Find IntegerTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 982 Accepted Submission(s): 229 Special Judge Problem Description people in USSS love math very much, and there is a famous math problem .
Input one line contains one integer T ;(1≤T≤1000000)
Output print two integers b ,c if b ,c exits;(1≤b,c≤1000 ,000 ,000) ;
Sample Input 1 2 3
Sample Output 4 5
Source
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【题意】
题目给出n和a,找出a^n+b^n=c^n,若存在这样的b和c则输出。
【解题思路】
5555还是太年轻被这个special judge坑了。
根据费马大定理:
当n>=3 || n==0时 不存在这样的b和c
所以只需讨论n=2和n=1的情况,n=2时,只需满足勾股数即可。
【代码】
#include<bits/stdc++.h>
using namespace std;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,a;
scanf("%d%d",&n,&a);
if(n>2||n==0)printf("-1 -1\n");
else if(n==1)//输出一组就好
printf("%d %d\n",1,1+a);
else{
int k=a/2;
if(a&1)//a为大于1的奇数
printf("%d %d\n",2*k*k+2*k,2*k*k+2*k+1);
else //a为大于4的偶数2n
printf("%d %d\n",k*k-1,k*k+1);
}
}
return 0;
}