【数位dp】【回文】

1205 - Palindromic Numbers

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
4 1 10 100 1 1 1000 1 10000	Case 1: 9 Case 2: 18 Case 3: 108 Case 4: 198

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
    
int T;
typedef long long ll;
int bits[20];
ll dp[20][20];

ll dfs(int len,int l,int r,bool flag,bool ok)
{
	if(r > l) return !flag || (flag && ok);
	if(!flag && ~dp[len][l])  return dp[len][l];
	int end = flag ? bits[l] : 9;
	ll res = 0;
	for(int i=0;i<=end;i++)
	{
		if(l == len && i == 0) continue;
		bool g = ok;
		if(ok) g = bits[r] >= i;
		else g = bits[r] > i;
		res += dfs(len,l-1,r+1,flag&&(i==end),g);
	}
	return flag ? res : dp[len][l] = res;

}


ll solve(ll x)
{
	if(x < 0) return 0;
	if(x == 0) return 1;
	ll tt = x;
	int cnt = 0;
	while(tt >0 )
	{
		bits[++cnt] = tt % 10;
		tt /= 10;
	}
	ll ret = 1;
	for(int i=cnt;i>=1;i--)
		ret += dfs(i,i,1,i==cnt,true);
	return ret;
}
int main()
{
	int C = 1;
    scanf("%d",&T);
	memset(dp,-1,sizeof(dp));
	while(T --)
	{
		ll l,r;
		scanf("%I64d%I64d",&l,&r);
		if(l > r) swap(l,r);
		printf("Case %d: %I64d\n",C++,solve(r)-solve(l-1));
	}
    return 0;
}