1059 Prime Factors （25 分）素数 “筛选法”

1059 Prime Factors （25 分）

Given any positive integer NNN, you are supposed to find all of its prime factors, and write them in the format NNN = p1k1×p2k2×⋯×pmkm{p_1}^{k_1} imes {p_2}^{k_2} imes cdots imes {p_m}^{k_m}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 NNN in the range of long int.

Output Specification:

Factor NNN in the format NNN = p1p_1p​1​​^k1k_1k​1​​*p2p_2p​2​​^k2k_2k​2​​*…*pmp_mp​m​​^kmk_mk​m​​, where pip_ip​i​​'s are prime factors of NNN in increasing order, and the exponent kik_ik​i​​ is the number of pip_ip​i​​ -- hence when there is only one pip_ip​i​​, kik_ik​i​​ is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291
思路：
　　1不是素数

#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cmath>
#include<climits>
#include<sstream>
#include<cstdio>
#include<string.h>
#include<unordered_map>
using namespace std;
unordered_map<string,set<int>>mp;
int main()
{
    long long int n;
    scanf("%lld",&n);
    int num=sqrt(n);
    vector<int> primes(num,1);
    for(int i=2;i<=sqrt(num);i++)
    {
        for(int j=i+i;j<=sqrt(num);j+=i)
            primes[j]=0;
    }
    if(n==1)
        printf("1=1");
    else
    {
        printf("%lld=",n);
        int state=0;
        for(int i=2;i<=num;i++)
        {
            int cnt=0;
            bool flag=false;
            while(primes[i]==1&&n%i==0)
            {
                n/=i;
                cnt++;
                flag=true;
            }
            if(flag)
            {
                if(state!=0) printf("*");
                state=1;
                printf("%d",i);
                if(cnt>1)
                    printf("^%d",cnt);
            }
            if(n==1)
                break;
        }
        if(n!=1)
        {
            if(state!=0) printf("*");
            printf("%d",n);
        }
    }
    return 0;
}