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 <stdio.h> #include <string.h> #include <stdbool.h> 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; }
