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求非空格的符号种类数,附简洁代码char *table[7]={"ffi","ffl","fi","fl","ff","``","\'\'"}; int use[155]; char buf[10005]; int main() { use[' ']=1; int ans=0,i; while(gets(buf)) { for(char *p=buf;*p;) { for(i=0;i<7;i++)if(strncmp(table[i],p,strlen(table[i]))==0)break; if(i<7)p+=strlen(table[i]); else i=*p++; if(!use[i]){use[i]=1;ans++;} } } printf("%d\n",ans); } Followed by: Post your reply here: |
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