AtCoder Beginner Contest 066 B - ss

题目链接：http://abc066.contest.atcoder.jp/tasks/abc066_b

Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

We will call a string that can be obtained by concatenating two equal strings an even string. For example, xyzxyz and aaaaaa are even, while ababab and xyzxyare not.

You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.

Constraints

• 2≤|S|≤200
• S is an even string consisting of lowercase English letters.
• There exists a non-empty even string that can be obtained by deleting one or more characters from the end of S.

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Input

Input is given from Standard Input in the following format:

S

Output

Print the length of the longest even string that can be obtained.

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Sample Input 1

Copy

abaababaab

Sample Output 1

Copy

6

• abaababaab itself is even, but we need to delete at least one character.
• abaababaa is not even.
• abaababa is not even.
• abaabab is not even.
• abaaba is even. Thus, we should print its length, 6.

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Sample Input 2

Copy

xxxx

Sample Output 2

Copy

2

• xxx is not even.
• xx is even.

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Sample Input 3

Copy

abcabcabcabc

Sample Output 3

Copy

6

The longest even string that can be obtained is abcabc, whose length is 6.

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Sample Input 4

Copy

akasakaakasakasakaakas

Sample Output 4

Copy

14

The longest even string that can be obtained is akasakaakasaka, whose length is 14.

题解：还是比较有趣 如果前一半字符和后一半字符一样就输出总字符数 否则字符串尾删除一个字符

　　　那就用栈吧 不满足的话就用栈删除很方便

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <queue>
#include <stack>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll gcd(ll a,ll b){
    return b?gcd(b,a%b):a;
}
const int N=100005;
const int mod=1e9+7;

int main()
{
    std::ios::sync_with_stdio(false);
    char a[201];
    scanf("%s",a);
    stack<char> s;
    int len=strlen(a);
    for(int i=0;i<len;i++)
        s.push(a[i]);
    s.pop();
    int t,flag;
    while(1){
        if(s.size()%2==1) s.pop();
        else {
            t=s.size();
            t/=2;
            flag=1;
            for(int i=0;i<t;i++){
                if(a[i]!=a[i+t]){
                    flag=0;
                    break;
                }
            }
            if(flag){
                cout<<t*2<<endl;
                break;
            }
            else s.pop();
        }
    }
    return 0;
}