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