HDOJ 4143

A Simple Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2670    Accepted Submission(s): 672

Problem Description

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

223

 

Sample Output

-11

 

一道思维题 。

由于数据范围很大(1<=n <= 10^9)，所以直接暴力必TLE，因式分解y^2 = n +x^2得：(y-x)*(y+x)=n ，分解n ， 一个因子y + x ， 另一个 y - x ， 因为（y + x) - (y - x) = 2 * y ,必为偶数  ， 且两因子不同 。枚举其中一个因子从sqrt（n）到1。输出第一个符合条件的解即可！

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int main(){    //freopen("in.txt" , "r" , stdin);    int T , n;    cin >> T;    while(T--)    {        cin >> n;        int ok = 1;        for(int i  = (int)sqrt(n) ; i >= 1 ; i--)        {            if(n % i == 0 && i != n / i && abs(i - n / i) % 2 == 0)            {                long long x = abs(i - n / i)  / 2;                cout << x << endl;                ok = 0;                break;            }        }        if(ok)printf("-1\n");    }    return 0;}