话，一次最小永远都是最小。
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
const int N=1e4+52;
int n,m;
int a[N];
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
n=read(),m=read();
int minn=INF;
int pos=0;
rep(i,1,n+1) a[i]=read();
rep(i,m+1,n+1)//非常严重的错误，INF要在内部更新，因为每次都是在里面挑选最小的然后更新，但是在第一次最小之后，所有的就没有比他更小的了，那么就直接失败了
{
int minn=INF;
rep(j,1,m+1)
{
if(minn>a[j]) minn=a[j],pos=j;
}
a[pos]+=a[i];
}
int maxn=-INF;
rep(i,1,m+1) maxn=max(maxn,a[i]) ;
cout<<maxn<<'\n';
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1232
特殊的情况先列出来，然后两种选择去递归，每次只能运一个人，运到最后两个人一起运。要么运输时间最短，要么回来时间最短，能者多劳。
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
//#include <iostream>
//#include <cstdio>
//#include <windows.h>
//#include <cstdlib>
//#include <ctime>
//using namespace std;
//int main()
//{
// int ok = 0;
// int n = 50;
// for (int i = 1; i <= n; ++i)
// {
// system("make.exe > make.txt");
// system("std.exe < make.txt > std.txt");
// double begin = clock();
// system("baoli.exe < make.txt > baoli.txt");
// double end = clock();
//
// double t = (end - begin);
// if (system("fc std.txt baoli.txt"))
// {
// printf("测试点#%d Wrong Answer\n", i);
// }
// else if (t > 1000) //1秒
// {
// printf("测试点#%d Time Limited Exceeded 用时 %.0lfms\n", i, t);
// }
// else
// {
// printf("测试点#%d Accepted 用时%.0lfms\n", i, t);
// ok++; //AC数量+1
// }
// }
// printf("\n");
// double res = 100.0 * ok / n;
// printf("共 %d 组测试数据，AC数据 %d 组。 得分%.1lf。", n, ok, res);
//
// Sleep(1000); //休眠1秒，为了节约对拍次数。
//}
const int N=1e5+52;
int t,n;
int a[N];
int cross(int n)
{
int t1,t2;
if(n==1) return a[1];
else if (n==2 ) return a[2];
else if(n==3) return a[1]+a[2]+a[3];
else
{
t1=2*a[1]+a[n-1]+a[n];//两种选择
t2=a[1]+2*a[2]+a[n];
return min(t1,t2)+cross(n-2);//每次少一个人，只要到特殊情况直接退出。
}
}
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
#ifdef my_dp
while(1)
{
system("data.exe > in.txt");
system("brute.exe < in.txt > brute.txt");
system("std.exe < in.txt > std.txt");
if(system("fc std.txt brute.txt"))
break;
}
#endif
#ifdef my_dp2
while (1) //一直循环，直到找到不一样的数据
{
system("data.exe");
system("baoli.exe");
system("std.exe");
if (system("fc std.txt baoli.txt")) //当 fc 返回1时，说明这时数据不一样
break; //不一样就跳出循环
}
#endif
clock_t start,end;
start=clock();
t=read();
while(t--)
{
n=read();
rep(i,1,n+1) a[i]=read();//两个思路要么过河时间尽量少，要么回来的时间尽量少，那么就是要么大的捆绑，要么就是能者多劳，全让a去运。
sort(a+1,a+1+n);
printf("%d\n",cross(n));
}
end=clock();
// printf("%lfms\n",(double)(end-start)/CLOCKS_PER_SEC*1000);
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1230
实际上就是从后往前选取界限点，对于最大的超出界限点作为新的界限点，这里只要排序一下，然后选取即可。
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
//#include <iostream>
//#include <cstdio>
//#include <windows.h>
//#include <cstdlib>
//#include <ctime>
//using namespace std;
