1059 Prime Factors (25分)

题目

Given any positive integer $N$, you are supposed to find all of its prime factors, and write them in the format $N={p_1}^{k_1}\times {p_2}^{k_2}\times \cdots {p_m}^{k_m}$.

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 $N={p_1}^{k_1}\times {p_2}^{k_2}\times \cdots \times {p_m}^{k_m}$, 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

题目大意

输出一个数的所有质因子。

思路

求质数还是采用常规方法，只是判断的范围可以动态变化，有这样的规律，当前因子为k则计算下一个因子从k开始，并且定义一个临时变量t，t等于已求出因子的乘积，那么判断的上限就是n/t。

代码

#include 
#include 
#include 
#include 
using namespace std;

bool isPrime(int x){
    for(int i=2; i<=sqrt(x); i++){
        if(x%i == 0)
            return false;
    }
    return true;
}

int main(){
    map<int,int> mp;
    int n;
    scanf("%d", &n);
    int cnt = 2, t = n;
    while(t > 1){
        for(int i=cnt; i<=t; i++){ // 上下限动态变化
            if(isPrime(i) && t%i == 0){
                t /= i;
                cnt = i;
                //printf("%d\n", i);
                map<int,int>::iterator it = mp.find(i);
                if(it != mp.end())
                    mp[i]++;
                else
                    mp[i] = 1;
                break;
            }
        }
    }
    printf("%d=", n);
    for(map<int,int>::iterator it=mp.begin(); it!=mp.end(); it++){
        if(it != mp.begin())
            printf("*");
        printf("%d", it->first);
        if(it->second > 1)
            printf("^%d", it->second);
    }
    if(mp.empty())
        printf("%d", n);
    printf("\n");
    return 0;
}