A1059. Prime Factors

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

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 typedef long long ll;
 typedef struct pt{
     int base, p;
     pt(){
         base = ;
         p = ;
     }
 }info;
 ll isPrime(ll N){
     ll sqr = (ll)sqrt(N * 1.0);
     if(N == )
         return ;
     for(int i = ; i <= sqr; i++){
         if(N % i == )
             return ;
     }
     return ;
 }
 ll primeTB[];
 info num[];
 ll findPrime(ll tb[], ll maxNum){
     int index = ;
     for(ll i = ; i <= maxNum; i++){
         if(isPrime(i)){
             tb[index++] = i;
         }
     }
     return index;
 }
 int main(){
     ll N, sqr, N2;
     scanf("%lld", &N);
     N2 = N;
     sqr = (int)sqrt(1.0 * N) + ;
     ll len = findPrime(primeTB, sqr);
     int pi = , index = , tag = ;
     for(ll i = ; N !=  && i < len; i++){
         tag = ;
         while(N % primeTB[i] == ){
             N = N / primeTB[i];
             num[index].base = primeTB[i];
             num[index].p++;
             tag = ;
         }
         if(tag == )
             index++;
     }
     if(N != ){
         num[index].base = N;
         num[index++].p = ;
     }
     if(index == ){
         num[].base = N;
         num[].p = ;
     }
     printf("%lld=", N2);
     printf("%d", num[].base);
     if(num[].p > ){
         printf("^%d", num[].p);
     }
     for(int i = ; i < index; i++){
         printf("*%d", num[i].base);
         if(num[i].p > ){
             printf("^%d", num[i].p);
         }
     }
     cin >> N;
     return ;
 }

总结：

1、判断素数的时候，循环条件为 i <= sqr。

2、生成的质数表可以范围到10^5, 也可以生成到 根号N的范围。

3、15 = 3 * 5，找因子只用找到根号N的范围，如果循环结束N仍然不等于1时，说明它就是大于根号N的一个因子，或者是N本身。