Codeforces 371C Hamburgers 【二分】

题目链接;http://blog.csdn.net/csdn364988181/article/details/48253267

 

题意：

有一个字符串表示制作1个汉堡的菜单。第二行给出拥有的每种材料的个数。第三行给出每种材料的加钱。第四行给出有的钱。问至多能弄到多少个汉堡。

 

题解：

二分。注意要开longlong，上界也要开的足够大。

 

代码：

 

#include 
#include 
#include 

typedef long long LL;
char menu[105];
int nb, ns, nc;
int pb, ps, pc;
int hb, hs, hc, len;
LL have;

LL l = 1LL, r = 2000000000000LL;

bool isok(LL num) {
	LL tb, ts, tc;
	tb = (num*hb-nb)*pb; ts = (num*hs-ns)*ps; tc = (num*hc-nc)*pc;

	if(tb < 0) tb = 0; if(ts < 0) ts = 0; if(tc < 0) tc = 0;

	return tb+ts+tc <= have;
}

int main() {
	int i = 0;
	scanf("%s %d %d %d %d %d %d %I64d", menu, &nb, &ns, &nc, &pb, &ps, &pc, &have);
	len = strlen(menu);

	for ( i = 0; i < len; i ++ ) {
		if(menu[i] == 'B') hb++;
		else if(menu[i] == 'S') hs ++;
		else if(menu[i] == 'C') hc ++;
	}

	int ans = -1;
	while ( l < r ) {
		LL mid = l+r >> 1;
		if( isok(mid) ) {
			l = mid+1;
		} else r = mid;
	}
	printf("%I64d\n", l-1);

	return 0;
}