HDU 6441 2018中国大学生程序设计竞赛 - 网络选拔赛 1004 Find Integer 数学知识毕达哥拉斯三元组模板

1004 ​​Find Integer​​

Problem
people in USSS love math very much, and there is a famous math problem .
give you two integers , ,you are required to find integers , such that a^n+b^n=c^n
Input
one line contains one integer T;(1<=T<=1000000)
next 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

解析：

费马大定理可知，n>2或n<=0,不存在正整数解，n=1的时候好构造，n=2的时候就不好构造了。

这里有一个毕达哥拉斯三元组解析​​点这里​​

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        ll a, n;

        scanf("%lld%lld", &n, &a);
        if (n > 2)printf("-1 -1\n");
        else
        {
            if (n == 0)printf("-1 -1\n");
            else if (n == 1)printf("%lld %lld\n", 1, a + 1);
            else if (n == 2)
            {
                if (a & 1)
                {
                    ll x = (a - 1) / 2;
                    ll c = x*x + (x + 1)*(x + 1);
                    printf("%lld %lld\n", c - 1, c);
                }
                else
                {
                    ll x = a / 2;
                    printf("%lld %lld\n", x*x - 1, x*x + 1);
                }
            }
        }
    }
}

模板2：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int main() {
    int t,a,n;
    ll b,c;
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d",&n,&a);
        if(n>2 || n==0)
            puts("-1 -1");
        else if(n==1) {
            printf("1 %d",a+1);
        } else {
            if(a&1) {
                c=(a*a+1)/2;
                b=c-1;
            } else {
                c=(a*a/2+2)/2;
                b=c-2;
            }
            printf("%lld %lld\n",b,c);
        }
 
    }
}