九度OJ--1006-ZOJ问题

#include <stdio.h>
#include <string.h>
 
int main(int argc, char* argv[])
{
    char str[1000];
    int front, middle, behind, N;
    int count_z = 0, count_j = 0;//记录zj的个数
    int index_z = -1, index_j = -1;//记录zj的位置
    int i;
 
    while(EOF != scanf("%s",str))
    {
        //查找ZJ的位置，并记录其前中后有多少个o字符
        index_j = -1;
        index_z = -1;
        count_j = 0;
        count_z = 0;
 
        for(i = 0; i < strlen(str); ++i)
        {
            if (str[i] == 'z')
            {
                index_z = i;
                count_z++;
            }
            if (str[i] == 'j')
            {
                index_j = i;
                count_j++;
            }
        }
        //如果count>1或者index中有一个为-1，则WA
        if (count_j > 1 | count_z > 1 | index_z == -1 | index_j == -1 | index_j < index_z)
        {
            printf("Wrong Answer\n");
            continue;
        }
        else
        {//计算前中后的‘o’的个数
            front  = index_z;
            middle = index_j - index_z - 1;
            behind = strlen(str) - index_j - 1;
 
            if (middle >= 1 && front*middle == behind)
            {
                printf("Accepted\n");
                continue;
            }
            else
            {
                printf("Wrong Answer\n");
                continue;
            }      
        }
    }
    return 0;
}
/**************************************************************
    Problem: 1006
    User: tenyee
    Language: C
    Result: Accepted
    Time:30 ms
    Memory:912 kb
****************************************************************/