POJ 2773 欧拉函数， 素数表

这题很很很不懂，谁能详细讲解下。

素数表第一次用，用了不熟练，欧拉函数求了好久出来，不太懂。

#include <stdio.h>#include <string.h>#include <math.h>#define MAXN 1000010int prime[MAXN], phi[MAXN];void Init(){int i, j, k = ceil(sqrt(1000010));//求素数表memset(prime, 0, sizeof(prime));prime[0] = prime[1] = 1;for (i = 2; i < k; i++){if (prime[i] == 0){for (j = 2 * i; j < MAXN; j += i)prime[j] = 1;}}//求欧拉函数值for (i = 1; i < MAXN; i++)phi[i] = i;for (i = 2; i < MAXN; i++){if (prime[i] == 0){for (j = i; j < MAXN; j += i)//phi(n) = n * (1 - 1/p1) * (1 - 1/p2)....[pi是与n的质因素]phi[j] = phi[j] / i * (i - 1);}}}int GCD(int a, int b){int t;while (b){t = b;b = a % b;a = t;}return a;}int main(){int m, K, cnt, i, flag, mul;Init();while (scanf("%d%d", &m, &K) != EOF){//判断m内第几个与m互素cnt = K % phi[m];//判断分为了几组mul = K / phi[m];//正好分为整数组if (cnt == 0){//m的最后一个与其互素的cnt = phi[m];//分组减一mul--;}flag = 0;for (i = 1; i < m; i++)//判断是否互素if (GCD(m, i) == 1){flag++;if (flag == cnt)break;}//printf("phi[m] = %d i = %d mul = %d\n", phi[m], i, mul);printf("%d\n", mul * m + i);}return 0;}

这个是我第n次RE后写的，超时了，有什么特别好的优化方法吗？ 请教下大家

#include <stdio.h>int Coprime (int a, int b){int t;while (b){t = b;b = a % b;a = t;}if (a == 1)return 1;return 0;}int main(){int i, m, cnt, K, flag, phi_m, mul;while (scanf("%d%d", &m, &K) != EOF){phi_m = flag = 0;for (i = 1; i <= m; i++)if (Coprime(m, i))phi_m++;cnt = K % phi_m;mul = K / phi_m;if (cnt == 0){cnt = phi_m;mul--;}for (i = 1; i <= m;  i++)if (Coprime(m, i)){flag++;if (flag == cnt)break;}printf("%d\n", mul * m + i);}return 0;}

今天又练了到素数表的题, 不过时间好慢200+MS 不知道别人的0MS怎么弄出来的

有谁知道怎么优化比较好, 发个链接, 谢谢。

POJ 3006 筛素数

#include <stdio.h>#include <string.h>int prime[1000010];void init(){int i, j;memset(prime, 0, sizeof(prime));prime[0] = prime[1] = 1;for (i = 2; i < 1000; i++){if (prime[i] == 0)for (j = 2 * i; j < 1000000; j += i)prime[j] = 1;}}int main(){int a, d, n, num, i;init();while (scanf("%d %d %d", &a, &d, &n) != EOF){if (!(a || d || n)) break;num = i = 0;while (num != n){if (prime[a + i * d] == 0) num++;i++;}printf("%d\n", a + (i - 1) * d);}return 0;}