程序代码：

#include <time.h>
#include <math.h>
#include <stdio.h>
#include <string.h>

int a[240] = {0};

void InitArray(int n)
{
    int i, temp = 2, j;

    for (i = a[0];i > 0;a[i--] = 0);
    for (i = 2, j = 1;n != 1;++i)
    {
        if (n % i)    continue;
        if (temp != i)
        {
            j += !!a[j];
            temp = i;
        }
        n /= i--, ++a[j];
    }
    a[0] = j;
}
int main()
{
    clock_t t2, t1 = clock();

    int j, temp;
    for (int i = 1000000;i > 240;--i)
    {
        InitArray(i);
        for (temp = j = 1;j <= a[0];++j)
            temp *= a[j] + 1;
        if (temp == 240)    printf("%d\t", i);
    }
    puts("");

    t2 = clock();
    printf("%.3f s\n", (t2-t1)/1000.0);
    return 0;
}

程序代码：

#include <time.h>
#include <math.h>
#include <stdio.h>

int a[240] = {0};
int ss[1230] = {2, 3, 5, 7, 11, 13, 17, 19};

void Init()
{    //10000以内质数表，1229个
    int i, j, k = 8, temp;
    for (i = 23;k != 1230;i += 2)
    {
        temp = (int)sqrt(i);
        for (j = 0;ss[j] <= temp;++j)
        {
            if (i % ss[j])    continue;
            break;
        }
        if (ss[j] <= temp)    continue;
        ss[k++] = i;
    }
}
void InitArray(int n)
{   //把 n分解质因数，指数部分依次存入数组 a，a[0]存储有效数位
    // 例如 240 = 2 ^ 4 * 3 * 5，则 数组 a：3 4 1 1
    int i, temp = 2, j;
    int s = (int)sqrt(n);
    
    for (i = a[0];i > 0;a[i--] = 0);    //数组 a清 0
    for (i = 0, j = 1;n != 1 && ss[i] <= s;++i)
    {
        if (n % ss[i])    continue;
        if (temp != ss[i])
        {
            j += !!a[j];
            temp = ss[i];
        }
        n /= ss[i--], ++a[j];
    }
    if (1 != n)    ++a[j += !!a[j]];
    a[0] = j;
}
int main()
{
    clock_t t2, t1 = clock();

    int j, temp;Init();
    for (int i = 100000000;i >= 240;--i)
    {
        InitArray(i);

        //按数组 a求出因数个数，240的因数个数为 (4+1)*(1+1)*(1+1) = 20
        for (temp = j = 1;j <= a[0];++j)
            temp *= a[j] + 1;

        if (temp >= 240)    printf("%10d\t", i);
    }
    puts("");

    t2 = clock();
    printf("%.3f s\n", (t2-t1)/1000.0);
    return 0;
}