HDU 1164 Eddy's research I

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1164

Eddy's research I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8736    Accepted Submission(s): 5383

Problem Description

Eddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can't write program, so Eddy has
to ask intelligent you to help him, he asks you to write a program which can do the number to divided into the multiply of prime number factor .

 

Input

The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.

 

Output

You have to print a line in the output for each entry with the answer to the previous question.

 

Sample Input

11
9412

 

Sample Output

11
2*2*13*181

 

Author

eddy
 

Recommend

JGShining

思路：简单的算术基本定理，不解释，直接上代码。

附上AC代码：

#include <bits/stdc++.h>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
const int maxn = 1<<8;
const int maxm = 54;
bool is_prime[maxn] = {true, true};
int prime[maxm];
int n;

void filter(){
int cnt = 0;
for (int i=2; i<maxn; ++i)
if (!is_prime[i]){
prime[cnt++] = i;
for (int j=i*i; j<maxn; j+=i)
is_prime[j] = true;
}
// printf("%d\n", cnt);
}

int main(){
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
filter();
while (~scanf("%d", &n)){
bool ok = false;
for (int i=0; i<maxm&&prime[i]*prime[i]<=n; ++i)
if (n%prime[i] == 0){
if (ok)
printf("*");
else
ok = true;
printf("%d", prime[i]);
n /= prime[i];
while (n%prime[i] == 0){
printf("*%d", prime[i]);
n /= prime[i];
}
}
if (n > 1){
if (ok)
printf("*");
printf("%d", n);
}
printf("\n");
}
return 0;
}