PAT 1059 Prime Factors （素数筛）

1059 Prime Factors (25分)

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1k1×p2k2×⋯×pmkm.

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​​​*​​​p​2​​​^​​​k​2​​​*​​​…​​*​​​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

之前写的这个题很像，好像写过很多次zhe'zhon 

 需要注意的是测试点3:1 不是素数，因为我刚开始写的没有考虑 1 所以 测试点 3 一直过不去 特判一下就好了

#include <bits/stdc++.h>
#define ll long long
#define Max 1110000
using namespace std;
map<ll,int> cnt; //prime的次数
bool isprime[Max];
int prime[Max];
ll a; 
int isPrime(ll a){
  if(a == 1) return 0;
  for(int i = 2; i <= sqrt(a); i++){
    if(a % i == 0) return 0;
  }
  return 1;
}
void getPrime(){
  int ind = 0;
  memset(isprime, true, sizeof(isprime));
  for(int i = 2; i <= 1000000; i++){
    if(isprime[i]){
      prime[ind++] = i;
    }
    for(int j = 0; j < ind && prime[j] * i <= 1000000; j++){
      isprime[prime[j] * i] = false;
      if(i % prime[j] == 0) {
        break;
      }
    }
  }
}
int main(){
  set<ll> s;
  s.clear(); cnt.clear();
  getPrime();
  cin >> a;
  if(isPrime(a)) {
    cout << a << "=" << a << endl;
  }else{
    cout << a << "=";
    int ind = 0;
    ll temp = a;
    while(true){
      if(temp % prime[ind] == 0) {
        temp /= prime[ind];
        s.insert(prime[ind]);
        cnt[prime[ind]]++;
      }else {
        ind++;
      }
      if(temp == 1) break;
    }
    if(a == 1) {
      s.insert(1);
      cnt[1]++; 
    } 
    int num = s.size();
//    cout << "--->: " << num << endl;
    set<ll>::iterator it;
    for(it = s.begin(); it != s.end(); it++){
      if(cnt[(*it)] >= 2) {
        cout << (*it) << "^" << cnt[(*it)];
      }else {
        cout << (*it);
      }
      num --;
      if(num != 0) cout << "*";
    }
    cout << endl;
  }
  return 0;
}