poj 2282 The Counting Problem & 3286 How many 0's?

这是一个组合数学题：

我就拿3124来做例子：我们要举出2的个数；

我们是从地位向高位列举：当我们拿个位时，我们就把个位变成2，那么前面有多少个数，那么就有多少个2,前面有0~312共有313个数；

我们在列举十位：2前面有0~30个数可取，个位就可以取0~10，所以共有31*10个，当前面是31时，2可以为0,1,2,3，共有4个数；总共有31*10 + 4；

我们在列举百位：1前面可以去0~2共有3个数，这是百位可以取2，（总是比3124小）后面可以去任意的数前面可以取3*100个，当取3时，那么后面不要算了，总共为3*100，因为1小于2；

有些人就会问一个为题就是2122是当取个位时有这个数，取十位，百位也有这个数，2122有3个2那么我们也是计算3次呀，因此，就没重复计算；

这里要注意就是取0是，前面不能全部为0；

How many 0's?：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;
__int64 num[12] = {1LL,10LL,100LL,1000LL,10000LL,100000LL,1000000LL,10000000LL,100000000LL,1000000000LL,10000000000LL,100000000000LL};
LL Solve( LL n )
{
    LL sum = 0,left,a;
       for( int i = 1 ; i < 12 ; i ++ )
       {
         left = n / num[i] - 1;
         sum += left * num[i-1];
         a = ( n % num[i] - n % num[i-1] )/num[i-1];
         if( a > 0 ) sum +=  num[i-1];
         else if( a == 0 ) sum += n%num[i-1] + 1;     
         if( n < num[i] ) break;    
    }
    return sum;
}
int main(  )
{
    LL n,m;
    while( scanf( "%I64d %I64d",&m,&n ),n!=-1||m!=-1 )
    {
       printf( "%I64d\n",Solve( n ) - Solve( m - 1 ) );       
    }
    //system( "pause" );
    return 0;
}

The Counting Problem：

View Code 
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;
LL num[12]={1LL,10LL,100LL,1000LL,10000LL,100000LL,1000000LL,10000000LL,100000000LL,1000000000LL,10000000000LL,100000000000LL};
LL Solve( LL n , int d )
{
   LL sum = 0,left,a;
   for( int i = 1 ; i < 12 ; i ++ )
   {
        left = n / num[i] - ( d == 0 );
        sum += left*num[i-1];
        a = ( n % num[i] - n%num[i-1] )/num[i-1];
        if( a > d ) sum += num[i-1];
        else if( a == d ) sum += n%num[i-1] +1;
        if( n < num[i] ) break;        
   }    
   return sum;
}
int main(  )
{
    LL n,m;
    while( scanf( "%I64d %I64d",&n,&m ),n||m )
    {
           if( m > n ) swap( n ,m );
           for( int i = 0; i <= 9 ; i ++ )
           printf( "%I64d ",Solve( n , i ) - Solve( m - 1 ,i ) );
           puts( "" );    
    }
    //system( "pause" );
    return 0;
}