A. PizzaForces
打表就会神奇的发现规律(当时想错了错好几发)
#include "bits/stdc++.h"
#define pll __builtin_popcount
#define all(a) a.begin(),a.end()
#define pb push_back
using namespace std;
using ll = long long ;
inline void solve()
{
ll n;cin>>n;
if(n<=6){
cout<<15<<"\n";
}else{
if(n%2==0)n--;
n-=4;
n=n/2+1;
cout<<n*5+10<<"\n";
}
}
int main()
{
//freopen("out.txt","w",stdout);
//int a[]={15,15,15,15,15,15,20,20,25,25,30,30,35,35,40,40,45,45,50,50,55,55,60,60,65,65,70,70,75,75,80,80,85,85,90,90,95,95,100,100,105,105,110,110,115,115,120,120,125,125,130,130,135,135,140,140,145,145,150,150,155,155,160,160,165,165,170,170,175,175,180,180,185,185,190,190,195,195,200,200,205,205,210,210,215,215,220,220,225,225,230,230,235,235,240,240,245,245,250,250,255,255,260,260,265,265,270,270,275,275,280,280,285,285,290,290,295,295,300,300,305,305,310,310,315,315,320,320,325,325};
//for(int i=1;i<=130;i++)mp[i]=a[i-1];
ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
solve();
return 0;
}
B. Two Tables
尽可能把能移动的对称到右上角不能移动的放左下角
#include "bits/stdc++.h"
#define pll __builtin_popcount
#define all(a) a.begin(),a.end()
#define pb push_back
using namespace std;
using ll = long long ;
void solve()
{
int w,h;
cin>>w>>h;
int a,b,c,d;
cin>>a>>b>>c>>d;
if(a<w-c){
int p=w-c-a;
a+=p;
c+=p;
}
if(b<h-d){
int p=h-d-b;
b+=p;
d+=p;
}
int u,v;
cin>>u>>v;
if(a<u&&b<v){
int ans=0x3f3f3f3f;
int p=u-a;
if(c+p<=w)ans=p;
p=v-b;
if(d+p<=h)ans=min(ans,p);
if(ans==0x3f3f3f3f){
cout<<"-1\n";return ;
}
cout<<(double)ans<<"\n";
return ;
}
cout<<(double)0<<"\n";
}
int main()
{
cout<<fixed<<setprecision(9);
ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
solve();
return 0;
}
C. Coin Rows
题意Alice和Bob想走到2,m且Alice走过的路的权值都会变成0，Alice想让Bob走的权值的和最小，Bob与之相反问答案在二者最优的情况是多少（前缀和模拟下就行了）
#include "bits/stdc++.h"
#define pll __builtin_popcount
#define all(a) a.begin(),a.end()
#define pb push_back
using namespace std;
using ll = long long ;
ll a[3][100002];
void solve()
{
int n;
cin>>n;
for(int i=1;i<=2;i++)
for(int j=1;j<=n;j++)cin>>a[i][j];
for(int i=1;i<=2;i++)
for(int j=1;j<=n;j++)
a[i][j]+=a[i][j-1];
ll ans=1e18;
for(int i=1;i<=n;i++){
ans=min(ans,max(a[1][n]-a[1][i],a[2][i-1]-a[2][0]));
}
cout<<ans<<"\n";
}
int main()
{
//cout<<fixed<<setprecision(9);
ios::sync_with_stdio(false);
int T=1;
cin>>T;
while(T--)
solve();
return 0;
}