ZOJ 3816 Generalized Palindromic Number

Generalized Palindromic Number

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

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

41212312241122

Sample Output

111211221
1121
题意：给出一个数n，求小于n的一个最大的广义回文数。
搜索学的不够啊！写了好久
#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#define ll long longusing namespace std;const int maxn=100;int num[maxn],cnt,le[maxn],ri[maxn];ll n;ll get_num(int l,int r){    ll ret=0;    for(int i=0;i<l;i++)  ret=ret*10+le[i];    for(int i=r;i>=1;i--)  ret=ret*10+ri[i];    return ret;}ll dfs(int l,int r,bool flag){    ll ans=-1;    if(l+r>cnt) return -1;    if(l+r==cnt)    {        ans=get_num(l,r);        if(ans>=n) return -1;       // cout<<ans<<endl;        return ans;    }    int m=flag ? num[l]:9;    for(int i=m;i>=0;i--)    {        le[l]=i;        if((l==0 || le[l]!=le[l-1]) && (l!=0 || i!=0) && (l+r+1!=cnt))        {             for(int j=1;j+l+r+1<=cnt;j++)             {                  ri[r+j]=i;                  ans=max(ans,dfs(l+1,r+j,flag && (i==m)));             }        }        else  ans=max(ans,dfs(l+1,r,flag && (i==m)));        if(ans>0)  return ans;    }    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%lld",&n);        cnt=0;        ll tp=n;        while(tp)        {            num[cnt++]=tp%10;            tp=tp/10;        }        reverse(num,num+cnt);        printf("%lld\n",dfs(0,0,1));    }    return 0;}