bzoj 1685: [Usaco2005 Oct]Allowance 津贴（贪心）

1685: [Usaco2005 Oct]Allowance 津贴

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 264 Solved: 195
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Description

As a reward for record milk production, Farmer John has decided to start paying Bessie the cow a small weekly allowance. FJ has a set of coins in N (1 <= N <= 20) different denominations, where each denomination of coin evenly divides the next-larger denomination (e.g., 1 cent coins, 5 cent coins, 10 cent coins, and 50 cent coins). Using the given set of coins, he would like to pay Bessie at least some given amount of money C (1 <= C <= 100,000,000) every week. Please help him compute the maximum number of weeks he can pay Bessie.

作为对勤勤恳恳工作的贝茜的奖励，约翰已经决定开始支付贝茜一个小的每周津贴．约翰有n(1≤N≤20)种币值的硬币，面值小的硬币总能整除面值较大的硬币．比如说，币值有如下几种：1美分，5美分，10美分，50美分…．．

利用给定的这些硬币，他将要每周付给贝茜一定金额的津贴C(1≤C≤10^8)．

请帮他计算出他最多能给贝茜发几周的津贴．

Input

第1行：2个用空格隔开的整数n和C.

第2到n+1行：每行两个整数表示一种币值的硬币．第一个整数V(I≤y≤10^8)，表示币值．

第二个整数B(1≤B≤10^6)，表示约翰拥有的这种硬币的个数．

Output

一个整数，表示约翰付给贝茜津贴得最多的周数．

Sample Input

3 6

10 1

1 100

5 120

Sample Output

111

不错的贪心题

思路很简单，所有货币按面值排序，每次都尽量选择面值大的就好了，但是有很多细节要处理

use[i]表示当前周为了凑津贴第i种货币需要用use[i]个

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef struct Coin
{int x, y;bool operator < (const Coin &b) const{if(x>b.x)return 1;return 0;}
}Coin;
Coin s[23];
int use[23];
int main(void)
{int n, c, i, now, ans;scanf("%d%d", &n, &c);for(i=1;i<=n;i++)scanf("%d%d", &s[i].x, &s[i].y);sort(s+1, s+n+1);ans = 0;while(1){now = c;memset(use, 0, sizeof(use));for(i=1;i<=n;i++){if(now>s[i].x){use[i] = min(now/s[i].x, s[i].y);now -= s[i].x*use[i];}}for(i=n;i>=1;i--){if(now!=0 && use[i]+1<=s[i].y){now = 0;use[i]++;break;}else if(now!=0){now += use[i]*s[i].x;use[i] = 0;}}if(now!=0)break;for(i=1;i<=n;i++)s[i].y -= use[i];ans++;}printf("%d\n", ans);return 0;
}