HDU 4143 A Simple Problem 题解

题目

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

输入

The first line is an integer \(T\), which is the the nuaber of cases.

Then T line followed each containing an integer n (\(1 \le n \le 10^9\)).

输出

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

样例输入

2
2
3

样例输出

-1
1

题解

\(\because y^2 = n +x^2\)

\(\therefore n=y^2-x^2 = (y+x)(y-x)\)

所以找到两个数字\(a, b\), 满足

1. \(a \cdot b = n\)

2. \((a+b)%2==0\) , 因为\((a+b)\%2=(y+x+y-x)\%2=(2\cdot y)\%2=0\)

3. \(a>b\) , 因为\(x>0\)

4. \((a-b)\%2==0\), 因为\((a-b)\%2=(y+x-y+x)\%2=(2\cdot x)\%2=0\)

5. \(a-b=2 \cdot x\)最小, 即\(x\)最小

因为求的是\(a-b\)最小且\(a \cdot b=n\), 所以从\(\sqrt n\) 开始遍历, 若满足条件, 更新最小值, 如果没找到, 就输出\(-1\)

代码

#include <cmath>
#include <cstdio>
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int n, minv = 0x7f7f7f7f, flag = 0;
        scanf("%d", &n);
        for (int i = 1; i < sqrt(n); i++) {
            if (n % i == 0 && (i + n / i) % 2 == 0 && (n / i - i) % 2 == 0 && (n / i - i) > 0) {
                flag = 1;
                if (n / i - i < minv) minv = n / i - i;
            }
        }
        printf("%d\n", flag ? minv / 2 : -1);
    }
    return 0;
}