hdoj A Simple Problem 4143 （数学求方程最小解）

A Simple Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4910    Accepted Submission(s): 1254

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
 
//将方程变换为：y^2-x^2=n;
    =>(y-x)*(y+x)=n;
所以枚举所有的（y-x）然后不断更新最小值。
#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;int main(){int t,n,i;scanf("%d",&t);while(t--){scanf("%d",&n);int ans=INF;for(i=1;i<=sqrt(n);i++){if(n%i==0){int a=i;int b=n/i;if(b-a>0&&(b-a)%2==0){int x=(b-a)/2;if(x<ans)ans=x;}}}if(ans==INF)printf("-1\n");elseprintf("%d\n",ans);}return 0;}