提示：是玄武纪oj上的题目

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 10, mod = 1e9 + 7;
int n, primes[N], cnt;
bool st[N];
void gp(int n) {
    for (int i = 2; i <= n; i++) {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] * i <= n; j++) {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main() {
    cin >> n;
    gp(n);
    ll res = 1;
    for (int i = 0; i < cnt; i++) {
        int p = primes[i];
        if (p == 0) continue; 
        int s = 0;
        for (int j = n; j > 0; j /= p) s += j / p;
        res = (ll)(2 * s + 1) * res % mod;
    }
    cout << res << endl;
    return 0;
}

据说只有1/7244的玄武纪oj用户能够猜对！ 答案在这里.