CodeForces 707C Pythagorean Triples (数论)

题意：给定一个数n，问你其他两边，能够组成直角三角形。

析：这是一个数论题。

如果 n 是奇数，那么那两边就是 (n*n-1)/2 和 (n*n+1)/2。

如果 n 是偶数，那么那两边就是 (n/2*n/2-1) 和 (n/2*n/2+1)。
那么剩下的就很简单了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 8;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
    LL x;
    while(cin >> x){
        if(x < 3){ printf("-1
");  continue; }

        if(x & 1)  printf("%I64d %I64d
", (x*x-1)/2, (x*x+1)/2);
        else{
            x /= 2;
            printf("%I64d %I64d
", x*x-1, x*x+1);
        }
    }
    return 0;
}