pat1059. Prime Factors (25)

1059. Prime Factors (25)

时间限制

50 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HE, Qinming

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p₁^k₁ * p₂^k₂ *…*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 = 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

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提交代码

 

思路：N = p₁^k₁ * p₂^k₂ *…*p_m^k(p_1<p2_<..pm-1<pm)  则从i=3开始循环，i依次增大，nn除以自己的质因数，不断减少。这里要注意，N最大的质因数可能只有一个，就是最后剩下的nn，这个需要判断。

#include<cstdio>
#include<stack>
#include<cstring>
#include<iostream>
#include<stack>
#include<set>
#include<map>
using namespace std;
map<long long,int> fac;
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    long long i,nn,n;
    scanf("%lld",&nn);
    if(nn==1){
        printf("1=1\n");
        return 0;
    }
    n=nn;
    if(n%2==0){
        fac[2]=1;
        n/=2;
        while(n%2==0){
            fac[2]++;
            n/=2;
        }
    }
    for(i=3;i<n;i+=2){
        if(n%i==0){
            n/=i;
            fac[i]=1;
            while(n%i==0){
                n/=i;
                fac[i]++;
            }
        }
    }
    if(n!=1){
        fac[n]=1;
    }
    printf("%lld=",nn);
    map<long long,int>::iterator it=fac.begin();
    printf("%lld",it->first);
    if(it->second>1){
        printf("^%d",it->second);
    }
    it++;
    for(;it!=fac.end();it++){
        printf("*%lld",it->first);
        if(it->second>1){
            printf("^%d",it->second);
        }
    }
    printf("\n");
    return 0;
}