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 *…*pm^km.
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:97532468Sample Output:
97532468=2^2*11*17*101*1291
思路
按指定格式打印一个数的所有素(质)数因子。
1.先初始化建立一个素数表。
2.从2开始,不断用素数i除输入的数num,能够整除说明满足条件,用cnt记录被整除的次数。
3.对于第一个除数的输出进行判断来决定在除数前面是否加"*"。
4.cnt > 2就在当前除数后加"^cnt"。
5.重复2.3.4直至num < 2。
注:对于1要进行特殊处理,不然输出为"1=",正确输出应该是"1=1"。
代码
#include<iostream> #include<vector> using namespace std; vector<int> isPrime(666666,1); void Init() { for(int i = 2; i * i < 666666; i++) { for(int j = 2; j*i < 666666; j++) { isPrime[i * j] = 0; } } } int main() { Init(); long long num; cin >> num; cout << num <<"="; if(num == 1) cout << 1; bool first = true; for(int i = 2; num >= 2; i++) { int cnt = 0; if(isPrime[i] == 1 && num % i == 0) { while(num % i == 0) { cnt++; num = num/i; } if(first) first = false; else cout << "*"; cout << i; if(cnt > 1) cout << "^" << cnt; } } }