数位DP

int a,b,shu[20],dp[20][11];
int dfs(int len,int pre, bool limit)
{
    if(len==0) return 1;
    if(!limit&&dp[len][pre]!=-1) return dp[len][pre];  //有限制的情况会不一样 
    int cnt=0,maxx=(limit?shu[len]:9);   //是否有限制 
    for(int i=0;i<=maxx;i++)
    {
        if(pre==4&&i==9)
        continue;
        cnt+=dfs(len-1,i,i==maxx&&limit);
    }
    return limit?cnt:dp[len][pre]=cnt;
}
int solve(int x)
{
    memset(shu,0,sizeof(shu));
    int k=0;
    while(x)
    {
        shu[++k]=x%10;
        x/=10;
    }
    return dfs(k,-1,true);
}
int main()
{
    while(scanf("%d%d",&a,&b),b!=0||a!=0)
    {
        printf("%d\n",solve(b)-solve(a-1));
    }
}