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?=?p?1???k?1????×p?2???k?2????×?×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??*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

解析:素數篩選，預處理一下就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; }

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