HDOJ 1039 Easier Done Than Said？ 简单的字符串处理

#include"stdio.h" #include"string.h" int vo(char *psw,int i) {return (psw[i]=='a'||psw[i]=='e'||psw[i]=='i'||psw[i]=='o'||psw[i]=='u');} int main() { char psw[21],ans[2][15]={{"not acceptable"},{"acceptable"}}; while(gets(psw)!=NULL) { if(psw[0]=='e'&&psw[1]=='n'&&psw[2]=='d')break; int f1=0,f2=1,f3=1; for(int i=0;i<strlen(psw);i++) { if(vo(psw,i)) { f1=1; if(i<strlen(psw)-2&&vo(psw,i+1)&&vo(psw,i+2))f2=0; if(i<strlen(psw)-1&&psw[i]==psw[i+1]&&(psw[i]!='e'&&psw[i]!='o'))f3=0; } else { if(i<strlen(psw)-2&&!vo(psw,i+1)&&!vo(psw,i+2))f2=0; if(i<strlen(psw)-1&&psw[i]==psw[i+1])f3=0; } } printf("<%s> is %s.\n",psw,ans[f1&&f2&&f3]); } return 0; }