Codeforces Round #368 (Div. 2) (勾股数)

传送门:Codeforces Round #368 (Div. 2)

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题意:给你一条边L,求输出另外两条能与它构成直角三角形的边。如不存在输出-1
(L<=1e9)

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思路:

1. 当L为大于1的奇数,令这条边为直角边,则另外两条边为b,c则,(c−b)(c+b)=L∗L−>令c−b=1−>c=(L∗L+1)/2

2. 当L为大于2的偶数2∗n时,b=n2−1,c=n2+1

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 1e18#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))vector<int>tmp;int main(){    long long n;    scanf("%lld",&n);    if(n==1||n==2)        printf("-1\n");    else{        if(n==4){            printf("3 5\n");            return 0;        }        if(n&1)            printf("%lld %lld\n",(n*n+1)/2,(n*n+1)/2-1);        else            printf("%lld %lld\n",n*n/4-1,n*n/4+1);    }    return 0;}