求范围内连续奇数个数是偶数，连续偶数个数是奇数的数的个数

For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).

Input

First line a t,then t cases.every line contains two integers L and R.

Output

Print the output for each case on one line in the format as shown below.

Sample Input

2    
1 100    
110 220 

Sample Output

Case #1: 29
Case #2: 36

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
ll dp[20][2][2];//dp[len][pre][e] 代表第几位，之前一位是奇数还是偶数，e表示是否还要添加一个与pre同为奇数或者偶数的数。
int a[20];
ll dfs(int len,int pre,int e,bool flag,bool s)//pre 表示上一个数，0表示偶数，1表示奇数。e=0不需要添加。e=1需要添加。s记录前缀和，s=false前面全是0，
{
    if(len<0)
        {
            return e==1?0:1;
        }
        if(flag==false&&dp[len][pre][e]!=-1&&s)
        {
            return dp[len][pre][e];
        }
        ll ans=0;
        int cnt=flag?a[len]:9;
        for(int i=0;i<=cnt;i++)
        {
            if(s==false)
            {
                if(i==0)
                {
                    ans+=dfs(len-1,0,0,flag&&i==cnt,false);
                }
                else
                {
                    ans+=dfs(len-1,i%2,i%2==1?1:0,flag&&i==cnt,true);
                }
            }
            else if(i%2==1)
            {
               if(pre%2==1)
               {
                   ans+=dfs(len-1,1,e^1,flag&&i==cnt,true);
               }
               else
                {
                    if(e==0)
                    {
                        ans+= dfs(len-1,1,1,flag&&i==cnt,true);
                    }
                }
            }
            else
            {
                if(pre%2==0)
                ans+=dfs(len-1,0,e^1,flag&&i==cnt,true);
                else
                {
                    if(e==0)
                    {
                        ans+=dfs(len-1,0,0,flag&&i==cnt,true);
                    }
                }
            }
        }

        if(!flag&&s)
            dp[len][pre][e]=ans;
        return ans;
}
ll solve(ll x)
{
    int len=0;
     memset(a,0,sizeof(0));
    while(x>0)
    {
        a[len++]=x%10;
        x/=10;
    }
    return dfs(len-1,0,0,true,false);
}
int main()
{
     int t,ca=1;
        scanf("%d",&t);
        memset(dp,-1,sizeof(dp));
        while(t--)
        {
            long long int l,r;
            scanf("%I64d%I64d",&l,&r);
            printf("Case #%d: %I64d\n",ca++,solve(r)-solve(l-1));
        }
        return 0;
}