【CF-371C】Hamburgers

题目链接：http://codeforces.com/problemset/problem/371/C

赤果果的大水题！！

题目大意：给你一个汉堡的配料，and现有的每种原料的个数，and每种原料的价格，and你有的money；问最多能做几个汉堡。

obviously 这道题具有二分性质 Thus 直接二分答案判断买x个汉堡时原料以及钱是否is enough；

第一眼看这道题，“哇，这是神题？”是的，我没戴眼镜；

Get down to the business，上代码

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<iostream>
 5 #include<cstdlib>
 6 #include<string>
 7 #include<cstring>
 8 #include<queue>
 9 #include<deque>
10 #include<stack>
11 #define LL long long
12 using namespace std;
13 char pl[105];
14 int nb,ns,nc,pb,ps,pc,b,s,c;
15 LL r;
16 bool check(LL x){
17     LL cb = max(((LL)b*x-nb)*pb,(LL)0);
18     LL cs = max(((LL)s*x-ns)*ps,(LL)0);
19     LL cc = max(((LL)c*x-nc)*pc,(LL)0);
20     if(cb+cs+cc <= r) return true;
21     else return false;
22 }
23 int main(){
24     scanf("%s",pl);
25     for(int i = 0;i < strlen(pl);i++){
26         if(pl[i] == 'B') b++;
27         else if(pl[i] == 'S') s++;
28         else if(pl[i] == 'C') c++;
29     }
30     scanf("%d%d%d",&nb,&ns,&nc);
31     scanf("%d%d%d",&pb,&ps,&pc);
32     scanf("%lld",&r);
33     LL l = 0,r = pow(10,12)*2,mid;
34     while(l < r){
35         mid = (l + r)>>1;
36         if(check(mid))
37            l = mid+1;
38         else r = mid;
39     }
40     printf("%lld
",l-1);
41     return 0;
42 }