Codeforces Round #454 (Div. 2, based on Technocup 2018 Elimination Round 4) C - Shockers
2017-12-24 01:58
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如果在!中没有出现过,那么这些字母没有嫌疑,如果在?或.中出现过,那么这些字母没有嫌疑。当最后只剩一个字母有嫌疑时,就确定了假设的字母,这时候我们只需要判别之后出现的?和!的数目即可。
#include <iostream> #include <string> using namespace std; bool notis[30]={false},mayis[30]={false}; string str; int main(){ int n,flag=0,cnt=0,tem,i,j;char p; std::ios::sync_with_stdio(false); cin>>n; for(i=0;i<n;i++){ cin>>p>>str; if(flag&&(p=='!'||p=='?'))cnt++; else{ tem=str.length(); if(p=='!'){ for(j=0;j<tem;j++)mayis[str[j]-'a']=true; for(j=0;j<26;j++){ if(!mayis[j])notis[j]=true; mayis[j]=false; } }else for(j=0;j<tem;j++)notis[str[j]-'a']=true; for(tem=j=0;j<26;j++)if(!notis[j])tem++; if(tem==1)flag=1; } } printf("%d",cnt>0?(cnt-1):0); }
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