这套题最后一题不会，然后先放一下，最后一题应该是大数据结构题
A:求连续最长严格递增的的串，O（n）简单dp
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 1e5+;
const int INF = 0x3f3f3f3f;
int n,dp[N],a[N],cnt,mx;
int main(){
scanf("%d",&n);a[]=INF;
for(int i=;i<=n;++i){
scanf("%d",&a[i]);
dp[i]=;
if(a[i]>a[i-])dp[i]=dp[i-]+;
mx=max(mx,dp[i]);
}
printf("%d\n",mx);
return ;
}
B:水题，map乱搞
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 1e5+;
const int INF = 2e9;
map<int,int>mp;
int a[],cnt;
int main(){
for(int i=;;++i){
if((1ll<<i)>INF)break;
a[++cnt]=(<<i);
}
LL ret=;
int x,n;scanf("%d",&n);
for(int i=;i<=n;++i){
scanf("%d",&x);
for(int j=cnt;j>;--j){
if(a[j]<=x)break;
int tmp=a[j]-x;
if(mp.find(tmp)!=mp.end())
ret+=mp[tmp];
}
if(mp.find(x)==mp.end())mp[x]=;
++mp[x];
}
printf("%I64d\n",ret);
return ;
}
C:一个典型的二分题，judge如何判断全被覆盖？只要用一下离线求和数组非0就好
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 1e5+;
const int INF = 2e9;
LL a[N],b[N];
int n,m,c[N];
bool judge(LL r){
memset(c,,sizeof(c));
for(int i=;i<=m;++i){
int x=lower_bound(a+,a++n,b[i]-r)-a;
int y=upper_bound(a+,a++n,b[i]+r)-a;
++c[x];--c[y];
}
for(int i=;i<=n;++i){
c[i]+=c[i-];
if(!c[i])return false;
}
return true;
}
int main(){
scanf("%d%d",&n,&m);
for(int i=;i<=n;++i)
scanf("%I64d",&a[i]);
sort(a+,a++n);
n=unique(a+,a++n)-a-;
for(int i=;i<=m;++i)
scanf("%I64d",&b[i]);
sort(b+,b++m);
m=unique(b+,b++m)-b-;
LL l=,r=INF;
while(l<r){
LL mid=(l+r)>>;
if(judge(mid))r=mid;
else l=mid+;
}
printf("%I64d\n",(l+r)>>);
return ;
}
D：一个简单的分类讨论，因为最多走k，以k为周期即可
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 1e5+;
const int INF = 2e9;
LL d,k,a,b,t;
int main(){
scanf("%I64d%I64d%I64d%I64d%I64d",&d,&k,&a,&b,&t);
if(d<=k){
printf("%I64d\n",d*a);
return ;
}
LL ret=k*a;d-=k;
if(d<=k){
ret+=min(d*a+t,d*b);
printf("%I64d\n",ret);
return ;
}
LL t1=k*a+t,t2=k*b;
if(t2<=t1){
printf("%I64d\n",ret+d*b);
return ;
}
else {
ret+=d/k*t1;
d-=d/k*k;
if(d==){printf("%I64d\n",ret);return ;}
t1=t+d*a,t2=d*b;
ret+=min(t1,t2);
printf("%I64d\n",ret);
}
return ;
}
E:求从每个点出发路径长度为k的边权和以及边权最小值，刚开始还以为是快速幂，结果发现发现这条路唯一确定，直接倍增即可
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 1e5+;
const int INF = 2e9;
struct Node{
int v,mn;
LL sum;
}f[N][];
LL retsum[N],k;
int n,retmin[N],cur[N];
void solve(){
for(int j=;(1ll<<j)<=k;++j)if(k&(1ll<<j)){
for(int i=;i<n;++i){
retsum[i]+=f[cur[i]][j].sum;
if(retmin[i]==-)retmin[i]=f[cur[i]][j].mn;
else retmin[i]=min(retmin[i],f[cur[i]][j].mn);
cur[i]=f[cur[i]][j].v;
}
}
for(int i=;i<n;++i)
printf("%I64d %d\n",retsum[i],retmin[i]);
}
int main(){
scanf("%d%I64d",&n,&k);
for(int i=;i<n;++i)scanf("%d",&f[i][].v),cur[i]=i,retmin[i]=-;
for(int i=;i<n;++i)scanf("%d",&f[i][].mn),f[i][].sum=f[i][].mn;
for(int j=;(1ll<<j)<=k;++j){
for(int i=;i<n;++i){
f[i][j].v=f[f[i][j-].v][j-].v;
f[i][j].sum=f[i][j-].sum+f[f[i][j-].v][j-].sum;
f[i][j].mn=min(f[i][j-].mn,f[f[i][j-].v][j-].mn);
}
}
solve();
return ;
}
F：不会，看了看别人的代码，并不能看懂，还是太弱