Problem Description
people in USSS love math very much, and there is a famous math problem .
give you two integers n,a,you are required to find 2 integers b,c such that an+bn=cn.
give you two integers n,a,you are required to find 2 integers b,c such that an+bn=cn.
Input
one line contains one integer T;(1≤T≤1000000)
next T lines contains two integers n,a;(0≤n≤1000,000,000,3≤a≤40000)
next T lines contains two integers n,a;(0≤n≤1000,000,000,3≤a≤40000)
Output
print two integers b,c if b,c exits;(1≤b,c≤1000,000,000);
else print two integers -1 -1 instead.
else print two integers -1 -1 instead.
Sample Input
1
2 3
Sample Output
4 5
解题思路：①费马大定理：当整数n>2时，关于x, y, z的方程 x^n + y^n = z^n 没有正整数解。
②智慧数：一个自然数若能表示为两个自然数的平方差，则这个自然数为“智慧数”。形如2k+1或4k的形式必为智慧数，k≥0。举个栗子：验证2687是否为智慧数，∵2687为奇数，∴设2687=2k+1（k为正整数），∴k=1343，∴2687=1344²-1343²，∴2687是智慧数。验证16是否为智慧数，∵4|16，∴k=4，∴16=52-32，∴16为智慧数。
AC代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 int t,n,a; 4 int main(){ 5 while(~scanf("%d",&t)){ 6 while(t--){ 7 scanf("%d%d",&n,&a); 8 if(!n||n>2)printf("-1 -1 ");//无解 9 else if(n==1)printf("%d %d ",1,a+1);//输出最小即可 10 else{ 11 a*=a; 12 if(a&1)printf("%d %d ",a/2,a/2+1);//奇数 13 else{ 14 if(a%4)printf("-1 -1 ");//无解 15 else printf("%d %d ",a/4-1,a/4+1); 16 } 17 } 18 } 19 } 20 return 0; 21 }