题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1102
正着做9种情况：
in
input
inputon
inputone
out
output
outputon
outputone
one
反着做貌似更简单一些，情况更少一点。
我是正着做的。
1 #include <cstdio> 2 3 const int MAXN = 10000000 + 5; 4 5 char str[MAXN]; 6 7 bool Judge() 8 { 9 if ( !(str[0] == 'o' || str[0] == 'p' || str[0] == 'i') ) return false; 10 int i = 0; 11 while ( str[i] != '\0' ) 12 { 13 switch( str[i] ) 14 { 15 case 'o': 16 if ( str[i+1] == 'u' && str[i+2] == 't' ) 17 { 18 if ( str[i+3] == 'p' && str[i+4] == 'u' && str[i+5] == 't' ) 19 { 20 if ( str[i+6] == 'o' && str[i+7] == 'n' ) 21 { 22 i += 8; 23 if ( str[i] == 'e' ) ++i; 24 25 } 26 else i += 6; 27 } 28 else i += 3; 29 } 30 else if ( str[i+1] == 'n' && str[i+2] == 'e' ) 31 i += 3; 32 else return false; 33 break; 34 case 'i': 35 if ( str[i+1] == 'n' ) 36 { 37 if ( str[i+2] == 'p' && str[i+3] == 'u' && str[i+4] == 't' ) 38 { 39 if ( str[i+5] == 'o' && str[i+6] == 'n' ) 40 { 41 i += 7; 42 if ( str[i] == 'e' ) ++i; 43 } 44 else i += 5; 45 } 46 else i += 2; 47 } 48 else return false; 49 break; 50 case 'p': 51 if ( str[i+1] == 'u' && str[i+2] == 't' && str[i+3] == 'o' && str[i+4] == 'n' ) i += 5; 52 else return false; 53 break; 54 default :return false; 55 } 56 } 57 return true; 58 } 59 60 int main() 61 { 62 int n; 63 scanf( "%d", &n ); 64 while ( n-- ) 65 { 66 scanf( "%s", str ); 67 if ( Judge() ) puts("YES"); 68 else puts("NO"); 69 } 70 return 0; 71 }