九度OJ--1006-ZOJ问题

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#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****************************************************************/
时间有点长了，后面想想哪占了这么多时间吧
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