因式分解(数学)(未完成 还需完善)

因式分解

 

# include <map>
# include <stack>
# include <queue>
# include <math.h>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
using namespace std;

int a[100000010];
int b[100000010];

void run()
{
    int t, num, q, n;
    scanf("%d", &n);
    t = sqrt(n);
    num = 0;
    for(int i = 2; i <= t; i++)
    {
        if(n % i == 0)
        {
            num++;
            a[num] = i;
        }
    }
    q = num;
    if(t*t == n)
        num--;
    for(int i = q; i >= 1; i--)
    {
        num++;
        a[num] = n/a[i];
    }
    num++;
    a[num] = n;
    b[1] = 1;
    for(int i = 2; i <= num; i++)
    {
        b[i] = 1;
        for(int j = i-1; j >= 1; j--)
        {
            if(a[i]%a[j] == 0)
                b[i] += b[j];
        }
    }
    printf("%d", b[num]);
}

int main(void)
{
    run();

    return 0;
}
因式分解的个数

 

 

# include <map>
# include <stack>
# include <queue>
# include <math.h>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
using namespace std;

int t = 0;
int p;
int s[100000010];

void f(int m, int n)
{
    int i,j;
    i = m;
    j = n;
    if(i < j)
    {
        if(j % i == 0)
        {
            t = t+1;
            s[t] = i;
            for(int k = 1; k <= t; k++)
            {
                printf("%d*", s[k]);
            }
            printf("%d=%d\n", j/i, p);
            f(2, n/i);
            t = t-1;
            f(i+1, j);
        }
        else
        {
            f(i+1, j);
        }
    }
}

void run()
{
    while(~scanf("%d", &p))
    {
        printf("%d=%d\n", p, p);
        f(2, p);
    }
}

int main(void)
{
    run();

    return 0;
}
所有因式分解的情况

 

posted @ 2013-10-22 20:54  GLSilence  阅读(232)  评论(0编辑  收藏  举报