uva 106 - Fermat vs. Pythagoras(素勾股数）

来源：互联网发布：java点击发送邮件源码编辑：程序博客网时间：2024/05/16 08:45

题目大意：uva 106 - Fermat vs. Pythagoras

题目大意：给出n，计算n以内有多少对素勾股数，并计算出n以内有多少数可以用来组成勾股数。

解题思路：暴力应该是会超时，本题肯定是考查勾股数的性质，上维基查了一下勾股数，上面讲的很清楚，只要将构造方法实现就好了。

#include <stdio.h>#include <string.h>#include <math.h>const int N = 1000010;bool vis[N];long long gcd(long long a, long long b) {return b == 0 ? a : gcd(b, a % b);}int main () {long long a, b, c, n, cntans, cntuse;while (scanf("%lld", &n) == 1) {cntans = cntuse = 0;memset(vis, 0, sizeof(vis));long long m = (long long)sqrt(n + 0.5);for (long long t = 1; t <= m; t += 1) {for (long long s = t + 1; s * t <= n; s += 2) {if (gcd(s, t) == 1) {a = s * t * 2;b = (s * s - t * t);c = (s * s + t * t);if (c <= n) {cntans++;if (!vis[a]) { cntuse++; vis[a] = 1; }if (!vis[b]) { cntuse++; vis[b] = 1; }if (!vis[c]) { cntuse++; vis[c] = 1; }}for (int i = 2; c * i <= n; i++) {if (!vis[a * i]) { cntuse++; vis[a * i] = 1; }if (!vis[b * i]) { cntuse++; vis[b * i] = 1; }if (!vis[c * i]) { cntuse++; vis[c * i] = 1; }}}}}printf("%lld %lld\n", cntans, n - cntuse);}return 0;}