//int main()
//{
// int ok = 0;
// int n = 50;
// for (int i = 1; i <= n; ++i)
// {
// system("make.exe > make.txt");
// system("std.exe < make.txt > std.txt");
// double begin = clock();
// system("baoli.exe < make.txt > baoli.txt");
// double end = clock();
//
// double t = (end - begin);
// if (system("fc std.txt baoli.txt"))
// {
// printf("测试点#%d Wrong Answer\n", i);
// }
// else if (t > 1000) //1秒
// {
// printf("测试点#%d Time Limited Exceeded 用时 %.0lfms\n", i, t);
// }
// else
// {
// printf("测试点#%d Accepted 用时%.0lfms\n", i, t);
// ok++; //AC数量+1
// }
// }
// printf("\n");
// double res = 100.0 * ok / n;
// printf("共 %d 组测试数据，AC数据 %d 组。 得分%.1lf。", n, ok, res);
//
// Sleep(1000); //休眠1秒，为了节约对拍次数。
//}
const int N=1e5+52;
int n;
struct point
{
int x1,y1;
bool operator <(const point &w) const
{
return x1<w.x1||(x1==w.x1&&y1<w.y1);
}
}a[N];
PII ans[N];
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
#ifdef my_dp
while(1)
{
system("data.exe > in.txt");
system("brute.exe < in.txt > brute.txt");
system("std.exe < in.txt > std.txt");
if(system("fc std.txt brute.txt"))
break;
}
#endif
#ifdef my_dp2
while (1) //一直循环，直到找到不一样的数据
{
system("data.exe");
system("baoli.exe");
system("std.exe");
if (system("fc std.txt baoli.txt")) //当 fc 返回1时，说明这时数据不一样
break; //不一样就跳出循环
}
#endif
clock_t start,end;
start=clock();
n=read();
rep(i,1,n+1) a[i].x1=read(),a[i].y1=read();
sort(a+1,a+1+n);
int t=a[n].y1;
int cnt=0;
ans[cnt++]={a[n].x1,a[n].y1};
per(i,n,1)
if(a[i].y1>t) t=a[i].y1,ans[cnt++]={a[i].x1,a[i].y1};
per(i,cnt,0)
printf("(%d,%d)%c",ans[i].x,ans[i].y,(i==0?'\n':','));
end=clock();
// printf("%lfms\n",(double)(end-start)/CLOCKS_PER_SEC*1000);
return 0;
}
8
http://ybt.ssoier.cn:8088/problem_show.php?pid=1228
简单的奶牛堆积问题
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
//#include <iostream>
//#include <cstdio>
//#include <windows.h>
//#include <cstdlib>
//#include <ctime>
//using namespace std;
//int main()
//{
// int ok = 0;
// int n = 50;
// for (int i = 1; i <= n; ++i)
// {
// system("make.exe > make.txt");
// system("std.exe < make.txt > std.txt");
// double begin = clock();
// system("baoli.exe < make.txt > baoli.txt");
// double end = clock();
//
// double t = (end - begin);
// if (system("fc std.txt baoli.txt"))
// {
// printf("测试点#%d Wrong Answer\n", i);
// }
// else if (t > 1000) //1秒
// {
// printf("测试点#%d Time Limited Exceeded 用时 %.0lfms\n", i, t);
// }
// else
// {
// printf("测试点#%d Accepted 用时%.0lfms\n", i, t);
// ok++; //AC数量+1
// }
// }
// printf("\n");
// double res = 100.0 * ok / n;
// printf("共 %d 组测试数据，AC数据 %d 组。 得分%.1lf。", n, ok, res);
//
// Sleep(1000); //休眠1秒，为了节约对拍次数。
//}
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
const int N=1e6+52;
int n,b;
int a[N];
