[暑假集训--数位dp]hdu5898 odd-even number

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

题意是要找x，x当中所有连续的奇数段长度是偶数，连续的偶数段长度是奇数，比如1357246

记一下当前这段是奇数还是偶数，这段已经有多长，有没有前导零。如果有前导零那么0可以也出现奇数次

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<LL,LL>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int len;
 LL l,r;
 LL f[][][][];
 int d[];
 inline LL dfs(int now,int dat,int lst,int lead,int fp)
 {
     if (now==)return (lst&)^(dat&);
     if (!fp&&f[now][dat][lst][lead]!=-)return f[now][dat][lst][lead];
     LL ans=;
     int mx=fp?d[now-]:;
     for (int i=;i<=mx;i++)
     {
         if (i==)
         {
             if (lead)
             {
                 if (lead&&now-!=)ans+=dfs(now-,,,,fp&&mx==);
                 else if (lead&&now-==)ans+=dfs(now-,,,,fp&&mx==);
                 continue;
             }
         }
         if ((i&)==(dat&))ans+=dfs(now-,i,lst+,,fp&&i==mx);
         else if ((dat&)^(lst&)||lead)
         {
             ans+=dfs(now-,i,,,fp&&i==mx);
         }
     }
     if (!fp)f[now][dat][lst][lead]=ans;
     return ans;
 }
 inline LL calc(LL x)
 {
     if (x==-)return ;
     if (x==)return ;
     LL xxx=x;
     len=;
     while (xxx)
     {
         d[++len]=xxx%;
         xxx/=;
     }
     LL sum=;
     sum+=dfs(len,,len==,len!=,d[len]==);
     for (LL i=;i<=d[len];i++)
     {
         sum+=dfs(len,i,,,i==d[len]);
     }
     return sum;
 }
 int main()
 {
     LL T=read();int cnt=;
     memset(f,-,sizeof(f));
     while (T--)
     {
         l=read();
         r=read();
         if (r<l)swap(l,r);
         printf("Case #%d: %lld\n",++cnt,calc(r)-calc(l-));
     }
 }

hdu 5898

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