[暑假集训--数位dp]LightOj1205 Palindromic Numbers

A palindromic number or numeral palindrome is a 'symmetrical' number like 16461 that remains the same when its digits are reversed. In this problem you will be given two integers i j, you have to find the number of palindromic numbers between i and j (inclusive).

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing two integers i j (0 ≤ i, j ≤ 10¹⁷).

Output

For each case, print the case number and the total number of palindromic numbers between i and j (inclusive).

Sample Input

4

1 10

100 1

1 1000

1 10000

Sample Output

Case 1: 9

Case 2: 18

Case 3: 108

Case 4: 198

问 l 到 r 有多少回文

枚举回文数的长度，然后数位dp。记一下当前位置是啥

 #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 zhan[];
 int d[];

 inline LL dfs(int now,int p,int dat,int fp)
 {
     if (now==(p+)/)
     {
         if (!fp)return ;
         for (int i=now-;i>=;i--)
         {
             if (d[i]>zhan[p+-i])return ;
             if (d[i]<zhan[p+-i])return ;
         }
         return ;
     }
     if (!fp&&f[now][p][dat]!=-)return f[now][p][dat];
     LL ans=;
     int mx=fp?d[now-]:;
     for (int i=;i<=mx;i++)
     {
         zhan[now-]=i;
         ans+=dfs(now-,p,i,fp&&i==mx);
         zhan[now-]=-;
     }
     if (!fp)f[now][p][dat]=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=;
     for (int i=;i<=len;i++)
     {
         for (int j=;j<=(i==len?d[len]:);j++)
         {
             zhan[i]=j;
             sum+=dfs(i,i,j,len==i&&j==d[len]);
             zhan[i]=-;
         }
     }
     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-));
     }
 }

LightOJ 1205