如题，我用整除分块做的，第四个点tle了，本地跑出来答案是对的，但是跑了半分钟。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 4e6 + 10;
int n, m;
ll dp[N], pre[N];

int main() {
    scanf("%d%d", &n, &m);
    dp[1] = 1; pre[1] = 1;
    for(int i = 2; i <= N - 10; i ++) {
        for(int j = 2; j <= i; j ++) {
            int p = i / j, r = i / p;
            dp[i] = (dp[p] * (r - j + 1) % m + dp[i]) % m;
            j = r; 
        }
        dp[i] = (dp[i] + pre[i - 1]) % m;
        pre[i] = (pre[i - 1] + dp[i]) % m;
    }
    printf("%lld", dp[n]);
}

求大佬帮忙看看，万分感谢！