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  • AtCoder Beginner Contest 066 B

    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.

    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.


    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.

    Sample Input 2

    Copy
    xxxx
    

    Sample Output 2

    Copy
    2
    
    • xxx is not even.
    • xx is even.

    Sample Input 3

    Copy
    abcabcabcabc
    

    Sample Output 3

    Copy
    6
    

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


    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

     1 #include <iostream>
     2 #include <algorithm>
     3 #include <cstring>
     4 #include <cstdio>
     5 #include <vector>
     6 #include <cstdlib>
     7 #include <iomanip>
     8 #include <cmath>
     9 #include <ctime>
    10 #include <map>
    11 #include <set>
    12 #include <queue>
    13 #include <stack>
    14 using namespace std;
    15 #define lowbit(x) (x&(-x))
    16 #define max(x,y) (x>y?x:y)
    17 #define min(x,y) (x<y?x:y)
    18 #define MAX 100000000000000000
    19 #define MOD 1000000007
    20 #define pi acos(-1.0)
    21 #define ei exp(1)
    22 #define PI 3.141592653589793238462
    23 #define INF 0x3f3f3f3f3f
    24 #define mem(a) (memset(a,0,sizeof(a)))
    25 typedef long long ll;
    26 ll gcd(ll a,ll b){
    27     return b?gcd(b,a%b):a;
    28 }
    29 const int N=100005;
    30 const int mod=1e9+7;
    31 
    32 int main()
    33 {
    34     std::ios::sync_with_stdio(false);
    35     char a[201];
    36     scanf("%s",a);
    37     stack<char> s;
    38     int len=strlen(a);
    39     for(int i=0;i<len;i++)
    40         s.push(a[i]);
    41     s.pop();
    42     int t,flag;
    43     while(1){
    44         if(s.size()%2==1) s.pop();
    45         else {
    46             t=s.size();
    47             t/=2;
    48             flag=1;
    49             for(int i=0;i<t;i++){
    50                 if(a[i]!=a[i+t]){
    51                     flag=0;
    52                     break;
    53                 }
    54             }
    55             if(flag){
    56                 cout<<t*2<<endl;
    57                 break;
    58             }
    59             else s.pop();
    60         }
    61     }
    62     return 0;
    63 }
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  • 原文地址:https://www.cnblogs.com/wydxry/p/7290311.html
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