HDU 3788 and 9du OJ 1006 test data are different, hduoj

HDU 3788 and 9du OJ 1006 test data are different, hduoj
ZOJ ProblemsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 2935 Accepted Submission (s): 889


Problem Description determines whether a given string (only contains 'Z', 'O', and 'J') can be AC.

The AC rules are as follows:
1. zoj can communicate with each other;
2. If the string format is xzojx, it can also be AC, where x can be N 'O' or is empty;
3. If azbjc can be AC, azbojac can also be AC, where a, B, and c are N 'O' or empty;
The Input contains multiple groups of test cases. Each line has a string that only contains 'Z', 'O', and 'J'. the string length is less than or equal to 1000;
Output outputs the string "Accepted" if it can be an AC string. Otherwise, Output "Wrong Answer ".
Sample Input

zojozojoozoojoooozoojoooozoojozojooooozojozojoooo

 
Sample Output

AcceptedAcceptedAcceptedAcceptedAcceptedAcceptedWrong AnswerWrong Answer

 

# Include <stdio. h> # include <string. h> char s [2001]; char s2 [2001]; void check () {int cnt = 0; memset (s2, 0, sizeof (s2); int flag1, flag2, flag3; flag1 = flag2 = flag3 =-1; int len = strlen (s); if (strcmp ("zoj", s) = 0) {printf ("Accepted \ n"); return;} else {for (int I = 0; I <len; ++ I) {if (s [I]! = 'Z' & s [I]! = 'O' & s [I]! = 'J') {flag3 = 1; break ;}} for (int I = 0; I <len; ++ I) {if (s [I] = 'Z') {flag1 = I; break;} for (int I = len-1; I> = 0; -- I) {if (s [I] = 'J') {flag2 = I; break ;}} if (flag1> flag2 | flag1 =-1 | flag2 =-1 | flag3 = 1) {printf ("Wrong Answer \ n ");} else {for (int I = flag1 + 1; I <flag2; I ++) {if (s [I] = 'O') {cnt ++ ;} // count the number of z and j intermediate o} if (cnt! = Flag2-flag1-1 | cnt = 0) {// judge whether any character except o exists in the middle of z and j; cnt = 0 is the zoj center without o, illegal; printf ("Wrong Answer \ n ");} else {// Number of o after j divided by the number of o before z equal to the number of o between z and j if (flag1 * cnt = len-flag2-1) {printf ("Accepted \ n");} else {printf ("Wrong Answer \ n") ;}}} int main (int argc, char * argv []) {while (~ Scanf ("% s", s) {check ();} return 0 ;}