1059 Prime Factors

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 = 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

解析:素数筛选，预处理一下就o了

#include<set>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<cmath>
#include<stack>
#include<cstdio>
#include<string>
#include<bitset>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<bits/stdc++.h>
using namespace std;#define e exp(1)
#define p acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b) {return b?gcd(b,a%b):a;
}const int maxn=5e5+5;
int cnt=0;
int prime[maxn],f[maxn+5];
void init()
{for(int i=2; i<maxn; i++){if(!f[i]){prime[cnt++]=i;for(int j=i; j<maxn; j+=i){f[j]=1;}}}
}
int main()
{init();ll n;scanf("%lld",&n);printf("%lld=",n);if(n==1){puts("1");return 0;}int count=0,flag=0;for(int i=0; i<cnt; i++){if(n==1)break;if(n%prime[i]==0){printf("%d",prime[i]);n/=prime[i];count=1;flag=0;while(n%prime[i]==0){if(n==0)break;if(!flag)printf("^");n/=prime[i];count++;flag=1;}if(flag)printf("%d",count);if(n!=1)printf("*");}}puts("");return 0;
}