Poj 3602 typographical ligatures

[Description]: the meaning of the question is to enter a string and ask how many different types of characters we have. That is to say, once a character appears, it will not be counted in the next time, especially ff, fi, FL, FFI, FFL.

[Analysis]: After briefly explaining the meaning of the question, we can solve the problem.

Code:

// 196K 0Ms#include<cstdio>#include<cstring>#include<iostream>using namespace std;char s[1000];int count1[1000];int ff[7];int len;void solve(){int ans = 0;memset(count1,0,sizeof(count1));memset(ff,0,sizeof(ff));for(int i = 0;i<len;i++){if(s[i] == 'f'){if(s[i+1] == 'f' && (s[i+2] == 'i' || s[i+2]=='l')){if(s[i+2] == 'i')ff[0]=1;if(s[i+2] == 'l')ff[1]=1;i+=2;}else if(s[i+1]=='f'||s[i+1]=='i'||s[i+1]=='l'){if(s[i+1]=='f')ff[2]=1;if(s[i+1]=='i')ff[3]=1;if(s[i+1]=='l')ff[4]=1;i++;}elsecount1[s[i]]=1;}else if(s[i]=='`'&&s[i+1]=='`'){ff[5]=1;i++;}else if(s[i]=='\''&&s[i+1]=='\''){ff[6]=1;i++;}elsecount1[s[i]]=1;}for(int i = 0;i<=256;i++)if(count1[i]) ans++;for(int i = 0;i<7;i++)if(ff[i]) ans++;cout<<ans<<endl;}int main(){char c;len = 0;while(cin>>c){if(c == EOF) break;if(c == ' '||c == '\n') continue;s[len++] = c;}solve();return 0;}

Poj 3602 typographical ligatures