标签:数组 project return gcc bug ace break problem get
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 << "\n"; #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; }
Project Euler 75: Singular integer right triangles
标签:数组 project return gcc bug ace break problem get
原文地址:https://www.cnblogs.com/widsom/p/10425908.html