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
#ifdef my_dp
while(1)
{
system("data.exe > in.txt");
system("brute.exe < in.txt > brute.txt");
system("std.exe < in.txt > std.txt");
if(system("fc std.txt brute.txt"))
break;
}
#endif
#ifdef my_dp2
while (1) //一直循环，直到找到不一样的数据
{
system("data.exe");
system("baoli.exe");
system("std.exe");
if (system("fc std.txt baoli.txt")) //当 fc 返回1时，说明这时数据不一样
break; //不一样就跳出循环
}
#endif
clock_t start,end;
start=clock();
n=read(),b=read();
int cnt=0;
rep(i,1,n+1) a[i]=read();
sort(a+1,a+1+n);
int sum=0;
per(i,n+1,1)
{
sum+=a[i];cnt++;
if(sum>=b) break;
}
printf("%d\n",cnt);
end=clock();
// printf("%lfms\n",(double)(end-start)/CLOCKS_PER_SEC*1000);
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1227
简单的题目，但是有点巧妙，大概是cf第二题的难度，需要把握题目，关键就是同步骑行，首先如果先出发，那么不用管，一定不会同步运行，因为比他慢，我有更快的，如果比他快，永远不会相遇。那么只要关注后面的即可，这是一定会相遇的，但是不一定是4500内相遇。问题实质上，是他会和他相遇到的最快的一同到达最后。然后只要把0算进去，那么即使没有相遇，也有一个基础的速度，也即在所有 的可能性中找一个时间最短的。
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
//#include <iostream>
//#include <cstdio>
//#include <windows.h>
//#include <cstdlib>
//#include <ctime>
//using namespace std;
//int main()
//{
// int ok = 0;
// int n = 50;
// for (int i = 1; i <= n; ++i)
// {
// system("make.exe > make.txt");
// system("std.exe < make.txt > std.txt");
// double begin = clock();
// system("baoli.exe < make.txt > baoli.txt");
// double end = clock();
//
// double t = (end - begin);
// if (system("fc std.txt baoli.txt"))
// {
// printf("测试点#%d Wrong Answer\n", i);
// }
// else if (t > 1000) //1秒
// {
// printf("测试点#%d Time Limited Exceeded 用时 %.0lfms\n", i, t);
// }
// else
// {
// printf("测试点#%d Accepted 用时%.0lfms\n", i, t);
// ok++; //AC数量+1
// }
// }
// printf("\n");
// double res = 100.0 * ok / n;
// printf("共 %d 组测试数据，AC数据 %d 组。 得分%.1lf。", n, ok, res);
//
// Sleep(1000); //休眠1秒，为了节约对拍次数。
//}
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
#ifdef my_dp
while(1)
{
system("data.exe > in.txt");
system("brute.exe < in.txt > brute.txt");
system("std.exe < in.txt > std.txt");
if(system("fc std.txt brute.txt"))
break;
}
#endif
#ifdef my_dp2
while (1) //一直循环，直到找到不一样的数据
{
system("data.exe");
system("baoli.exe");
system("std.exe");
if (system("fc std.txt baoli.txt")) //当 fc 返回1时，说明这时数据不一样
break; //不一样就跳出循环
}
#endif
clock_t start,end;
start=clock();
int v,t,n;
while(n=read(),n)
{
double time,ans=9999999;
rep(i,1,n+1)
{
v=read(),t=read();
if(t<0)continue ;
time=(4500*1.0/v*3.6);
time=ceil(time)+t;//很巧妙的一题， 大概是cf的第二题的难度，抓住同步行走，先走的永远不会同步，后走的一定会相遇。
//
这里有个注意点，就是第一次同步的时候一定要计算，因为后走的有可能在后面才会遇上不会提供加速
if(ans>time) ans=time;
}
cout<<ans<<'\n';
}
end=clock();
// printf("%lfms\n",(double)(end-start)/CLOCKS_PER_SEC*1000);
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1229
