ABC 044 C - Tak and Cards

今天做了AtCoder Beginner Contest 044的C题：

Problem Statement

Tak has $N$ cards. On the $i$-th $(1\le i\le N)$ card is written an integer ${\mathrm{xi}}^{}$. He is selecting one or more cards from these $N$ cards, so that the average of the integers written on the selected cards is exactly $A$. In how many ways can he make his selection?

Constraints

• $\mathrm{1\le N\le 50}$

• $\mathrm{1\le A\le 50}$

• $\mathrm{1\le xi\; \le 50}$

• $\mathrm{N,A,xi}$

are integers.

大意就是从n个数中选i个数，并使它们的和为ni，求选的方法数。

显然，这是一道dp相关的题目（要求方案数），设dp[i][j][k]是在前i个数中选j（j>=1）个数、其和为k的方案总数。第i个数有选与不选2种可能，由此得出转移方程dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k-x[i]](j>=1)

下面是代码

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#define maxn 55
#define ll long long
using namespace std;
ll dp[maxn][maxn][maxn*maxn];
int x[maxn];
int n,A;
ll ans;
int main()
{
    scanf("%d%d", &n, &A);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", x + i);
    }
    for (int i = 0; i <= n; i++)
    {
        dp[i][0][0] = 1;
    }
    
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= i; j++)
        {
            for (int k = 0; k <= n * A; k++)
            {
                
                if (k >= x[i] &&j>=1)
                    dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - x[i]];//注意j此时要大于等于1，否则会访问错误的地方，我在这就卡了一阵子
                else
                    dp[i][j][k] = dp[i - 1][j][k];
            }
        }
    }
    for (int t = 1; t <= n; t++)
        ans += dp[n][t][t*A];
    printf("%lld", ans);
    return 0;
}