CF707C Pythagorean Triples（數論）

Pythagorean Triples

題意：給定一個數，求出另外兩個數，使這三個數構成勾股數
分析：
$a^2=b^2-c^2,a^2=(b-c)*(b+c)$
當a^2爲奇數，令(b-c)=1,(b+c) = a^2;
當a^2爲偶數，令(b-c)=2,(b+c) = a^2/2;
求解方程即可
a<=2時，不可能構成勾股數

Code：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a));
#define lowbit(x) (x & -x)
#define lrt nl, mid, rt << 1
#define rrt mid + 1, nr, rt << 1 | 1
template <typename T>
inline void read(T& t) {
    t = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        t = t * 10 + ch - '0';
        ch = getchar();
    }
    t *= f;
}
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const ll Inf = 9223372036854775807;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;

int main(void) {
    ll a;
    read(a);
    if (a <= 2)
        printf("-1\n");
    else {
        ll aa = a * a;
        if (aa & 1)
            printf("%lld %lld\n", (aa - 1) / 2, (aa + 1) / 2);
        else
            printf("%lld %lld\n", aa / 4 - 1, aa / 4 + 1);
    }
    return 0;
}