PAT--1059. 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 *…*pm^km.

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 *…*pm^km, where pi’s are prime factors of N in increasing order, and the exponent ki is the number of pi – hence when there is only one pi, ki is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

题解

整数分解，算术基本定理。
素数打表预处理。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 10000;
long n;
bool vis[maxn];
vector<int> p;
vector< pair<int, int> > e;

void init(){
    for(int i = 2; i < maxn; ++i){
        if(!vis[i]) p.push_back(i);
        for(int j = i; j < maxn; j += i) vis[j] = true;
    }
}

int main(){

    init();
    cin >> n;

    if(n == 1){
        printf("1=1\n");
        return 0;
    }

    long nn = n;

    for(int i = 0; i < p.size() && nn; ++i){
        if(nn % p[i] == 0){
            int k = 0;
            while(nn % p[i] == 0){
                k++;
                nn /= p[i];
            }
            e.push_back(make_pair(p[i], k));
        }
    }
    printf("%ld=", n);
    for(int i = 0; i < e.size(); ++i){
        if(i) printf("*");
        printf("%d", e[i].first);
        if(e[i].second != 1) printf("^%d", e[i].second);
    }
    printf("\n");

    return 0;
}