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
Source
题目的意思是已知 n、 a 求解方程 a^n + b^n = c^n;
从而首先由费马大定律
n>2 时 an+bn=cnan+bn=cn 没有整数解,只需要计算 n=0,1,2 这三种情况:
1、n=0,任何的正整数 b,cb,c 都无法使等式成立。
2、n=1,任意取。
3、n=2,a2=(c+b)(c−b),分两种情况讨论:
若 a为奇数,则 a2 也为奇数,则取 b=(a*a-1)/2,c=(a*a+1)/2
若 a 为偶数,则 a2 必然是 4 的倍数,则取 b=(a*a-4)/4,c=(a*a+4)/4
#include<iostream> #include<cstdio> using namespace std; #define fi first #define se second #define mp make_pair #define pb push_back #define sz(x) ((int)(x).size()) #define all(x) (x).begin(), (x).end() #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int i=n-1;i>=a;i--) typedef long long ll; typedef pair<int,int> pii; //head int main() { ll T,a,n; cin>>T; while(T--){ scanf("%lld %lld",&n,&a); if(n==0||n>2) printf("-1 -1\n"); else if(n==1){ printf("1 %lld",a+1); } else if(a&1){ printf("%lld %lld\n",(a*a-1)/2,(a*a+1)/2); } else printf("%lld %lld\n",(a*a-4)/4,(a*a+4)/4); } return 0; }
不过还是以后可以输出输入尽量都用scanf和printf。。。