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
Output for Sample Input
Case 1: 9
Case 2: 18
Case 3: 108
Case 4: 198
代码：
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 int a[40],tmp[40]; 5 ll dp[40][100][100]; 6 ll dfs(int start,int pos,int ok,bool limit) 7 { 8 if(pos<0) 9 return ok; 10 if(!limit&&dp[pos][ok][start]!=-1) 11 return dp[pos][ok][start]; 12 ll ans=0; 13 int up=limit?a[pos]:9; 14 for(int d=0; d<=up; ++d) 15 { 16 tmp[pos]=d; 17 if(start==pos&&d==0) 18 ans+=dfs(start-1,pos-1,ok,limit&&d==up); 19 else if(ok&&pos<(start+1)/2) 20 ans+=dfs(start,pos-1,tmp[start-pos]==d,limit&&d==up); 21 else 22 ans+=dfs(start,pos-1,ok,limit&&d==up); 23 } 24 if(!limit) 25 dp[pos][ok][start]=ans; 26 return ans; 27 } 28 ll solve(ll x) 29 { 30 memset(a,0,sizeof(a)); 31 int cnt=0; 32 while(x!=0) 33 { 34 a[cnt++]=x%10; 35 x/=10; 36 } 37 return dfs(cnt-1,cnt-1,1,1); 38 } 39 int main() 40 { 41 memset(dp,-1,sizeof(dp)); 42 int t,cnt=1; 43 scanf("%d",&t); 44 while(t--) 45 { 46 ll x,y; 47 scanf("%lld%lld",&x,&y); 48 if(x>y) 49 swap(x,y); 50 printf("Case %d: %lld ",cnt++,solve(y)-solve(x-1)); 51 } 52 return 0; 53 }