PAT (Advanced Level) Practise 1059 Prime Factors (25)

1059. Prime Factors (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HE, Qinming

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 *…*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 = p₁^k₁ * p₂^k₂ *…*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

题意：给你一个数，将它拆成几个素数的乘积

解题思路：素数筛找出所有素数，然后暴力找出哪些素数能被整除

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int visit[1000009], prime[1000009];LL n;void init(){memset(visit, 1, sizeof visit);visit[0] = visit[1] = 0;int cnt = 0;for (int i = 2; i <= 1000005; i++){if (!visit[i]) continue;prime[cnt++] = i;for (int j = i * 2; j <= 1000005; j += i) visit[j] = 0;}}int main(){init();while (~scanf("%lld", &n)){printf("%d=", n);if (n == 1) { printf("%d", n); continue; }int flag = 0;for (int i = 0; prime[i] <= n; i++){if (n%prime[i]) continue;int sum = 0;while (n%prime[i] == 0) { sum++; n /= prime[i]; }if (flag) printf("*");flag = 1;printf("%d", prime[i]);if (sum > 1) printf("^%d", sum);}printf("\n");}return 0;}