Allowance - POJ 3040 贪心

Allowance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1554		Accepted: 644

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：组合金币，使得每个星期都有不少于C的奖励，问最多能组合多少个星期。

思路：如果恰好能组成C，那么就是尽可能从大往小取，如果不能组成C的话，就在之前的基础上，从小往大取。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int val,num;
}coin[25];
bool cmp(node a,node b)
{
    return a.val<b.val;
}
int n,c,num[25],ans;
bool solve()
{
    int i,j,k,ret=0;
    for(i=n;i>=1;i--)
       if(c-ret>=coin[i].val)
       {
           k=min((c-ret)/coin[i].val,coin[i].num);
           num[i]+=k;
           ret+=coin[i].val*k;
       }
    if(ret==c)
      return true;
    for(i=1;i<=n;i++)
       if(ret<c)
       {
           k=min((c-ret)/coin[i].val+1,coin[i].num-num[i]);
           num[i]+=k;
           ret+=coin[i].val*k;
       }
    if(ret<c)
      return false;
    else
      return true;
}
int main()
{
    int i,j,k,maxn;
    while(~scanf("%d%d",&n,&c))
    {
        for(i=1;i<=n;i++)
           scanf("%d%d",&coin[i].val,&coin[i].num);
        sort(coin+1,coin+1+n,cmp);
        while(true)
        {
            memset(num,0,sizeof(num));
            if(!solve())
              break;
            maxn=1e9;
            for(i=1;i<=n;i++)
               if(num[i]>0)
                 maxn=min(maxn,coin[i].num/num[i]);
            ans+=maxn;
            for(i=1;i<=n;i++)
               coin[i].num-=maxn*num[i];
        }
        printf("%d\n",ans);
    }
}