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

(ecause y^2 = n +x^2)

( herefore 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=(2cdot 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=(2cdot 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
", flag ? minv / 2 : -1);
    }
    return 0;
}