PAT甲级 1059. Prime Factors (25)

题目：

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

思路：

1.先构建质数表，根据N的大小，把1~sqrt(N)之间的质数找出来；

2.根据质数表，从小到大把N的因子找出，这里利用map比较好做；

3.注意N=1的情况

代码：

#include<iostream>#include<cstring>#include<map>#include<math.h>using namespace std;int main(){//ifstream cin;//cin.open("case1.txt");long N;cin >> N;if (N == 1)printf("1=1\n");else{//构建质数表bool *prime = (bool *)malloc(sizeof(bool) * sqrt(N));memset(prime, 0, sizeof(bool)*sqrt(N));int j, i = 2;for (; i <= sqrt(N); ++i){if (!prime[i]){for (j = 2 * i; j <= sqrt(N); j += i){prime[j] = 1;}}}//根据质数表进行分解Nlong NN = N;map<int, int> p;for (i = 2; i <= N;){if ((!prime[i]) && (N%i == 0)){//遇到质数，且该质数是N的因数p[i]++;N /= i;}else{//跳过++i;}}//outputmap<int, int>::iterator iter = p.begin();printf("%ld=", NN);while (1){printf("%d", iter->first);if (iter->second > 1){printf("^%d", iter->second);}iter++;if (iter != p.end()){printf("*");}else{break;}}printf("\n");}    system("pause");return 0;}