洛谷 P1356 数列的整除性（DP，数学）

题目链接

https://www.luogu.com.cn/problem/P1356

思路

$n$的大小只有$10^4$，$k$的大小只有$100$，我们令$dp[i][j]$表示前$i$个数字构成的所有表达式，表达式的值模$k$为$j$是否存在。

则转移方程为：

$dp[i][mod(j+a[i])]|=dp[i-1][j]$和$dp[i][mod(j-a[i])]|=dp[i-1][j]$

时间复杂度：$O(mnk)$

代码

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double

typedef long long i64;
typedef unsigned long long u64;
typedef pair<int, int> pii;

const int N = 1e4 + 5, M = 1e2 + 5;
const int mod = 10007;
const int inf = 0x3f3f3f3f3f3f3f3f;

std::mt19937 rnd(time(0));

int n, k;
int a[N];
int MOD(int x)
{
	return (x % k + k) % k;
}
void solve(int test_case)
{
	cin >> n >> k;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	vector<vector<int>>dp(n + 1, vector<int>(k, 0));
	dp[1][MOD(a[1])] = 1;
	for (int i = 2; i <= n; i++)
	{
		for (int j = 0; j < k; j++)
		{
			int res = MOD(j + a[i]);
			dp[i][res] |= dp[i - 1][j];
			res = MOD(j - a[i]);
			dp[i][res] |= dp[i - 1][j];
		}
	}
	if (dp[n][0])
	{
		cout << "Divisible" << endl;
	}
	else cout << "Not divisible" << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int test = 1;
	cin >> test;
	for (int i = 1; i <= test; i++)
	{
		solve(i);
	}
	return 0;
}