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==){
         printf("1=1\n");
         return ;
     }
     n=nn;
     if(n%==){
         fac[]=;
         n/=;
         while(n%==){
             fac[]++;
             n/=;
         }
     }
     for(i=;i<n;i+=){
         if(n%i==){
             n/=i;
             fac[i]=;
             while(n%i==){
                 n/=i;
                 fac[i]++;
             }
         }
     }
     if(n!=){
         fac[n]=;
     }
     printf("%lld=",nn);
     map<long long,int>::iterator it=fac.begin();
     printf("%lld",it->first);
     if(it->second>){
         printf("^%d",it->second);
     }
     it++;
     for(;it!=fac.end();it++){
         printf("*%lld",it->first);
         if(it->second>){
             printf("^%d",it->second);
         }
     }
     printf("\n");
     return ;
 }