2013
12-09

# The Last Practice

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 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.

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.

60
12
-1

Case 1.
2 2 3 1 5 1

Case 2.
2 2 3 1
Hint

60=2^2*3^1*5^1

#include<iostream>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<algorithm>
using namespace std;
#define M 100000
bool visit[101000];
int prime[101000];
int num2[14000];
struct node
{
int num;
int a;
} unit[7000];
void table()
{
memset(visit,true,sizeof(visit));
int num = 0;
for (int i = 2; i <= M; ++i)
{
if (visit[i] == true)
{
num++;
prime[num] = i;
}
for (int j = 1; ((j <= num) && (i * prime[j] <= M));  ++j)
{
visit[i * prime[j]] = false;
if (i % prime[j] == 0) break; //µã¾¦Ö®±Ê
}
}
}
/*
4
Case 1.
2 2
23

Case 2.
23 1
-1

*/
/*
4
Case 1.
2 2
23

Case 2.
23 1
-1
*/
int main()
{
memset(prime, 0, sizeof(prime));
table();
int n;
int i,j,k;
int y=1;
//for(int i=1;i<7000;i++)
//printf("%d\n",prime[i]);
for(i=1; i<7000; i++)
{

unit[i].a=prime[i];
unit[i].num=0;
}
while(scanf("%d",&n),n>0)
{

if(y!=1)
printf("\n");
printf("Case %d.\n",y);
//for(i=1;i<7000;i++)
//printf("%d\n",unit[i].a);
for(i=1; i<7000; i++)
{
unit[i].num=0;
}
for(i=1; i<7000; i++)
{
if(n<0)
break;
while(n%unit[i].a==0)
{

unit[i].num++;
n=n/unit[i].a;
//printf("%d %d %d\n",unit[i].a,n,unit[i].num);
}

}
int t=0;
for(i=1; i<7000; i++)
{
if(unit[i].num)
{
num2[t]=unit[i].a;
num2[t+1]=unit[i].num;
t=t+2;
}
}
//printf("%d\n",t);

for(i=0; i<t-1; i++)
{
printf("%d ",num2[i]);
}
printf("%d \n",num2[t-1]);
//printf("\n");
y++;
}
return 0;
}

1. 老实说，这种方法就是穷举，复杂度是2^n，之所以能够AC是应为题目的测试数据有问题，要么数据量很小，要么能够得到k == t，否则即使n = 30，也要很久才能得出结果，本人亲测