标签:ati clu ons let cstring ios 位置 ring and
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\n",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\n",i,i+1,s[i],s[i+1]); } if(len<=1)break; } if(k%2==0)printf("No"); else printf("Yes"); return 0; }
1.23 div2 b-字符串处理(string 与 char 字符串
标签:ati clu ons let cstring ios 位置 ring and
原文地址:https://www.cnblogs.com/-ifrush/p/10308643.html