ZOJ - 3816 Generalized Palindromic Number dfs

Generalized Palindromic Number

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

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

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9
 10 #define N 55
 11 #define M 15
 12 #define mod 6
 13 #define mod2 100000000
 14 #define ll long long
 15 #define maxi(a,b) (a)>(b)? (a) : (b)
 16 #define mini(a,b) (a)<(b)? (a) : (b)
 17
 18 using namespace std;
 19
 20 int T;
 21 ll n;
 22 char s[N];
 23 ll ans;
 24 int len;
 25 char leftnum[N];
 26 char rightnum[N];
 27
 28 void ini()
 29 {
 30 //    scanf("%lld",&n);
 31     cin>>n;
 32     ans=-1;
 33     //memset(left,0,sizeof(left));
 34     //memset(right,0,sizeof(right));
 35     sprintf(s+1,"%lld",n);
 36     //sprintf(s+1,"%I64d",n);
 37     len=strlen(s+1);
 38     for(int i=1;i<=len;i++){
 39         s[i]-=‘0‘;
 40     }
 41 }
 42
 43 void out()
 44 {
 45     //printf("%lld\n",ans);
 46     cout<<ans<<endl;
 47 }
 48
 49 ll getans(int l,int r)
 50 {
 51     ll re=0;
 52     int i;
 53     for(i=1;i<=l;i++){
 54         re=re*10+leftnum[i];
 55     }
 56     for(i=r;i>=1;i--){
 57         re=re*10+rightnum[i];
 58     }
 59     return re;
 60 }
 61
 62 ll dfs(int l,int r,int flag)
 63 {
 64     ll re=-1;
 65     int i,k,m;
 66     if(l+r-1>len) return -1ll;
 67     if(l+r-1==len){
 68         re=getans(l-1,r);
 69         if(re>=n) return -1ll;
 70         return re;
 71     }
 72
 73     m= (flag==1) ? s[l] : 9;
 74     for(i=m;i>=0;i--)
 75     {
 76         leftnum[l]=i;
 77         if( (l==1 || leftnum[l]!=leftnum[l-1]) && !(l==1 && i==0) && (l+r!=len) )
 78         {
 79             for(k=1;k+l+r<=len;k++){
 80                 rightnum[k+r]=i;
 81                 re=max(re,dfs(l+1,r+k,flag && (m==i) ) );
 82             }
 83         }
 84         else{
 85             re=max(re,dfs(l+1,r,flag && (m==i) ));
 86         }
 87         if(re>0){
 88             return re;
 89         }
 90     }
 91     return re;
 92 }
 93
 94 int main()
 95 {
 96     //freopen("data.in","r",stdin);
 97     scanf("%d",&T);
 98     for(int cnt=1;cnt<=T;cnt++)
 99    // while(T--)
100     //while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)
101     {
102         ini();
103         ans=dfs(1,0,1);
104         out();
105     }
106
107     return 0;
108 }