HDU 4206 Treasure Map

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4206

大意：给你一个数n，问b^2-a^2 == n,求出满足条件的a 和b，若有多个输出最小的那个，若没，输出IMPOSSIBLE

 

b^2-a^2 = (b-a)*(b+a) = p * q = n

因为1<=n<=10^9,现在我们只要枚举p从1到根号10^9,

a = (q-p)/2;
b = (q+p)/2;其中(q-p)和(q+p)都为偶数,n%p == 0（用于剪枝）

然后枚举的时候取最小的a

#include <iostream>
#include <cmath>
using namespace std;

const int INF = 1<<30;
int main()
{
    int t;
    int n;
    int i, j;
    int p, q;
    int a, b;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        a = INF;
        for (p = 1; p <= (int)sqrt(double(n)); p++)
        {
            if (n % p == 0 && (n/p - p) % 2 == 0)
            {
                q = n/p;
                if (a > (q-p)/2)
                {
                    a = (q-p)/2;
                    b = (q+p)/2;
                }
            }
        }
        if (a == INF)
        {
            puts("IMPOSSIBLE");
        }
        else
        {
            printf("%d %d\n", a, b);
        }
    }
    return 0;
}