lightOJ 1205(Palindromic Numbers数位DP)

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

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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
Source
Problem Setter: Jane Alam Jan

题意比较简单，我就直接上代码了，一些步骤我有记在代码里；

/* *********************************************** Author :xdlove Created Time :2015年08月18日 星期二 13时18分54秒 File Name :xd.cpp ************************************************ */

/* * lightOJ 1205 (数位DP) * dp[i][j] 表示以j开头的i位数 * 处理比较烦。。。 */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

/******************************** please don't hack me!! /(ToT)/~~ __------__ /~ ~\ | //^\\//^\| /~~\ || T| |T|:~\ | |6 ||___|_|_||:| \__. / o \/' | ( O ) /~~~~\ `\ \ / | |~~\ | ) ~------~`\ /' | | | / ____ /~~~)\ (_/' | | | /' | ( | | | | \ / __)/ \ \ \ \ \/ /' \ `\ \ \|\ / | |\___| \ | \____/ | | /^~> \ _/ < | | \ \ | | \ \ \ -^-\ \ | ) `\_______/^\______/ ************************************/

#define clr(a) memset(a,0,sizeof(a));
typedef long long ll;
ll dp[20][10];
ll haha;
int tol;
int bit[20],nm[20];

bool check(ll x)
{
    ll n = x;
    ll res = 0;
    while(n)
    {
        res = res * 10 + n % 10;
        n /= 10;
    }
    return x == res;
}

ll fuck(ll l,ll r,int t)
{
    //除了第一位不能取0外，其他位可以取0，用t来进行标记
    ll ans = 0;
    if(l == r) 
    {
        ll tp = 0;
        nm[l] = nm[r] = bit[r];
        for(int i = tol; i >= 1; i--)
            tp = tp * 10 + nm[i];
        if(tp <= haha) //因为我是一位一位的确定，所以最后需要与原先的数进行比较，因为这个坑了我好久 
            return bit[l] + 1;
        return bit[l];
    }
    if(l > r) 
    {
        ll tp = 0;
        for(int i = tol; i >= 1; i--)
            tp = tp * 10 + nm[i];
        if(tp <= haha) return 1;
        return 0;
    }
    for(int i = bit[r] - 1; i >= t; i--)
        ans += dp[r - l + 1][i];
    nm[l] = nm[r] = bit[r];
    return ans + fuck(l + 1,r - 1,0);
}

ll solve(ll n)
{
    ll ans = 1; //0也是回文数
    haha = n;
    if(n < 0) return 0;
    if(n < 10) return n + 1;
    tol = 0;
    while(n) //将n进行拆位
    {
        bit[++tol] = n % 10;
        n /= 10;
    }
    for(int i = tol - 1; i >= 1; i--) // 预先算出每一个位为0的时候所有的回文数
        for(int j = 1; j < 10; j++)
            ans += dp[i][j];
    ans += fuck(1,tol,1); //递归处理，一次确定每一个位的数
    return ans;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    for(int i = 0; i < 10; i++)
        dp[1][i] = dp[2][i] = 1;
    for(int i = 3; i <= 18; i++)
        for(int j = 0; j < 10; j++)
            for(int k = 0; k < 10; k++)
                dp[i][j] += dp[i - 2][k];
    int T,cnt = 0;
    ll l,r;
    cin>>T;
    while(T--)
    {
        cin>>l>>r;
        if(l > r) swap(l,r);
        printf("Case %d: ",++cnt);
        cout<<solve(r) - solve(l - 1)<<endl;
    }
    return 0;
}