#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
const int maxn = 10 ;
int dp[maxn][3];//0什么都没有，1,有6,2有62或4
int bit[maxn] ;
int dfs(int pos , int flag ,int lim)
{
if(pos == 0)
return (flag !=2) ;
if(dp[pos][flag] != -1 && !lim)
return dp[pos][flag] ;
int num = lim ? bit[pos] : 9;
int ans = 0;
for(int i = 0;i <= num ;i++)
{
int flag_x = flag;
if(flag == 0 && i == 6)
flag_x = 1;
else if(flag == 1 && i == 2)
flag_x = 2;
else if(i == 4)
flag_x = 2;
else if(flag == 1 && i != 6)
flag_x = 0;
ans+=dfs(pos - 1 ,flag_x , lim&&(i==num)) ;
}
if(!lim) dp[pos][flag] = ans ;
return ans;
}
int solve(int n)
{
memset(dp , -1 ,sizeof(dp)) ;
int pos = 0;
while(n)
{
bit[++pos] = n%10;
n/=10 ;
}
return dfs(pos , 0 , 1) ;
}
int main()
{
int n, m;
while(scanf("%d%d",&n ,&m) && (n+m))
{
printf("%d\n" , solve(m) - solve(n-1)) ;
}
return 0;
}
hdu2089 数位dp水题
原文：http://blog.csdn.net/cq_pf/article/details/45483573
