poj1787_完全背包（个数不是无限个）

题目链接：http://poj.org/problem?id=1787

题目大意：有1， 5， 10， 25面值的硬币， 求要用最多的硬币个数组成面值p?

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <vector>
using namespace std;

int max(int a, int b)
{
    return a > b ? a : b;
}
int dp[10010], num[5][10010], used[10010];
int main()
{
    int a[5], p, c[5], i, j;
    a[1] = 1, a[2] = 5, a[3] = 10, a[4] = 25;
    while(~scanf("%d %d %d %d %d", &p, &c[1], &c[2], &c[3], &c[4]))
    {
        if(p + c[1] + c[2] + c[3] + c[4] == 0)
            break;
        memset(dp, -1, sizeof(dp));
        memset(num, 0, sizeof(num));        
        dp[0] = 0;
        for(i = 1; i <= 4; i++)
        {
            memset(used, 0, sizeof(used));
            for(j = a[i]; j <= p; j++)
            {
                if(dp[j - a[i]] != -1 && used[j - a[i]] + 1 <= c[i] && dp[j] < dp[j - a[i]] + 1)
                {            
                    dp[j] = dp[j - a[i]] + 1;      //组成p用的硬币总数              
                    used[j] = used[j - a[i]] + 1;//组成P用的a[i]面值硬币的个数
                    num[i][j] = used[j];
                }
            }
        }
       // cout << used[p] ;
        if(dp[p] == -1)
            printf("Charlie cannot buy coffee.
");
        else
        {
            int t4 = used[p];
            int t3 = num[3][p - t4 * 25];
            int t2 = num[2][p - t4 * 25 - t3 * 10];
            int t1 = num[1][p - t4 * 25 - t3 * 10 - t2 * 5]; 
            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.
", t1, t2, t3, t4);
        }
    }
    return 0;
}