3E A Simple Problem —— 数论

题目

For a given positive integer \(n\), please find the smallest positive integer \(x\) that we can find an integer \(y\) such that \(y^2 = n +x^2\).

Input

The first line is an integer \(T\), which is the the number of cases.
Then \(T\) line followed each containing an integer \(n (1<=n <= 10^9)\).

Output

For each integer \(n\), please print output the \(x\) in a single line, if \(x\) does not exit , print \(-1\) instead.

Sample Input

2
2
3

Sample Output

-1
1

题解

解题思路

我们先来推导一下
\(y^2 = n +x^2\)
\(n = y^2 - x^2\)
\(n = (y - x)(y + x)\)
\(n = ab\)
\(a = y - x, b = y + x\)
\(x = (b - a) / 2\)
所以，我们枚举n的因数，再找x的最小值

代码

#include <cstdio>
#include <algorithm>
using namespace std;
const int M = 0x7fffffff;
int t, n, ans;
int main() {
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        ans = M;
        for(int i = 1; i * i <= n; i++)
            if (!(n % i) && !((n/i - i) % 2) && (n/i - i) / 2)
                ans = min(ans, (n/i - i) / 2);
        if (ans == M) puts("-1");
        else printf("%d\n", ans);
    }
    return 0;
}