Can I use Chinese? ↵
这个题是一道二分题目 ↵
he question is a two-point question. ↵
题意很简单，不需要解释了 ↵
It is very simple, do not need to explain. ↵
但是我一直没过样例，原因是我的变量类型搞错了 ↵
But I haven't had the sample because my variables are wrong. ↵
应该是`long long`，但是我写成了int ↵
It should be `long long`, but I use the `int`. ↵
下面是代码 ↵
Next is my code. ↵
↵
~~~~~↵
#include <iostream>↵
#include <cstring>↵
#include <cstdio>↵
using namespace std;↵
int q[5]; // B C S↵
int c[5];↵
int m[5];↵
char s[110];↵
int B,C,S;↵
long long tot,ans;↵
int check(long long n)↵
{↵
long long tmp=tot;↵
if(n*c[1]>q[1]) tmp-=(n*c[1]-q[1])*m[1];↵
if(n*c[2]>q[2]) tmp-=(n*c[2]-q[2])*m[2];↵
if(n*c[3]>q[3]) tmp-=(n*c[3]-q[3])*m[3];↵
if(tmp>=0)↵
return 1;↵
else↵
return 0;↵
}↵
int main()↵
{↵
cin>>s;↵
for(int i=0;i<strlen(s);i++)↵
{↵
if(s[i]=='B') c[1]++;↵
if(s[i]=='S') c[2]++;↵
if(s[i]=='C') c[3]++;↵
}↵
cin>>q[1]>>q[2]>>q[3];↵
cin>>m[1]>>m[2]>>m[3];↵
cin>>tot;↵
int d=tot;↵
long long l=1,r=2333333333333;↵
long long mid;↵
while(l<=r)↵
{↵
mid=(l+r)/2;↵
if(check(mid))↵
l=mid+1,ans=mid;↵
else r=mid-1;↵
}↵
cout<<ans;↵
return 0;↵
↵
}↵
~~~~~↵
这个题是一道二分题目 ↵
he question is a two-point question. ↵
题意很简单，不需要解释了 ↵
It is very simple, do not need to explain. ↵
但是我一直没过样例，原因是我的变量类型搞错了 ↵
But I haven't had the sample because my variables are wrong. ↵
应该是`long long`，但是我写成了int ↵
It should be `long long`, but I use the `int`. ↵
下面是代码 ↵
Next is my code. ↵
↵
~~~~~↵
#include <iostream>↵
#include <cstring>↵
#include <cstdio>↵
using namespace std;↵
int q[5]; // B C S↵
int c[5];↵
int m[5];↵
char s[110];↵
int B,C,S;↵
long long tot,ans;↵
int check(long long n)↵
{↵
long long tmp=tot;↵
if(n*c[1]>q[1]) tmp-=(n*c[1]-q[1])*m[1];↵
if(n*c[2]>q[2]) tmp-=(n*c[2]-q[2])*m[2];↵
if(n*c[3]>q[3]) tmp-=(n*c[3]-q[3])*m[3];↵
if(tmp>=0)↵
return 1;↵
else↵
return 0;↵
}↵
int main()↵
{↵
cin>>s;↵
for(int i=0;i<strlen(s);i++)↵
{↵
if(s[i]=='B') c[1]++;↵
if(s[i]=='S') c[2]++;↵
if(s[i]=='C') c[3]++;↵
}↵
cin>>q[1]>>q[2]>>q[3];↵
cin>>m[1]>>m[2]>>m[3];↵
cin>>tot;↵
int d=tot;↵
long long l=1,r=2333333333333;↵
long long mid;↵
while(l<=r)↵
{↵
mid=(l+r)/2;↵
if(check(mid))↵
l=mid+1,ans=mid;↵
else r=mid-1;↵
}↵
cout<<ans;↵
return 0;↵
↵
}↵
~~~~~↵
