HDU 5974 2016ICPC大连 D: A Simple Math Problem

题意：

已知①x+y=a；②LCM(x, y)=b，给出a和b问是否有x和y满足条件（x<y）

假设x' = x/Gcd(x,y)，y' = y/Gcd(x,y)那么有

暴力枚举Gcd(x, y)，然后联立上面的第1个和第4个式子就可以求出x'，y'，乘回Gcd判断下是否合法就好

复杂度O(Tsqrt(a))

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int Gcd(int x, int y)
{if(y==0)return x;return Gcd(y, x%y);
}
int main(void)
{int a, b, i, y, x, p, temp;while(scanf("%d%d", &a, &b)!=EOF){for(i=1;i*i<=a;i++){p = i;if(a%p==0 && b%p==0){temp = sqrt(fabs(a*a/p/p-4.0*b/p))+1e-7;y = (temp+a/p)/2;x = a/p-y;if(Gcd(y, x)==1 && x*p+y*p==a && x*y*p==b){x *= p;y *= p;if(x>y)swap(x, y);printf("%d %d\n", x, y);break;}}p = a/i;if(a%p==0 && b%p==0){temp = sqrt(fabs(a*a/p/p-4.0*b/p))+1e-7;y = (temp+a/p)/2;x = a/p-y;if(Gcd(y, x)==1 && x*p+y*p==a && x*y*p==b){x *= p;y *= p;if(x>y)swap(x, y);printf("%d %d\n", x, y);break;}}}if(i*i>a)printf("No Solution\n");}return 0;
}

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