poj_1837

dp[i][j]
用前i个砝码，使"力矩"大小为j的情况数量
极端情况下，20个重量为25的砝码，都挂在距离中心点-15的位置，得到力矩是-7500
所以，让j=7500时为平衡状态

#include <cstdio>
#include <cstring>

#define MAXN 20

int C, G, dp[MAXN+1][2*MAXN*15*25+1], hook[MAXN+1], weight[MAXN+1];

int main(int argc, char const *argv[])
{
    // freopen("in", "r", stdin);
    scanf("%d%d", &C, &G);
    for(int i = 1; i <= C; ++i)
        scanf("%d", &hook[i]);
    for(int i = 1; i <= G; ++i)
        scanf("%d", &weight[i]);

    // initialize
    memset(dp, 0 ,sizeof(dp));
    dp[0][7500] = 1;

    // dp
    for(int i = 1; i <= G; ++i)
        for(int j = 0; j <= 15000; ++j)
            if(dp[i-1][j])
                for(int k = 1; k <= C; ++k)
                    dp[i][ j+hook[k]*weight[i] ] += dp[i-1][j];

    printf("%d
", dp[G][7500]);
    return 0;
}