素数 代码

按照《算法分析与设计》里的思路：

1.

// 输出 100 以内的素数

#include<iostream>

#include<math.h>

using namespace std;

const int NUM = 101;

int main(int argc,char *argv[])

{

                 int prime[NUM];

                 int i(0),j(0);

                 for(;i!=NUM;++i)prime[i] = i;

                 for(i = 2;i <= sqrt(NUM);++i){

                                 if(prime[i] == 0)continue ;

                                 for(j = i*i;j < NUM;j += i)

                                                prime[j] = 0;

                }

                

                 for(i=2;i!=NUM;++i)

                                 if(prime[i]!=0)cout<<prime[i]<<endl;

                getchar();

                 return 0;

}

2.

修改后，得到一个区间的素数

// 输出 low 到 high 之间的素数

#include<iostream>

#include<math.h>

using namespace std;

int main(int argc,char *argv[])

{

                 int low,high;

                cout<< "input your numbers like: 1 100" <<endl;

                cin>>low>>high;

                 if(low < 2 || low >= high){

                                cout<< "input error!"<<endl;

                                getchar();

                                 return -1;

                }

                 // 动态数组，初始化

                 int *prime = new int[high - low + 1];

                 int i;

                 for(i = 0;i != high - low + 1;++i)

                                prime[i] = low + i;

                 for(i = 2;i <= sqrt(high);++i){                                              // 外层

                                 if(i*i <= sqrt(low)){                                                               // i <= 小数的平方根

                                                 int j = 0;

                                                 while(prime[j] % i != 0)++j;               // 寻找第一个能被整除的值

                                                 for(;j < high - low + 1;j += i)

                                                                prime[j] = 0;

                                }

                                 else{                                                                                    // i > 小数的平方根，直接找到数值为 i*i 的下标，即 i*i - m

                                                 int j = i*i - low;

                                                 for(;j < high - low + 1;j += i)

                                                                prime[j] = 0;

                                }

                }

                 for(i = 0;i != high - low + 1;++i)

                                 if(prime[i] != 0)cout<<prime[i]<<endl;

delete[] prime;

                getchar();

                getchar();

                 return 0;

}