https://codeforces.com/contest/1542/problem/C

一些规则如下：
Some rules are as follows:

1 1到n中，a的倍数有n/a个
2 如果一个数是1,2,3,4,5,…,k的倍数，那么这个数是1,2,…,k的最小公倍数
3 如果一个数是1,2,3,…,k的最小公倍数的倍数，并且不是1,2,3,…,k+1的最小公倍数的倍数，那么这个数的函数值为k+1
4
1 From 1 to N, there are n / a multiples of A
2 If a number is a multiple of 1,2,3,4,5,…, K, then the number is the least common multiple of 1,2,…, K
3 If a number is a multiple of the least common multiple of 1,2,3,…, K, and is not a multiple of the least common multiple of 1,2,3,…, K + 1, then the function value of the number is K + 1

因此，我只需保证1到n中每个数，若为1的lcm的倍数，加一，若为1到2的lcm的倍数，再加一，若为1到3的lcm的倍数，再加一，依次下去。另外在最后每个数再加一即可。
So I just need to make sure that every number from 1 to N, if it’s a multiple of LCM of 1, add one, if it’s a multiple of LCM of 1 to 2, add one, if it’s a multiple of LCM of 1 to 3, add one, and so on. In addition, add one to each last number.

代码
Code:

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll t, n, lc[200], mod = 1e9 + 7;

void getlcm()
{
    lc[0] = 1;

    for(ll u = 1; lc[u - 1] <= 1e16; u ++) lc[u] = lcm(lc[u - 1], u);
}

int main()
{
    getlcm();

    cin >> t;

    while(t --)
    {
        cin >> n;

        ll res = n % mod;

        for(ll i = 1; lc[i]; i ++) (res += n / lc[i]) %= mod;

        cout << res << endl;
    }
}