Codeforces 371C Hanburgers

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;

}
        

还可以用贪心
贪心就比较好想了

尽可能地将原有材料耗光，然后买

#include<iostream>
#include<cstdio>
#include<cstring> 
#include<cmath>
using namespace std;
int nb,ns,nc;
int mb,ms,mc;
long long m;
char a[109];
int B,S,C;
long long ans;
int main()
{
    cin>>a;
    for(int i=0;i<strlen(a);i++)
    {
        if(a[i]=='B') B++;
        if(a[i]=='C') C++;
        if(a[i]=='S') S++;
    }
    cin>>nb>>ns>>nc;
    cin>>mb>>ms>>mc;
    cin>>m;
    
    int minn=233333333;
    if(B) minn=min(minn,nb/B);
    if(C) minn=min(minn,nc/C);
    if(S) minn=min(minn,ns/S);
    ans+=minn;
    ns-=S*minn;nb-=B*minn;nc-=C*minn;
    
    if(S==0) ns=0;
    if(C==0) nc=0;
    if(B==0) nb=0;
    
    int zz=B*mb+C*mc+S*ms;
    int tot=0;
    
    while(nb!=0||ns!=0||nc!=0)
    {
        tot=0;
        if(nb>=B)
            nb-=B;
        else
        {
            tot+=(B-nb)*mb;
            nb=0;
        }
        if(nc>=C)
            nc-=C;
        else
        {
            tot+=(C-nc)*mc;
            nc=0;
        }
        if(ns>=S)
            ns-=S;
        else
        {
            tot+=(S-ns)*ms;
            ns=0;
        }
        if(tot>m)
            break;
        else
            m-=tot,ans++; 
    }
    ans+=m/zz;
    cout<<ans;
    return 0;
}