poj 2282 数位统计

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数位统计其实就是将某一位假定为某一个数字情况下统计所有的组合情况然后求和，在我的博客中有详细的讲解０的做法的博客

poj　3286 统计０

下面是本题的代码，比较容易懂：

注意一个trick就是n,m大小关系是不确定的

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;typedef long long LL;LL res[2][11];void calc ( int num , int flag){LL temp1 = 1 , temp2 = 0;    for ( int i = 0 ; i <= 9 ; i++ )        res[flag][i] = 1;while ( num ){LL pos = num %10;num /= 10;if ( pos == 0 ) res[flag][0] += temp1*(num-1)+temp2+1;else res[flag][0] += temp1*num;for ( int i = 1 ; i <= 9 ; i++ )if ( pos > i ) res[flag][i] += temp1*(num+1);else if ( pos == i ) res[flag][i] += temp1*num+temp2+1;            else res[flag][i] += temp1*num; temp2 += temp1*pos;temp1 *= 10;}}int main ( ){LL n , m;while ( ~scanf ( "%lld%lld" , &n , &m ) , n+m )    {        if ( n > m ) swap ( n , m );        calc ( n-1 , 0);        calc ( m , 1 );        for ( int i = 0 ; i < 9 ; i++ )            printf ( "%lld " , res[1][i] - res[0][i] );        printf ( "%lld\n" , res[1][9]-res[0][9] );}}

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