poj 3040 Allowance

                                                                                                      Allowance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3008		Accepted: 1218

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 6
10 1
1 100
5 120

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.

 

题意：有各种面值的钞票，每种都有若干张，每个星期都要用这些钞票的一部分去支付C元及以上的费用，最终要使得支付的星期尽可能的多，那么最多可以支付多少星期

思路：首先按面值从大到小来排序，若钞票的面值大于等于C，只能一张一张先付掉，之后只剩下比C小的钞票，这些钞票进行各种组合使得sum大于C，具体组合方法：

先按面值从大到小循环，用一个数组来记录各种各种面值的钞票各用了多少张，使得总和sum尽可能接近C且比C小一点；第二步：按面值从小到大循环，继续用上面所述数组来记录各种面值的钞票还需要用多少张，最终使得总和sum超过C，但最多只能超过一点点。

组合出sum之后，看看这种组合最多能有几组。之后再进行下一次新的组合……直到所有的组合都用完了，结束。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
using namespace std;
const int V_MAX = 20;
struct money {
    int V;
    int num;
    bool operator <(const money&b) {
        return V > b.V;
    }
};
money mon[V_MAX];
int need[V_MAX];
int main() {
    int N, C;
    while (cin >> N >> C) {
        int week = 0;
        for (int i = 0;i < N;i++) {
            scanf("%d%d", &mon[i].V, &mon[i].num);
            if (mon[i].V >= C) { 
                week += mon[i].num;//面值大于C的钱只能一张一星期的给
                mon[i].num = 0;
            }
        }
        sort(mon,mon+N);//按面值从大到小排
        while (1) {//每一次循环将产生一种能超过sum的钞票组合
            int sum = C;//每一种组合要凑足sum
            memset(need, 0, sizeof(need));
            for (int i = 0;i < N;i++) {//面值从大到小
                if(sum > 0&&mon[i].num) {
                    int can_use = min(mon[i].num,sum/mon[i].V);//当前面值的钞票的最多能拿几张，尽量多，但不能超过sum
                    if (can_use>0) {
                        sum -= can_use*mon[i].V;
                        need[i] = can_use;
                    }
                }
            }
            for (int i = N - 1;i >= 0;i--) {//面值从小到大
                if (sum > 0 && mon[i].num) {
                    int can_use = min(mon[i].num-need[i],(sum+mon[i].V-1)/mon[i].V);//当前面值的钞票尽量多拿，但超过sum一点点就可以了，不要浪费
                    if (can_use>0) {
                        sum -= can_use*mon[i].V;
                        need[i] += can_use;
                    }
                }
            }

            if (sum > 0) {
                break;
            }
            //至此拼凑出sum需要各种面值的钞票的数量已经都存放在need[]数组中了
            int add = INT_MAX;
            for (int i = 0;i < N;i++) {
                if (need[i] == 0)continue;
                 add= min(add, mon[i].num / need[i]);//当前的拼凑出sum的组合最多能有几组
            }
            week += add;
            for (int i = 0;i < N;i++) {//将该组合所使用掉的钞票去除
                if (need[i] == 0)continue;
                mon[i].num -= need[i] * add;
            }

        }
        cout << week << endl;
    }
    return 0;
}