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    Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.

    For example, triples (3, 4, 5), (5, 12, 13) and (6, 8, 10) are Pythagorean triples.

    Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.

    Katya had no problems with completing this task. Will you do the same?

    Input

    The only line of the input contains single integer n (1 ≤ n ≤ 109) — the length of some side of a right triangle.

    Output

    Print two integers m and k (1 ≤ m, k ≤ 1018), such that nm and k form a Pythagorean triple, in the only line.

    In case if there is no any Pythagorean triple containing integer n, print  - 1 in the only line. If there are many answers, print any of them.

    Examples

    Input
    3
    Output
    4 5
    Input
    6
    Output
    8 10
    Input
    1
    Output
    -1
    Input
    17
    Output
    144 145
    Input
    67
    Output
    2244 2245


    构造直角三角形,设已知边为x,另两条边分别为x,y。
    则有a*a=y*y-x*x.即a*a=(y-x)(y+x),令A=y-x,B=y+x;
    解得 x=(A+B)/2,y=(A-B)/2.

    现在判断什么样的A,B可使x和y是整数。
    分类讨论,当a*a是偶数的时候,A=2,B=a*a/2.
    当a*a是奇数的时候,A=1,B=a*a.

    若解不出来,则输出-1.

     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<cstring>
     4 #include<string>
     5 #include<cmath>
     6 #include<algorithm>
     7 #include<queue>
     8 #include<stack>
     9 #include<deque>
    10 #include<map>
    11 #include<iostream>
    12 using namespace std;
    13 typedef long long  LL;
    14 const double pi=acos(-1.0);
    15 const double e=exp(1);
    16 const int N = 10;
    17 
    18 int main()
    19 {
    20     LL i,p,j;
    21     LL x,y,a;
    22     scanf("%lld",&a);
    23     if(a%2==0)
    24     {
    25         x=(a*a/2-2)/2;
    26         y=(a*a/2+2)/2;
    27         if(x<=0||y<=0||x==y)
    28             printf("-1
    ");
    29         else
    30             printf("%lld %lld
    ",x,y);
    31     }
    32     else
    33     {
    34         x=(a*a-1)/2;
    35         y=(a*a+1)/2;
    36         if(x<=0||y<=0||x==y)
    37             printf("-1
    ");
    38         else
    39             printf("%lld %lld
    ",x,y);
    40     }
    41     return 0;
    42 }
    View Code


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  • 原文地址:https://www.cnblogs.com/daybreaking/p/9691104.html
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