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 = 0;
        p = 0;
    }
}info;
ll isPrime(ll N){
    ll sqr = (ll)sqrt(N * 1.0);
    if(N == 1)
        return 0;
    for(int i = 2; i <= sqr; i++){
        if(N % i == 0)
            return 0;
    }
    return 1;
}
ll primeTB[99999999];
info num[10000];
ll findPrime(ll tb[], ll maxNum){
    int index = 0;
    for(ll i = 2; 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) + 1;
    ll len = findPrime(primeTB, sqr);
    int pi = 0, index = 0, tag = 0;
    for(ll i = 0; N != 1 && i < len; i++){
        tag = 0;
        while(N % primeTB[i] == 0){
            N = N / primeTB[i];
            num[index].base = primeTB[i];
            num[index].p++;
            tag = 1;
        }
        if(tag == 1)
            index++;
    }
    if(N != 1){
        num[index].base = N;
        num[index++].p = 1;
    }
    if(index == 0){
        num[0].base = N;
        num[0].p = 1;
    }
    printf("%lld=", N2);
    printf("%d", num[0].base);
    if(num[0].p > 1){
        printf("^%d", num[0].p);
    }
    for(int i = 1; i < index; i++){
        printf("*%d", num[i].base);
        if(num[i].p > 1){
            printf("^%d", num[i].p);
        }
    }
    cin >> N;
    return 0;
}

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。