bunoj 13124(数位dp)

数位dp每次都给我一种繁琐的感觉。。

A - Palindromic Numbers

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status

Description

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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <queue>
#include <sstream>
#include <iostream>
using namespace std;
#define INF 0x3fffffff

typedef __int64 LL;

LL save[];
int bit[];
int cnt;
LL dp[];

LL get(LL x)
{
    if(x==) return ;
    LL tx=x;
    cnt=;
    while(x)
    {
        bit[++cnt]=x%;
        x/=;
    }
    LL ans=;
    ans=dp[cnt-];
    for(int i=;i<bit[cnt];i++)
    {
        if(cnt->=)
            ans+=save[cnt-];
        else ans++;
    }
    if(cnt==) return ans+; //
    int tmp=;
    for(int i=;i<=(cnt+)/;i++)
    {
        if( cnt-tmp>=)
            ans += save[cnt-tmp]*bit[cnt-i+];
        else ans += bit[cnt-i+];
        tmp+=;
    }
    int bit1[];

    for(int i=;i<=(cnt+)/;i++)
    {
        bit1[cnt-i+]=bit[cnt-i+];
        bit1[i]=bit[cnt-i+];
    } //

    LL sum=;
    for(int i=cnt;i>=;i--)
        sum=sum*+bit1[i];
    if(sum<=tx) ans++;
    return ans;
}

void dfs(int s)
{
    LL sum=;
    if(s==)
    {
        dp[]=;
        return ;
    }
    dfs(s-);
    sum += dp[s-];
    sum += *(save[s-]);
    dp[s]=sum;
}

int main()
{
    //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","w",stdout);
    save[]=;
    for(int i=;i<=;i++)
    {
        LL tmp=;
        if(i%==)
        {
            for(int j=;j<i/;j++)
                tmp*=;
        }
        else
        {
            for(int j=;j<i/+;j++)
                tmp*=;
        }
        save[i]=tmp;
    }
    dp[]=;
    dfs();
    int T;
    scanf("%d",&T);
    int tt=;
    int kk=;
    for(int i=;i<=;i++)
        kk++;

    while(T--)
    {
        LL a,b;
        cin>>a>>b;
        if(a>b) swap(a,b);
        if(a==)
        {
            a=;
        }
        else
            a=get(a-);
        b=get(b);
        cout<<"Case "<<tt++<<": ";
        cout<<b-a<<endl;
    }
    return ;
}