Project Euler 75: Singular integer right triangles

题目链接

思路：

勾股数组，又称毕达格拉斯三元组。

公式：a = s*t　　b = (s^2 - t^2) / 2　　c = (s^2 + t^2) / 2　　s > t >=1 且为互质的奇数

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "
";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 1500000;
int cnt[N+1];
int main() {
    for (LL s = 1; ; s += 2) {
        if(s > N) break;
        for (LL t = 1; t < s; t += 2) {
            LL a = s*t, b = (s*s - t*t)/2, c = (s*s + t*t)/2;
            if(a > N || b > N || c > N) break;
            if(__gcd(s, t) > 1) continue;
            if(a+b+c > N) continue;
            LL tot = a+b+c;
            for (LL i = tot; i <= N; i += tot) cnt[i]++;
        }
    }
    int ans = 0;
    for (int i = 1; i <= N; ++i) if(cnt[i] == 1) ans++;
    cout << ans << endl;
    return 0;
}