【POJ】3040 - Allowance（贪心）

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Allowance

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2940 Accepted: 1189

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 ompute the maximum number of weeks he can pay Bessie.

Input

* Line 1: Two space-separated integers: N and C 

* Lines 2..N+1: Each line corresponds to a denomination of coin and contains two integers: the value V (1 <= V <= 100,000,000) of the denomination, and the number of coins B (1 <= B <= 1,000,000) of this denomation in Farmer John's possession.

Output

* Line 1: A single integer that is the number of weeks Farmer John can pay Bessie at least C allowance

Sample Input

3 610 11 1005 120

Sample Output

111

Hint

INPUT DETAILS: 
FJ would like to pay Bessie 6 cents per week. He has 100 1-cent coins,120 5-cent coins, and 1 10-cent coin. 

OUTPUT DETAILS: 
FJ can overpay Bessie with the one 10-cent coin for 1 week, then pay Bessie two 5-cent coins for 10 weeks and then pay Bessie one 1-cent coin and one 5-cent coin for 100 weeks.

Source

USACO 2005 October Silver

这个贪心题真心好，但是也挺麻烦。

如果钱数大于等于C的，那么就直接发出。如果小于 c 的，则从大往小拿，可以等于但是不能大于 c ，如果等于 c 就发出去，如果到最后也没有等于 c 的，再从小往大拿，能发就发，如果反向滚回来还不能发出去的话，剩下的钱就不能发了，直接退出就行。（实力滚键盘）

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node{__int64 v,m;}coin[22];bool cmp(node a,node b){return a.v > b.v;}int main(){__int64 n,c;__int64 ans;while (~scanf ("%I64d %I64d",&n,&c)){for (int i = 0 ; i < n ; i++)scanf ("%I64d %I64d",&coin[i].v,&coin[i].m);sort (coin , coin + n , cmp);ans = 0;bool flag = true;//是否能继续拿 for (int i = 0 ; i < n ; i++){if (coin[i].v >= c)ans += coin[i].m;else//首先从大往小拿 {while (coin[i].m){int t = 0;//已经拿的钱数 for (int j = i ; j < n ; j++){if (!coin[j].m)//没钱了就跳过 continue;while (coin[j].m && t < c){t += coin[j].v;coin[j].m--;}if (t == c)//如果正好发完就发了 {ans++;break;}else//否则就再往下搜 {t -= coin[j].v;coin[j].m++;}}if (t == c)continue;for (int j = n - 1 ; j >= i ; j--)//不够再从小往大拿 {if (!coin[j].m)continue;if (t + coin[j].m * coin[j].v >= c)//够拿就拿{ans++;coin[j].m -= (c - t + coin[j].v - 1) / coin[j].v;t = c;//管它后来是多少钱，就按c算 break;}else//不够就先拿完 {t += coin[j].m * coin[j].v;coin[j].m = 0;}}if (t < c)//如果还不够拿，则无法再发工资{flag = false; break;}}}if (!flag)break;}printf ("%I64d\n",ans);}return 0;}