poj 1365 Prime Land

题意：给定你一个以质数为底表示的数；问你这个数减1，之后用质数为底的表示法。

例如：

17 1  = 17^1 =17 - 1；
5 1 2 1 = 5^1*2^1=10 - 1；

2 4 = 2^4 = 16
3 2 = 3^2= 9;

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int prime[4024],cnt;
void Prime(  )
{
   bool hash[20000] = { 0 };
   int t = ( int )sqrt( 32768.0 ) + 1;  
   for( int i = 3 ; i <= t; i +=2 )
   {
       if( !hash[i>>1] ) 
       {
           int tt = i << 1;//  = i * 2;
           for( int j = i*i ; j <= 32768 ; j += tt )    
                hash[j>>1] = true;    
       }        
   }    
   cnt = 0;
   prime[cnt++] = 2;
   t = 32768/2;
   for( int i = 1 ; i <= t ; i ++ )
   {
        if( !hash[i] )
        {
           prime[cnt++] = ( i<<1 ) + 1;    
        }        
   }
}
int Input( char str[])
{
   int sum = 1,num[2]={0},s;
   int t = 0;
   int len = strlen( str );
   for( int i = 0; i < len ; i ++ )
   {
       if( str[i] ==' '  )
       {
           if( t==1 )
           {
               s = 1;
               for( int j = 0 ;j < num[1] ; j ++ )
               {
                   sum *= num[0];    
               }
               num[0] = num[1] = 0;
               t = 0;
          }
          else t++;
       }    
       else num[t] = num[t]*10 + str[i] - '0';
   }
   for( int j = 0 ;j < num[1] ; j ++ )
   {
       sum *= num[0];    
   }
   return sum;
}
void Solve( int num )
{
    int a = 0;
    for( int i = cnt - 1; i >= 0 ; i -- )
    {    
        if( num % prime[i] == 0 )
        {
           int t = 0; 
           while( num % prime[i] == 0 )
           {
                 num /= prime[i];
                 t++;
           }          
          if( a==0 )printf( "%d %d",prime[i] ,t );
          else printf( " %d %d",prime[i],t );
          a++;    
        }    
    }
    puts( "" );
}
int main( )
{
    char str[1024];
    Prime();
   while( 1 )
   {
       gets( str );
       if(  str[0] == '0' ) break;
       int t = Input( str ) ;   
       Solve( t - 1 );        
   }
   return 0;
}