Rikka with Subset HDU - 6092 （背包）

参考http://blog.youkuaiyun.com/qq_36368339/article/details/76934053

Sample Input
2
2 3
1 1 1 1
3 3
1 3 3 1

Sample Output
1 2
1 1 1
Hint

In the first sample, A is [1,2]. A has four subsets [],[1],[2],[1,2] and the sums of each subset are 0,1,2,3. So B=[1,1,1,1]

题意：接下来一行n和m，下面一行m+1个数字代表B（0~m）求A序列，Bi 代表A序列中的所有子集之和为i的有Bi个，A序列总和为m，n个

int a[max_n], b[max_n], dp[max_n], c[max_n];
//dp[i]表示：加和为i的子集个数；

int main() {
    int t, n, m;
    scanf("%d", &t);
    while(t--) {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        memset(dp, 0, sizeof(dp));
        scanf("%d %d", &n, &m);
        for(int i = 0; i <= m; i++)
            scanf("%d", &b[i]);
        dp[0] = 1; //初始化值
        int p = 0, sum = 0;
        for(int i = 1; i <= m; i++) {
            c[i] = b[i] - dp[i];//A序列中值为i的个数
            for(int j = 0; j < c[i]; j++) {
                a[p++] = i; //对A序列赋值
                for(int k = m; k>= i; k--) { //处理成01背包 
                    dp[k] += dp[k - i]; //和为k的A子集个数相加去更新B序列
                }
            }
        }