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ARC-060C - Tak and Cards - 动态规划 20
2018/05

ARC-060C - Tak and Cards - 动态规划

题目

Problem Statement

Tak has $N$ cards. On the $i$-th $(1≤i≤N)$ card is written an integer $x_i$. 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

$1≤N≤50$
$1≤A≤50$
$1≤x_i≤50$
$N, A, x_i$ are integers.

Partial Score

200 points will be awarded for passing the test set satisfying $1≤N≤16$.

Input

The input is given from Standard Input in the following format:

$N$ $A$
$x_1$ $x_2$ $…$ $x_N$

Output

Print the number of ways to select cards such that the average of the written integers is exactly $A$.

题意

给你N张卡片，每个卡片有一个权值，然你选任意多张卡片（不能为$0$），使得这些卡片权值的平均值严格等于A，问你有多少种选法。

思路

因为比较小，所以三维递推的复杂度是允许的。dp[i][j][k]代表前i个元素里严格选j个的和为k的方案数，很容易得到递推公式：$$dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - num[i]]$$

然后$50^4$的long long开不下，又发现每一层只会用到上一层的状态，用滚动数组优化一下第一维就可以。

实现

#include <bits/stdc++.h>
const int maxn = 57;
long long f[maxn][maxn * maxn], ans;
int n, num[maxn], a;
int main() {
scanf("%d%d", &n, &a);
for (int i = 1; i <= n; i++) scanf("%d", num + i);
f = f = 1;
for (int i = 1, cur = 1; i <= n; i++, cur ^= 1)
for (int k = 1; k <= i; k++)
for (int val = 0; val <= 2500; val++) {
f[cur][k][val] = f[cur ^ 1][k][val];
if (val >= num[i]) f[cur][k][val] += f[cur ^ 1][k - 1][val - num[i]];
}
for (int i = 1, cur = n & 1; i <= n; i++) ans += f[cur][i][a * i];
printf("%lld\n", ans);
return 0;
}

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