标签：dp   hdu   动态规划   多重背包   组合背包   

题意   给你n种面额不同的金币和每种金币的个数  求这些金币能组合成的面额在m内有多少种

还是明显的背包问题  d[i]表示这些金币在i内能组合成的最大面额  初始化d为负无穷  d[0]=0  这样就可以保证d[i]恰好为i时才能为正值

原因可以自己想想  然后就用背包背吧  直接多重背包也可以过  但是分成多重背包和完全背包要快一点

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 105, M = 100005;
int main()
{
    int n, m, val[N], num[N], d[M], ans;
    while (scanf ("%d%d", &n, &m), n && m)
    {
        memset (d, 0x8f, sizeof (d));
        d[0] = ans = 0;
        for (int i = 1; i <= n; ++i)
            scanf ("%d", &val[i]);
        for (int i = 1; i <= n; ++i)
            scanf ("%d", &num[i]);
        for (int i = 1; i <= n; ++i)
        {
            if (num[i]*val[i] >= m)
                for (int j = val[i]; j <= m; ++j)
                    d[j] = max (d[j], d[j - val[i]] + val[i]);
            else
            {
                for (int k = 1; k <= num[i]; k *= 2)
                {
                    for (int j = m; j >= k * val[i]; --j)
                        d[j] = max (d[j], d[j - k * val[i]] + k * val[i]);
                    num[i] -= k;
                }
                if (num[i] > 0)
                    for (int j = m; j >= num[i]*val[i]; --j)
                        d[j] = max (d[j], d[j - num[i] * val[i]] + num[i] * val[i]);
            }
        }
        for (int i = 1; i <= m; ++i)
            if (d[i] > 0) ++ans;
        printf ("%d\n", ans);
    }
    return 0;
}

Coins

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

8
4

HDU 2844 Coins (组合背包),布布扣,bubuko.com

HDU 2844 Coins (组合背包)

标签：dp   hdu   动态规划   多重背包   组合背包   

原文地址：http://blog.csdn.net/iooden/article/details/38469877