Symmetric Matrix 牛客网暑期ACM多校训练营（第一场） B dp 组合数学

Count the number of n x n matrices A satisfying the following condition modulo m.

* Ai, j ∈ {0, 1, 2} for all 1 ≤ i, j ≤ n.

* Ai, j = Aj, i for all 1 ≤ i, j ≤ n.

* Ai, 1 + Ai, 2 + ... + Ai, n = 2 for all 1 ≤ i ≤ n.

* A1, 1 = A2, 2 = ... = An, n = 0.

输入描述：

The input consists of several test cases and is terminated by end-of-file. Each test case contains two integers n and m.

输出描述：

For each test case, print an integer which denotes the result.

备注 * 1 ≤ n ≤ 10 5 * 1 ≤ m ≤ 10 9 * The sum of n does not exceed 10 7 .

示例1:

输入

3 1000000000

100000 1000000000

输出

507109376

还是没有想的很清楚，明天再贴题解吧。先贴着代码

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll dp[maxn];
int main() {
    ll n, m;
    dp[2] = dp[3] = 1;
    while( cin >> n >> m ) {
        for( ll i = 4; i <= n; i ++ ) {
            dp[i] = ( ( (i-1) * (dp[i-1]+dp[i-2]) ) % m
                     - ( (i-1)*(i-2)/2*dp[i-3]) % m + m ) % m;
        }
        cout << dp[n] << endl;
    }
    return 0;
}