ZOJ

Generalized Palindromic Number

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Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                            

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A number that will be the same when it is written forwards or backwards is known as a palindromic number. For example, 1234321 is a palindromic number.

We call a number generalized palindromic number, if after merging all the consecutive same digits, the resulting number is a palindromic number. For example, 122111 is a generalized palindromic number. Because after merging, 122111 turns into 121 which is a palindromic number.

Now you are given a positive integer N, please find the largest generalized palindromic number less than N.

Input

There are multiple test cases. The first line of input contains an integer T (about 5000) indicating the number of test cases. For each test case:

There is only one integer N (1 <= N <= 10¹⁸).

Output

For each test case, output the largest generalized palindromic number less than N.

Sample Input

4
12
123
1224
1122

Sample Output

11
121
1221
1121

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                            Author: LIN, Xi                                         Source: The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

转自：http://blog.csdn.net/u011345136/article/details/39122741

题意：求小于N的回文数，这个数的回文相同的数可以缩成一个数
思路：dfs(l, r, flag)：l代表左边侧长度，r代表右边的长度，flag代表是否处于边界，然后在搜索右边匹配左边的同时，枚举k，k代表右边连续相同的数都匹配左边的个数

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>

#define N 55
#define M 15
#define mod 6
#define mod2 100000000
#define ll long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

int T;
ll n;
char s[N];
ll ans;
int len;
char leftnum[N];
char rightnum[N];

void ini()
{
//    scanf("%lld",&n);
    cin>>n;
    ans=-1;
    //memset(left,0,sizeof(left));
    //memset(right,0,sizeof(right));
    sprintf(s+1,"%lld",n);
    //sprintf(s+1,"%I64d",n);
    len=strlen(s+1);
    for(int i=1;i<=len;i++){
        s[i]-='0';
    }
}

void out()
{
    //printf("%lld
",ans);
    cout<<ans<<endl;
}

ll getans(int l,int r)
{
    ll re=0;
    int i;
    for(i=1;i<=l;i++){
        re=re*10+leftnum[i];
    }
    for(i=r;i>=1;i--){
        re=re*10+rightnum[i];
    }
    return re;
}

ll dfs(int l,int r,int flag)
{
    ll re=-1;
    int i,k,m;
    if(l+r-1>len) return -1ll;
    if(l+r-1==len){
        re=getans(l-1,r);
        if(re>=n) return -1ll;
        return re;
    }

    m= (flag==1) ? s[l] : 9;
    for(i=m;i>=0;i--)
    {
        leftnum[l]=i;
        if( (l==1 || leftnum[l]!=leftnum[l-1]) && !(l==1 && i==0) && (l+r!=len) )
        {
            for(k=1;k+l+r<=len;k++){
                rightnum[k+r]=i;
                re=max(re,dfs(l+1,r+k,flag && (m==i) ) );
            }
        }
        else{
            re=max(re,dfs(l+1,r,flag && (m==i) ));
        }
        if(re>0){
            return re;
        }
    }
    return re;
}

int main()
{
    //freopen("data.in","r",stdin);
    scanf("%d",&T);
    for(int cnt=1;cnt<=T;cnt++)
   // while(T--)
    //while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)
    {
        ini();
        ans=dfs(1,0,1);
        out();
    }

    return 0;
}