PAT 1059 Prime Factors(质因数分解)

最新推荐文章于 2024-10-18 17:10:13 发布

原创最新推荐文章于 2024-10-18 17:10:13 发布 · 220 阅读

0 ·

CC 4.0 BY-SA版权

数据结构专栏收录该内容

46 篇文章

订阅专栏

本文介绍了一种质因数分解算法，用于将任意正整数N分解为其所有质因数的乘积形式。通过使用C++编程语言，详细展示了算法的实现过程，并提供了完整的代码示例。该算法能够有效地找出并打印出正整数的所有质因数及其指数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.

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 = p1^k1*p2^k2*…*pm^km, where pi's are prime factors of N in increasing order, and the exponent ki is the number of pi -- hence when there is only one pi, ki is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

代码

#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
typedef long long LL;
vector<int> factor;
vector<int> index;
void Divide(LL N)
{
    int Num,Temp=N;
    //质因数分解算法
    for(int i=2;i*i<=N;i++)
    {
        if(N%i==0)
        {
            factor.push_back(i);
            Num=0;
            while(N%i==0)
            {
                N/=i;
                Num+=1;
            }
            index.push_back(Num);
        }
    }
    if(N>1)
    {
        factor.push_back(N);
        index.push_back(1);
    }
    int Len=factor.size();
    printf("%d=",Temp);
    for(int i=0;i<Len;i++)
    {
        if(i==0)
        {
            if(index[i]!=1)
                printf("%d^%d",factor[i],index[i]);
            else
                printf("%d",factor[i]);
        }
        else
        {
            if(index[i]!=1)
                printf("*%d^%d",factor[i],index[i]);
            else
                printf("*%d",factor[i]);
        }
    }
    return ;
}
int main()
{
    LL N;
    scanf("%lld",&N);
    if(N==1)
        printf("1=1");
    else
        Divide(N);
    return 0;
}