Find Integer HDU - 6441
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
.
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)
Output
print two integers b,c if b,c exits;(1≤b,c≤1000,000,000)
;
else print two integers -1 -1 instead.
Sample Input
1
2 3
Sample Output
4 5
题意:
给定
无解输出-1 -1
分析:
观察可得n=0的时候无解
根据费马大定理可知 时无解
因此只需要考虑n=2的情况,即勾股数
这里有一个结论:
对于大于2的任意偶数2n(n>1),都可构成一组勾股数,三边分别是:
任意一个大于1的奇数2n+1(n>1)为边也可以构成勾股数,其三边分别是:
code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
int T;
scanf("%d",&T);
while(T--){
ll n,a;
scanf("%lld%lld",&n,&a);
if(n >= 3 || n == 0)
printf("-1 -1\n");
else if(n == 1)
printf("%lld %lld\n",a,a*2);
else{
ll tmp;
if(a & 1){
tmp = (a - 1) / 2;
printf("%lld %lld\n",tmp*tmp*2+tmp*2,tmp*tmp*2+tmp*2+1);
}
else{
tmp = a / 2;
printf("%lld %lld\n",tmp*tmp-1,tmp*tmp+1);
}
}
}
return 0;
}