• AtCoder Beginner Contest 044 B


    Time limit : 2sec / Memory limit : 256MB

    Score : 200 points

    Problem Statement

    Let w be a string consisting of lowercase letters. We will call w beautiful if the following condition is satisfied:

    • Each lowercase letter of the English alphabet occurs even number of times in w.

    You are given the string w. Determine if w is beautiful.

    Constraints

    • 1≤|w|≤100
    • w consists of lowercase letters (a-z).

    Input

    The input is given from Standard Input in the following format:

    w
    

    Output

    Print Yes if w is beautiful. Print No otherwise.


    Sample Input 1

    Copy
    abaccaba
    

    Sample Output 1

    Copy
    Yes
    

    a occurs four times, b occurs twice, c occurs twice and the other letters occur zero times.


    Sample Input 2

    Copy
    hthth
    

    Sample Output 2

    Copy
    No

    题解:很简单 但是没有考虑长度为一的时候哇了两发
     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 bool cmp(int x,int y)
    30 {
    31     return x>y;
    32 }
    33 const int N=10005;
    34 const int mod=1e9+7;
    35 int main()
    36 {
    37     std::ios::sync_with_stdio(false);
    38     char a[101];
    39     scanf("%s",a);
    40     int len=strlen(a);
    41     if(len==1) {
    42         cout<<"No"<<endl;
    43         return 0;
    44     }
    45     sort(a,a+len);
    46     int t=1,flag=1;
    47     for(int i=1;i<len;i++){
    48         if(a[i]==a[i-1]) t++;
    49         else{
    50             if(t%2==1) {
    51                 flag=0;
    52                 break;
    53             }
    54             else t=1;
    55         }
    56     }
    57     if(flag) cout<<"Yes"<<endl;
    58     else cout<<"No"<<endl;
    59     return 0;
    60 }
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  • 原文地址:https://www.cnblogs.com/shixinzei/p/7288486.html
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