HDU 1405 The Last Practice

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本文解析了一道关于素数分解的编程题，题目要求将输入的正整数分解为若干个素数的乘积，并按升序输出素数及其出现次数。文章提供了完整的C语言代码实现，包括筛选素数的方法及分解过程。

http://acm.hdu.edu.cn/showproblem.php?pid=1405

The Last Practice

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4309 Accepted Submission(s): 872

Problem Description

Tomorrow is contest day, Are you all ready?
We have been training for 45 days, and all guys must be tired.But , you are so lucky comparing with many excellent boys who have no chance to attend the Province-Final.

Now, your task is relaxing yourself and making the last practice. I guess that at least there are 2 problems which are easier than this problem.
what does this problem describe?
Give you a positive integer, please split it to some prime numbers, and you can got it through sample input and sample output.

Input

Input file contains multiple test case, each case consists of a positive integer n(1<n<65536), one per line. a negative terminates the input, and it should not to be processed.

Output

For each test case you should output its factor as sample output (prime factor must come forth ascending ), there is a blank line between outputs.

Sample Input

   60 12 -1
  

Sample Output

   Case 1. 2 2 3 1 5 1 Case 2. 2 2 3 1 Hint: 60=2^2*3^1*5^1
  

Author

lcy

Source

杭电ACM集训队训练赛（IV）

Recommend

Ignatius.L

分析：筛选素数，注意格式。。

代码如下：

#include<stdio.h>
#include<string.h>
bool prime[65540];
void fun()
{
int i,j;
memset(prime,1,sizeof(prime));
for(i=2;i<65540;i++)
{
for(j=2;j*i<65540;j++)
if(prime[i])
prime[i*j]=0;
}
prime[0]=prime[1]=0;
}
int main()
{
fun();
int i,j,n,count=0;
int z;
while(scanf("%d",&n)==1)
{
if(n<0) break;
if(count!=0) printf("\n");
printf("Case %d.\n",++count);
if(n==1||n==0) continue;
for(i=2;i<65540;i++)
{
z=0;
if(n%i==0&&prime[i])
{
while(n%i==0)
{
n=n/i;
z++;
}
printf("%d %d ",i,z);
}
if(n==1)
{
printf("\n");
break;
}
}
}
return 0;
}