异常的情况就是最大的超出了剩余的，那么无法两两匹配，反之，则成立，并且作为一个整体，两两匹配，通过对内部的合理分配实现完全消耗，奇数通过分配消耗，偶数，内部持平降低偏差，均衡化，最后完全消耗
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
//#include <iostream>
//#include <cstdio>
//#include <windows.h>
//#include <cstdlib>
//#include <ctime>
//using namespace std;
//int main()
//{
// int ok = 0;
// int n = 50;
// for (int i = 1; i <= n; ++i)
// {
// system("make.exe > make.txt");
// system("std.exe < make.txt > std.txt");
// double begin = clock();
// system("baoli.exe < make.txt > baoli.txt");
// double end = clock();
//
// double t = (end - begin);
// if (system("fc std.txt baoli.txt"))
// {
// printf("测试点#%d Wrong Answer\n", i);
// }
// else if (t > 1000) //1秒
// {
// printf("测试点#%d Time Limited Exceeded 用时 %.0lfms\n", i, t);
// }
// else
// {
// printf("测试点#%d Accepted 用时%.0lfms\n", i, t);
// ok++; //AC数量+1
// }
// }
// printf("\n");
// double res = 100.0 * ok / n;
// printf("共 %d 组测试数据，AC数据 %d 组。 得分%.1lf。", n, ok, res);
//
// Sleep(1000); //休眠1秒，为了节约对拍次数。
//}
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
#ifdef my_dp
while(1)
{
system("data.exe > in.txt");
system("brute.exe < in.txt > brute.txt");
system("std.exe < in.txt > std.txt");
if(system("fc std.txt brute.txt"))
break;
}
#endif
#ifdef my_dp2
while (1) //一直循环，直到找到不一样的数据
{
system("data.exe");
system("baoli.exe");
system("std.exe");
if (system("fc std.txt baoli.txt")) //当 fc 返回1时，说明这时数据不一样
break; //不一样就跳出循环
}
#endif
clock_t start,end;
start=clock();
//一个同样很巧妙的题，同上一题，是这样的，显然一堆电池，一定是两两配对消耗的，分类来看，如果有一个超出了剩余之和，那么显然无法全部匹配完
//反之，如果不存在这样的异常存在，那么作为一个整体，一定可以两两配对，对整体里面的合理分配即可。
//奇数看情况去掉，偶数则不断的降低偏差程度，均衡，到最后刚好一对一。
int n;
while(~scanf("%d",&n))
{
int k;
int sum=0,maxn=0;
rep(i,1,n+1)k=read(),maxn=max(maxn,k),sum+=k;
if(maxn>(sum-maxn)) printf("%.1lf\n",(sum-maxn)*1.0);
else printf("%.1lf\n",sum*1.0/2);
}
end=clock();
// printf("%lfms\n",(double)(end-start)/CLOCKS_PER_SEC*1000);
return 0;
}
http://ybt.ssoier.cn:8088/problem_show.php?pid=1231
这题很麻烦，如果直接去模拟的话，只有提高复杂度，来减少代码量，因为如果你使用直接去做的话，会面临位置变化控制问题，谁和谁比较，这意味着你无法使用flag来标记，必须执行删除操作，同时删完之后还要保证下一位的比较位置正确，其次，在删除的过程中，大小的变化，还有有可能数组越界，以及删完了，然后把界外数字移动进来了，所以这也是一个问题，比较的时候有可能会把界外数字也放进来了，这就很麻烦，因此你只有一次一次的去做，还有就是，如果有多个变量，一定不要全部用i，一定不要混用，一定不要用做，血泪的教训
点击查看代码
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=int(l);i<int(r);i++)
#define per(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define repg(i,l,r) for(i=int(l);i<int(r);i++)
#define perg(i,r,l) for(int i=int(r-1);i>=int(l);i--)
#define rept(i,t) for(int i=int(h[t]);~i;i=ne[i])
#define reph(i,t,h) for(int i=int(h[t]);~i;i=ne[i])
#define x first
#define y second
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef unsigned long long lll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF=0x3f3f3f3f,mod=1e9+7;
const ll INFF=1e18;
const double eps=1e-9;
