长沙理工 ACM 数位 DP 1488

题目链接  http://www.acmore.net/problem.php?id=1488

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#define M 1000000007
using namespace std;

long long arr[32],dp[32];

long long D2( long long num )
{
    long long ans = 1;
    for( int i = 1; i <= num; i++ )
    ans *= 10;
    return ans;
}

long long D1( long long num )
{
    long long t = 1,ans = 0;
    for( int i = 0; i < num; i++ )
    {
        ans += (arr[i]*t);
        t *=10;
    }
    return ans;
}

long long DFS( long long pos,long long limit )
{
    if( pos == -1 ){ return 0;}
    if( !limit && dp[pos] >= 0 ) return dp[pos];
    long long end = limit?arr[pos]:9;
    long long res = 0;
    for( int i = 0; i <= end; i++ )
    {
        long long ans = 0;
        if( i == 1 )
        {
            if( limit && ( i == end ) )
                 ans = ( ans + 1 + D1( pos ) )%M;
            else ans = ( ans + D2( pos ) )%M;
        }
        res = ( res + ans)%M;
        res = ( res + DFS( pos-1,limit && i == arr[pos] ) )%M;
    }
    if( !limit )dp[pos] = res;
    return res%M;
}

long long work( long long N )
{
    int k = 0;
    while( N )
    {
        arr[k++] = N%10;
        N /= 10;
    }
    memset( dp,-1,sizeof(dp) );
    return DFS( k-1,1 );
}
int main( )
{
    long long a,b;
    while( scanf("%lld %lld",&a,&b) != EOF )
        printf("%lld\n",(work(b) - work(a-1)+M)%M );
    return 0;
}

　　

转载于:https://www.cnblogs.com/wulangzhou/archive/2013/04/26/3044699.html