HDU1405 The Last Practice

                                                    The Last Practice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6065    Accepted Submission(s): 1235

Problem DescriptionTomorrow 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. 

InputInput 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. 

OutputFor 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

6012-1

 

Sample Output

Case 1.2 2 3 1 5 1Case 2.2 2 3 1Hint60=2^2*3^1*5^1 

 

Authorlcy 

Source杭電ACM集訓隊訓練賽（IV）
 

RecommendIgnatius.L

本題注意：格式

#include<stdio.h>#include<math.h>#include<string.h>int pri[65538];int ispri[65538];//32769int n;void prime()   //求出素數數組pri[]{    int i,j;    memset(ispri,true,sizeof(ispri));    for(i=2;i<32769;i++)        for(j=i+i;j<65538;j+=i)            ispri[j]=false;    i=0;    for(j=2;j<65538;j++)        if(ispri[j])            pri[i++]=j;    n=i;}int main(){    int m;    prime();    int a[n];    int x=1;    while(scanf("%d",&m)&&m>0)    {        memset(a,0,sizeof(a));        if(x!=1)     //資料群組中間要空行            printf("\n");        for(int i=0;m!=1&&i<n;)        {            if(m%pri[i]==0)            {                a[i]++;                m/=pri[i];            }            else                i++;        }        printf("Case %d.\n",x++);        for(int i=0;i<n;i++)        {            if(a[i]!=0)                printf("%d %d ",pri[i],a[i]);   //每個資料後面都要有空格        }        printf("\n");    }    return 0;}