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.

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

分析：根据费马大定理：n>2时无解

　　只要考虑n=0,1,2

　　n=0时，无解

　　n=1时，随意构造两个数满足a+b=c

　　n=2时，随意构造一组勾股数满足a*a+b*b=c*c

AC代码：

#include <map>

#include <set>

#include <stack>

#include <cmath>

#include <queue>

#include <cstdio>

#include <vector>

#include <string>

#include <bitset>

#include <cstring>

#include <iomanip>

#include <iostream>

#include <algorithm>

#define ls (r<<1)

#define rs (r<<1|1)

#define debug(a) cout << #a << " " << a << endl

using namespace std;

typedef long long ll;

const ll maxn = 1e6+10;

const ll mod = 998244353;

const double pi = acos(-1.0);

const double eps = 1e-8;

int main() {

    ios::sync_with_stdio(0);

    ll T;

    scanf("%lld",&T);

    while( T -- ) {

        ll n, a, b, c;

        scanf("%lld%lld",&n,&a);

        if( n >= 3 || n == 0 ) {

            printf("-1 -1\n");

        }

        else if( n == 1 ) {

            printf("1 %lld\n",a+1);

        }

        else {

            ll h = 1;

            while( a%2 == 0 ) {

                a = a/2;

                h = h*2;

            }

            ll s, t;

            s = a;

            t = 1;

            b = (s*s-t*t)/2;

            c = (s*s+t*t)/2;

            b = b*h;

            c = c*h;

            printf("%lld %lld\n",b,c);

        }

     }

    return 0;

}

巴特西

2018中国大学生程序设计竞赛 - 网络选拔赛 hdu Find Integer 数论

Find Integer

最新文章

热门文章