POJ 3040 - Allowance（贪心）

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

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思路：

贪心三个步骤：

1：面额大于等于c的硬币，无法节约，全部先发掉。

2：然后对硬币面额从大到小尽量凑得接近C，允许等于或不足C，但是不能超出C

3：接着按硬币面额从小到大凑满C（凑满的意思是允许超出一个最小面值，ps此处的最小面值指的是硬币剩余量不为0的那些硬币中的最小面值），凑满之后得出了最优解，发掉。再进入步骤2，直到剩下的钱数无法凑满c。

CODE：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;struct node {    int val, num;    bool operator < (const node &a) const{        return val > a.val;    }}coin[25];int need[25];int main(){//freopen("in", "r", stdin);    int n, c;    while(~scanf("%d %d", &n, &c)) {        int ans = 0, nj = 0;        for(int i = 0; i < n; ++i) {            int v, b;            scanf("%d %d", &v, &b);            if(v >= c) ans += b;            else {                coin[nj].val = v; coin[nj++].num = b;            }        }        sort(coin, coin + nj);        while(1) {            int res = c;            memset(need, 0, sizeof(need));            for(int i = 0; i < nj; ++i) {                if(res && coin[i].num) {                    int c = res / coin[i].val;                    c = min(c, coin[i].num);                    need[i] = c;                    res -= c*coin[i].val;                }            }            if(res) {                for(int i = nj-1; i >= 0; --i) {                    if(coin[i].val >= res && (coin[i].num - need[i])) {                        need[i]++;                        res = 0;                        break;                    }                }                if(res) break;            }            int d = 10000000;            for(int i = 0; i < nj; ++i) {                if(need[i]) {                    d = min(d, coin[i].num/need[i]);                }            }            ans += d;            for(int i = 0; i < nj; ++i) {                coin[i].num -= d*need[i];            }        }        printf("%d\n", ans);    }    return 0;}