PAT1059. Prime Factors

Sep 1, 2018

题目

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmk**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 = p1^k1*p2^k2*…*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=2^2*11*17*101*1291

分析

先生成十万以内的素数表，然后在素数表内枚举n的因数。

对于因子查找的结论：如果n存在[1,n]的因子，则在sqrt(n)的两边对称存在
对于质因数有一个结论：对一个正整数n来说，如果它存在[2,n]范围内的质因子，要么这些质因子全部小于等于sqrt(n)，要么只存在一个位于[sqrt(n),n]的质因子。

代码

#include<cstdio>
#include<cmath>
int primeList[100001];
int primeNum=0;

struct factor {
    int f;
    int e;
}factor[100];

bool isPrime(int n)
{
    int sqrtn = sqrt(n);
    for(int i=2;i<=sqrtn;i++)
        if(n%i==0) return 0;
    return 1;
}
void buildPrime()
{
    for(int i=2;i<100001;i++)
        if(isPrime(i))
            primeList[primeNum++]=i;
}




int main()
{

    long int n;
    scanf("%ld",&n);
    printf("%ld=",n);
    if(n==1)
        printf("1");
    else
    {
        buildPrime();
        int num=0;
        int sqrtn=sqrt(n);
        //find factor rang of [2...sqrtn]
        for(int i=0;primeList[i]<=sqrtn;i++)
        {
            if(n%primeList[i]==0)
            {
                factor[num].f=primeList[i];
                while(n%primeList[i]==0)
                {
                    n/=primeList[i];
                    factor[num].e++;
                }
                num++;
            }
        }

        if(n>1)
        {
            factor[num].f=n;
            factor[num].e=1;
            num++;
        }

        //print
        for(int i=0;i<num;i++)
        {
            if(i>0)printf("*");
            if(factor[i].e>1)
                printf("%d^%d",factor[i].f,factor[i].e);
            else
                printf("%d",factor[i].f);
        }
    }
    return 0;
}