关于整数的质因子分解和筛素数的分析

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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//12031230int 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 = 0; 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：

于2014.11.26修正：线性筛素数应该从第一个素数开始。

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