LightOJ 1205 Palindromic Numbers, lightojpalindromic

LightOJ 1205 Palindromic Numbers, lightojpalindromic

Digital DP ....

Palindromic Numbers Time Limit:2000 MS Memory Limit:32768KB 64bit IO Format:% Lld & % llu [Submit] [Go Back] [Status] Description A palindromic number or numeral palindrome is a 'invalid rical' number like 16461 that remains the same when its digits are reversed. In this problem you will be given two integersI j, You have to find the number of palindromic numbersIAndJ(Aggressive ). Input Input starts with an integerT (≤ 200), Denoting the number of test cases. Each case starts with a line containing two integersI j (0 ≤ I, j ≤ 1017). Output For each case, print the case number and the total number of palindromic numbersIAndJ(Aggressive ). 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 Source Problem Setter: Jane Alam Jan [Submit] [Go Back] [Status]

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long int LL;int a[70];LL dp[70][70];LL dfs(int len,int l,int r,bool limit,bool ok){if(l<r) return !limit||(limit&&ok);if(!limit&&~dp[len][l])return dp[len][l];LL ret=0;int mx=limit?a[l]:9;for(int i=0;i<=mx;i++){if(l==len-1&&i==0)continue;int g=ok;if(g) g=a[r]>=i;else g=a[r]>i;ret+=dfs(len,l-1,r+1,limit&&i==mx,g);}if(!limit)dp[len][l]=ret;return ret;}LL gaoit(LL n){if(n<0) return 0;if(n==0) return 1;int len=0;while(n){a[len++]=n%10;n/=10;}LL ret=1;for(int i=len;i>=1;i--)ret+=dfs(i,i-1,0,i==len,1);return ret;}int main(){int T_T,cas=1;cin>>T_T;memset(dp,-1,sizeof(dp));while(T_T--){LL x,y;cin>>x>>y;if(x>y) swap(x,y);printf("Case %d: %lld\n",cas++,gaoit(y)-gaoit(x-1));}return 0;}