Palindromic Numbers LightOJ - 1205

题目大意：

求区间内的回文数个数

 

题目思路：

　　数位dp，先枚举前一半数字，然后填上相应的后一半数字。

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define MAXSIZE 1000005
#define LL long long

using namespace std;

LL dp[55][55][2];
int temp[MAXSIZE],bit[MAXSIZE];

LL dfs(int pos,int st,int pre,int limit)
{
    if(pos < 1)
        return 1;
    if(!limit && dp[pos][st][pre]!=-1)
        return dp[pos][st][pre];
    int len = limit?bit[pos]:9;
    LL ans = 0;
    for(int i=0;i<=len;i++)
    {
        temp[pos] = i;
        if(pre==0 && i==0)
        {
            ans += dfs(pos-1,st-1,0,i==len&&limit);
        }

        else if(pos > st/2) //枚举前一半
        {
            ans += dfs(pos-1,st,1,i==len&&limit);
        }

        else if(temp[pos] == temp[st-pos+1]) //填上后一半
        {
            ans += dfs(pos-1,st,1,i==len&&limit);
        }
    }

    if(!limit)
        dp[pos][st][pre] = ans;
    return ans;
}

LL Solve(LL n)
{
    int pos = 0;
    memset(dp,-1,sizeof(dp));
    while(n)
    {
        bit[++pos] = n%10;
        n /= 10;
    }
    LL ans = dfs(pos,pos,0,1);
    return ans;
}

int main()
{
    int T,cns=1;
    LL n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&n,&m);
        if(n > m)
            swap(n,m);
        LL ans = Solve(m) - Solve(n-1);
        printf("Case %d: %lld\n",cns++,ans);
    }
    return 0;
}