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二分搜索 Codeforces Round #218 (Div. 2) C. Hamburgers

时间:2015-07-25 19:44:12      阅读:128      评论:0      收藏:0      [点我收藏+]

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题目传送门

 1 /*
 2     题意:一个汉堡制作由字符串得出,自己有一些原材料,还有钱可以去商店购买原材料,问最多能做几个汉堡
 3     二分:二分汉堡个数,判断此时所花费的钱是否在规定以内
 4 */
 5 #include <cstdio>
 6 #include <algorithm>
 7 #include <cmath>
 8 using namespace std;
 9 
10 typedef long long ll;
11 const int MAXN = 1e2 + 10;
12 const int INF = 0x3f3f3f3f;
13 char ham[MAXN];
14 ll nb, ns, nc;
15 ll pb, ps, pc;
16 ll b, s, c;
17 ll m;
18 
19 bool check(ll x)  {
20     ll cost = 0;
21     if (b * x > nb) cost += (b * x - nb) * pb;
22     if (s * x > ns) cost += (s * x - ns) * ps;
23     if (c * x > nc) cost += (c * x - nc) * pc;
24     return cost <= m;
25 }
26 
27 int main(void)  {       //Codeforces Round #218 (Div. 2) C. Hamburgers
28     scanf ("%s", &ham);
29     scanf ("%I64d%I64d%I64d", &nb, &ns, &nc);
30     scanf ("%I64d%I64d%I64d", &pb, &ps, &pc);
31     scanf ("%I64d", &m);
32     b = s = c = 0;
33     for (int i=0; ham[i]; ++i)    {
34         if (ham[i] == B)    b++;
35         else if (ham[i] == S)   s++;
36         else    c++;
37     }
38     ll l = 0, r = 1e13;
39     while (l + 1 < r)   {
40         ll mid = (l + r) >> 1;
41         if (check (mid))    l = mid;
42         else    r = mid;
43     }
44     printf ("%I64d\n", l);
45 
46     return 0;
47 }

 

二分搜索 Codeforces Round #218 (Div. 2) C. Hamburgers

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原文地址:http://www.cnblogs.com/Running-Time/p/4676325.html

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