mt19937 mrand(random_device{}());
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << " " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
rep(i, 0, sz(v)) {
if(i) out << " ";
out << v[i];
}
return out;
}
int rnd(int x){return mrand()%x;}
ll qmi(ll a,ll b){ll res=1;a%=mod;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
ll qmul(ll a,ll b){ll res=0;a%=mod;while(b){if(b&1) res=(res+a)%mod;a=(a+a)%mod;b>>=1;} return res;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
/*ll read()
{
ll x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')x=x*10+(c-'0'),c=getchar();
return f*x;
}*/
//#include <iostream>
//#include <cstdio>
//#include <windows.h>
//#include <cstdlib>
//#include <ctime>
//using namespace std;
//int main()
//{
// int ok = 0;
// int n = 50;
// for (int i = 1; i <= n; ++i)
// {
// system("data.exe > data.txt");
// system("std.exe < data.txt > std.txt");
// double begin = clock();
// system("brute.exe < data.txt > brute.txt");
// double end = clock();
//
// double t = (end - begin);
// if (system("fc std.txt brute.txt"))
// {
// printf("测试点#%d Wrong Answer\n", i);
// }
// else if (t > 1000) //1秒
// {
// printf("测试点#%d Time Limited Exceeded 用时 %.0lfms\n", i, t);
// }
// else
// {
// printf("测试点#%d Accepted 用时%.0lfms\n", i, t);
// ok++; //AC数量+1
// }
// }
// printf("\n");
// double res = 100.0 * ok / n;
// printf("共 %d 组测试数据，AC数据 %d 组。 得分%.1lf。", n, ok, res);
//
// Sleep(1000); //休眠1秒，为了节约对拍次数。
//}
inline int read()
{
int x=0;
char ch;
bool fx=false;
do ch=getchar();while(~ch&&ch!='-'&&(ch<48||ch>57));
if(ch=='-')fx=true,ch=getchar();
for(;ch>47&&ch<58;ch=getchar())
x=(x<<1)+(x<<3)+(ch^48);
return fx?-x:x;
}
int n,k;
int a[32];
bool flag[32];
int main()
{
#ifdef my_test
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
#ifdef my_dp
while(1)
{
system("data.exe > in.txt");
system("brute.exe < in.txt > brute.txt");
system("std.exe < in.txt > std.txt");
if(system("fc std.txt brute.txt"))
break;
}
#endif
#ifdef my_dp2
while (1) //一直循环，直到找到不一样的数据
{
system("data.exe");
system("baoli.exe");
system("std.exe");
if (system("fc std.txt baoli.txt")) //当 fc 返回1时，说明这时数据不一样
break; //不一样就跳出循环
}
#endif
clock_t start,end;
start=clock();
int t;t=read();
//很烦的一道题，题解给出的答案是提高复杂度，减少代码
//原因在于如果采用直接模拟，那么会有如下的问题，第一个删除移位，会不会越界，第二个如果连续删，会删光，删光之后怎么办，第三个删除了之后谁和谁比较，保持在哪个位置上。
//第四个删到什么程度停止，然后最后如果没有倒序的话，从哪里处理，最后的输出
while(t--)
{
int cnt=0,num=0;
n=read(),k=read();
while(n) a[cnt++]=n%10,n/=10;
rep(x,1,k+1)
{
per(i,cnt,1)
{
if(a[i]>a[i-1])
{
num++;
rep(j,i,cnt)a[j]=a[j+1];
cnt--;
break;
}
}
if(num>=k) break;
}
if(num<k)
{
int s=k-num;
rep(i,0,cnt) a[i]=a[i+s];
cnt-=s;
}
per(i,cnt,0) printf("%d",a[i]);
puts("");
}
end=clock();
// printf("%lfms\n",(double)(end-start)/CLOCKS_PER_SEC*1000);
return 0;
}