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?：

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;

__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

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;

}