HDU-2018中国大学生程序设计竞赛-网络选拔赛-1004-Find Integer
Find Integer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
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 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
提示:此题运用了费马大定理,如果n>2,不可能出现b、c对应a实现a^n+b^n=c^n;
如果n==2,则有当a为奇数时有规律,当a为偶数时有规律;
如果n==1,情况更好判断。
代码实现如下(g++):
#include <iostream> #include <cstdio> #define ll long long int using namespace std; int main() { int t; ll z; ll n,a; scanf("%d",&t); while(t--) { scanf("%lld %lld",&n,&a); if(n==1) { printf("1 %lld\n",a+1); } else if(n==2) { z=a*a; if(a%2) { printf("%lld %lld\n",z/2,z/2+1); } else { z = z/4; printf("%lld %lld\n",z-1,z+1); } } else printf("-1 -1\n"); } return 0; }