D. K-good(数学推导)

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#c++ #算法 #数据结构

本文介绍了一种算法，用于确定一个正整数n是否为k-good数，即能否由k个正整数相加得到，且这k个数除以k有不同的余数。通过将输入的数n分解并调整其形式，可以找到符合条件的最小k值。

We say that a positive integer nn is kk-good for some positive integer kk if nn can be expressed as a sum of kk positive integers which give kk distinct remainders when divided by kk.

Given a positive integer nn, find some k≥2k≥2 so that nn is kk-good or tell that such a kk does not exist.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1051≤t≤105) — the number of test cases.

Each test case consists of one line with an integer nn (2≤n≤10182≤n≤1018).

Output

For each test case, print a line with a value of kk such that nn is kk-good (k≥2k≥2), or −1−1 if nn is not kk-good for any kk. If there are multiple valid values of kk, you can print any of them.

Example

input

Copy

output

Copy

Note

66 is a 33-good number since it can be expressed as a sum of 33 numbers which give different remainders when divided by 33: 6=1+2+36=1+2+3.

1515 is also a 33-good number since 15=1+5+915=1+5+9 and 1,5,91,5,9 give different remainders when divided by 33.

2020 is a 55-good number since 20=2+3+4+5+620=2+3+4+5+6 and 2,3,4,5,62,3,4,5,6 give different remainders when divided by 55.

思路:

1,差值一定,某一段求和,可用等差数列求和

2,n=k*(k+1)/2+m*k(m>=0)==>2*n==k*(2*m+1+k);

3,因为右侧一奇一偶,提取2^i,根据奇数情况判断

代码:

#include<bits/stdc++.h>
using namespace std;
const int maxj =1e6+100,mod = 1e6+100;
int v[50],w[50],dp[55][3100];
int32_t main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        int te=1;
        while(n%2==0){
            n/=2;
            te*=2;
        }
        if(n==1)cout<<-1<<'\n';//全是奇数
        else cout<<min(te*2,n)<<'\n';//奇偶中最小作为k
    }
    return 0;
}