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:
97532468
Sample 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;
}
}
}