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最新推荐文章于 2022-05-29 14:03:39 发布

原创最新推荐文章于 2022-05-29 14:03:39 发布 · 194 阅读

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本文介绍了一种质因数分解算法，该算法能够将任意正整数N分解为其所有质因数的乘积形式，即N=p1^k1*p2^k2*...*pm^km，并提供了详细的算法实现代码。

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p
1

k
1

×p
2

k
2

×⋯×p
m

k
m

.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range of long int.

Output Specification:

Factor N in the format N = p
1
^k
1
*p
2
^k
2
*…*p
m
^k
m
, where p
i
's are prime factors of N in increasing order, and the exponent k
i
is the number of p
i
– hence when there is only one p
i
, k
i
is 1 and must NOT be printed out.

Sample Input:

97532468
Sample Output:

97532468=2^211171011291

代码

#include<stdio.h>
struct fac
{
    int x;
    int count=0;
}f[10];
const int max=1000001;
int p[max]={0};
int k=0;
bool fa[max]={false};
void pri()
{
    for(int i=2;i<max;i++)
    {
        if(fa[i]==false)
            {
                p[k++]=i;
            for(int j=i+i;j<max;j=j+i)
            {
                fa[j]=true;
            }
            }
    }
}
int main()
{
    void pri();
    pri();
    int n;
    int i=0;
    int t=0;
    scanf("%d",&n);printf("%d=",n);
    if(n==1)
    {
      printf("1");
       return 0;
    }
    while(n!=1)
    {
        if(n%p[i]==0)
        {
            f[t].x=p[i];
            while(n%p[i]==0)
            {
                f[t].count++;
                n=n/p[i];
            }
            t++;
        }
        else
                i++;
    }
    
    for(int j=0;j<t;j++)
    {
        printf("%d",f[j].x);
        if(f[j].count!=1)
            printf("^%d",f[j].count);
        if(j!=t-1)
        printf("*");
    }
}