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 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.
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 1Hint60=2^2*3^1*5^1
Author
lcy
Source
Recommend
Ignatius.L
本题注意：格式
#include<stdio.h>
#include<math.h>
#include<string.h>
int pri[65538];
int ispri[65538];//32769
int n;
void prime() //求出素数数组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) //数据组中间要空行
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]); //每个数据后面都要有空格
}
printf("\n");
}
return 0;
}