lightoj 1205 数位dp

1205 - Palindromic Numbers

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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

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PROBLEM SETTER: JANE ALAM JAN

题意：

求区间内回文串数的个数

//开个数组存一下回文串，还要处理前导零。sta表示到目前为止是否合法
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int t,bit[],num[];
ll f[][][],P,Q;
ll dfs(int pos,int s,int sta,bool limit)
{
    if(pos==) return sta;
    if(!limit&&f[pos][s][sta]!=-) return f[pos][s][sta];
    int max_b=(limit?bit[pos]:);
    ll ans=;
    for(int i=;i<=max_b;i++){
        num[pos]=i;
        if(pos==s&&i==)
            ans+=dfs(pos-,s-,sta,limit&&(i==max_b));
        else if(sta&&pos<=s/)
            ans+=dfs(pos-,s,num[s-pos+]==i,limit&&(i==max_b));
        else ans+=dfs(pos-,s,sta,limit&&(i==max_b));
    }
    if(!limit) f[pos][s][sta]=ans;
    return ans;
}
ll solve(ll x)
{
    if(x<) return ;
    int pos=;
    while(x){
        bit[++pos]=x%;
        x/=;
    }
    return dfs(pos,pos,,);
}
int main()
{
    memset(f,-,sizeof(f));
    scanf("%d",&t);
    for(int cas=;cas<=t;cas++){
        scanf("%lld%lld",&P,&Q);
        if(P>Q) swap(P,Q);
        printf("Case %d: %lld\n",cas,solve(Q)-solve(P-));
    }
    return ;
}