Number Theory

Time Limit: 1 Second      Memory Limit: 65536 KB

Given an integer , calculate the smallest positive integer that is not a sum of distinct divisors of .

Input

There are multiple test cases. The first line of the input contains an integer  (about 2000), indicating the number of test cases. For each test case:

The first line contains one integer  ().

Output

For each test case output one line indicating the answer.

Sample Input

5
102
28
8
5
1

Sample Output

13
57
16
2
2

Author: LIU, Mingrui
Source: ZOJ Monthly, June 2018
 

#include <bits/stdc++.h>
#define ll long long 
using namespace std;

const int mn = 1e6;

int cnt;
bool is[mn + 10];
int pm[mn + 10];
void prime()
{
    for (int i = 2; i <= 1e6 + 3; i++)
    {
        if(is[i])
            continue;
        pm[++cnt] = i;
        for (int j = i + i; j <= 1e6 + 3; j += i)
            is[j] = 1;
    }
}

int main()
{
    prime();
    
    int T;
    scanf("%d", &T);
    while (T--)
    {
        ll n;
        scanf("%lld", &n);
        
        ll ans = 1;
        for (int i = 1; i <= cnt && pm[i] * pm[i] <= n; i++)
        {
            if (ans + 1 < pm[i])
                break;
            int t = 0;
            while (n % pm[i] == 0)
            {
                t++;
                n /= pm[i];
            }
            
            ll tans = ans;
            ll tt = pm[i];
            for (int j = 1; j <= t; j++)
            {
                ans += tans * tt;
                tt *= pm[i];
                if (ans + 1 < tt)
                    break;
            }
        }
        if (n > 1 && ans + 1 >= n)
            ans *= (n + 1);
        
        printf("%lld\n", ans + 1);
    }
    
    return 0;
}