关于整数的质因子分解和筛素数的分析

本文出自：blog.csdn.net/svitter

筛素数筛不到10000000

#include <iostream>
#include <cstdio>
#include <string.h>
#include <cmath>
#include <time.h>
using namespace std;


#define N  1000000//12031230

int digit[N];    //记录素因子
int num[N];      //记录每个素因子个数 
int c;              //记录素因子的个数

void divide(int n) {
    memset(digit, 0, sizeof(digit));
    memset(num ,0, sizeof(num));

    int temp = n;
    for(long long i = 2; i <= n; i++) {
        if(temp % i== 0){
            digit[c] = i;
            num[c] = 0;
            c++;
        }
        while(temp % i== 0){
            num[c-1]++;
            temp /= i;
        }
    }
}

typedef long long ll;
ll prime[N], nprime, factor[N], numfactor[N], ct;
bool isprime[N];
void makeprime() {// 打 1~N 的素数表
    int i, j, temp;
    nprime = 0;
    memset(isprime, 1, sizeof(isprime));
    isprime[1] = 0;
    temp = sqrt(N + 0.0);
    for(i = 2;i <= temp; ++i) {
        if(isprime[i]) {
            ++nprime;
            prime[nprime] = i;
            for(j=i+i; j < N; j += i) {
                isprime[j] = 0;
            }
        }
    }
}

void makeprime2() { // 线性筛素数
    memset(isprime, 1, sizeof(isprime));
    int i, j;
    int num = 0;
    for(i = 2 ; i <= N; i++)
    {
        if(isprime[i])
            prime[num++] = i;
        for(j = 1; j < num && i * prime[j] <= N; j++)
        {
            isprime[i * prime[j]] = 0;
            if(i % prime[j] == 0) break;
        }
    }
}

void divide2(ll n) {// 进行素数分解
    int i;
    int temp = sqrt(n + 0.0);
    ct=0;
    memset(numfactor, 0, sizeof(numfactor));
    for(i = 1; i <= nprime; ++i) {
        if(prime[i] > temp) break;
        if(n % prime[i] == 0) {
            factor[++ct]=prime[i];
            while(n % prime[i] == 0) {
                numfactor[ct]++;
                n /= prime[i];
            }
        }
    }
    if(n != 1) {
        factor[++ct] = n;
        numfactor[ct]++;
    }
} 


int main() {
    
    // 针对单一的数字
    clock_t t1 = clock(),t2;
    divide(N);
    t2 = clock();
    printf("cost time is: %d\n", t2-t1);
    
    t1 = clock();
    makeprime();
    divide2(N);
    t2 = clock();
    printf("cost time is: %d\n", t2-t1);


    t1 = clock();
    makeprime2();
    divide2(N);
    t2 = clock();
    printf("cost time is: %d\n", t2-t1);
    
    
    // 复杂度也是o(n)， 因为sqrt
    t1 = clock();
    makeprime();
    t2 = clock();
    printf("cost time is: %d\n", t2-t1);

    // 所谓线性筛素数
    t1 = clock();
    makeprime2();
    t2 = clock();
    printf("cost time is: %d\n", t2-t1);
    
    return 0;

}

运行结果：