Find Integer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6597 Accepted Submission(s): 1852
Special Judge
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 a^n+b^n=c^n.
give you two integers n,a,you are required to find 2 integers b,c such that a^n+b^n=c^n.
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
题目大意：
已知 n、 a 求解方程 a^n + b^n = c^n;
解题思路：
根据费马大定理 n > 2 时无解；
n == 0 时 无解；
n == 1 时 a， 1， a+1；
n == 2 时，根据已知 a 构造一组勾股数
当 a 为 大于 4 的偶数 2*k 时： b = k^2 - 1, c = k^2 + 1;
当 a 为 大于 1 的奇数 2*k +1 时： b = 2*k^2 + 2*k, c = 2*n^2 + 2*n + 1;
code：
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <cmath> 5 #define mod 1000000007 6 #define INF 0x3f3f3f3f 7 using namespace std; 8 9 const int MAXN = 1e9; 10 int N, a; 11 int main() 12 { 13 int T, b, c, k; 14 scanf("%d", &T); 15 while(T--) 16 { 17 scanf("%d%d", &N, &a); 18 if(N>2 || N==0) printf("-1 -1 "); 19 else if(N==1) printf("%d %d ", 1, a+1); 20 if(N==2) 21 { 22 if(a%2){ 23 k = (a-1)/2; 24 b = 2*k*k+2*k; 25 c = 2*k*k+2*k+1; 26 printf("%d %d ", b, c); 27 } 28 else 29 { 30 k = a/2; 31 b = k*k-1; 32 c = k*k+1; 33 printf("%d %d ", b, c); 34 } 35 } 36 } 37 return 0; 38 }