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传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6441
题意:给你 n , a , 让你求 b , c ,满足 a ^ n + b ^ n = c ^ n.
思路:根据费马大定理内容:当n >2时,关于x, y, z的方程 x^n + y^n = z^n 没有正整数解。所以只需要考虑n为0,1,2时即可。
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n = 0时,无解
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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
直角三角形a^2+b^2=c^2奇数列a法则:
- a = 2n + 1
- b = n ^ 2 + (n + 1) ^ 2 - 1
- c = n ^ 2 + (n + 1) ^ 2
直角三角形a^2+b^2=c^2偶数列a法则:
- a = 2n
- b = n ^ 2 - 1
- c = n ^ 2 + 1
AC代码:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
int t;
ll z, 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;
}