HDU 3788 and nine degrees OJ 1006 test data is not the same

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 The given string (including only ' Z ', ' o ', ' J ' three characters) to infer if he can AC.

Whether the rules of the AC are for example the following:
1. Zoj can ac;
2. If the string form is XZOJX, it can also be AC, where x can be n ' o ' or null;
3. If the AZBJC can be AC, the Azbojac can also be AC. The a,b,c is n ' o ' or empty;
Input inputs include multiple sets of test cases, each with a string that contains only ' Z ', ' o ', ' J ' three characters, and the string length is less than or equal to 1000.
Output for the given string, assuming that you can AC, please export the string "Accepted", 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++; }//Statistics Z, j the number of Middle O} if (cnt!=flag2-flag1-1| | cnt==0) {//infer whether characters except o appear in the middle of Z and J. Cnt=0 is the middle of Zoj does not appear o.                 Not legal.             printf ("Wrong answer\n");                     } else{the number of O after//j divided by the number of O in front of Z equals 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;}

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HDU 3788 and nine degrees OJ 1006 test data is not the same