Light oj 1205 - Palindromic Numbers（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

/*求一个区间回文数的个数*/#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define bug printf("hihi\n")#define eps 1e-12typedef long long ll;using namespace std;#define N 20ll dp[20][N];int bit[N];int a[N];ll dfs(int pos,int len,bool bound){    if(pos==-1) return 1;    if(len&&!bound&&dp[len][pos]!=-1) return dp[len][pos];    int up=bound ? bit[pos]:9;    ll ans=0;    for(int i=0;i<=up;i++)    {        if(len==0)        {            a[pos]=i;            ans+=dfs(pos-1,i ? pos:0,bound&&i==up);        }        else        {            int t=(len+1)>>1;            bool ha=(len&1) ? pos>=t:pos>=t;            if(ha)            {                a[pos]=i;                ans+=dfs(pos-1,len,bound&&i==up);            }            else            {                a[pos]=i;                if(a[len-pos]==i)                    ans+=dfs(pos-1,len,bound&&i==up);            }        }    }    if(!bound&&len) dp[len][pos]=ans;    return ans;}ll solve(ll a){    if(a<0) return 0;    if(a==0) return 1;    int i,j;    int len=0;    while(a)    {       bit[len++]=a%10;       a/=10;    }    return dfs(len-1,0,true);}int main(){    int i,j,t,ca=0;    ll a,b;    scanf("%d",&t);    memset(dp,-1,sizeof(dp));    while(t--)    {        scanf("%lld%lld",&a,&b);        if(a>b) swap(a,b);        a=solve(a-1);        b=solve(b);        printf("Case %d: %lld\n",++ca,b-a);    }    return 0;}