1.23 div2 b-字符串处理(string 与 char 字符串

Posted 2021-02-05 -ifrush

B. Game with string
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Two people are playing a game with a string s

, consisting of lowercase latin letters.

On a player‘s turn, he should choose two consecutive equal letters in the string and delete them.

For example, if the string is equal to "xaax" than there is only one possible turn: delete "aa", so the string will become "xx". A player not able to make a turn loses.

Your task is to determine which player will win if both play optimally.
Input

The only line contains the string s
, consisting of lowercase latin letters (1≤|s|≤100000), where |s| means the length of a string s

.
Output

If the first player wins, print "Yes". If the second player wins, print "No".

开始用字符串模拟判断删除的整个过程，结果各种分类讨论，各种出错

而用string里自带的 erase删除字符 瞬间简洁

ps：c++ string的erase删除方法

1. 从位置pos=10处开始删除，直到结尾  str.erase(10);

2.  删除迭代器[first, last)区间的所有字符,返回一个指向被删除的最后一个元素的下一个字符的迭代器. str.erase(str.begin()+10);

string.size() 不能判断空字符串 删除字符前要用len保存字符长度

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
using namespace std;
 int main(){
     string s;
     int k=0;
     getline(cin,s);
     int len=s.size();
     for(int i=0;i<len;i++){
         while(i>=0&&s[i]==s[i+1]){
//             printf("%d %d %c %c
",i,i+1,s[i],s[i+1]);
             k++;
             s.erase(s.begin()+i,s.begin()+i+2);
             len-=2;
              if(len<=1)break;
             i--;
            if(i<0)i=0;
//             printf("%d %d %c %c
",i,i+1,s[i],s[i+1]);
         }
         if(len<=1)break;
     }
     if(k%2==0)printf("No");
     else printf("Yes");
     return 0;
 }