ZOJ 3311

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这道题关键就是题意，尤其是第三条。如果"aZbJc" is Accepted，那么"aZbOJcO" is Accepted。这里不要相当然的以为：既然"aZbJc" 可以Ac，那么"aZbJc"中，a就是'x' , b就是'O' , c就是"xO"了呗。所以"aZbJcO"就是"xZOOJxO"了。那么第三条想说的就是：除了"xZOJxO" is Accepted 之外，"xZOOJxOO" is also Accepted。

其实不然，因为以我Wa了近20次的经验告诉我，那么想是过不了的。。。

那么，应该怎么理解那就话呢？其实，题目说的很明白，就是如果"aZbJc" is Accepted,那么"aZbOJcO" is also Accepted。想一下，当"aZbJc"为"xZOJxO"时，"xZOOJxOO" is Accepted，因为"xZOJxO"符合第二条所以是Ac，那么按照第三条，既然"xZOOJxOO" is Accepted，那么"xZOOOJOOO" is also Accepted！对吧？

然后总结一下规律就是能Ac的字符串中'Z' 和 'J'分别只出现一次，并且'Z'在'J'前面，而且'Z' 和 'J'之间 'O'的个数等于'J'后面一直到字符串结束'O'的个数减去'Z'之前所有'O'的个数。

为了方便理解，我们假设变量Front是'Z'之前的'O'的个数,Middle是'Z' 和 'J'之间'O'的个数，Back是'J'之后'O'的个数。根据上一段说的Middle应该满足等式：Middle == Back - Front。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define MAXSIZE 110char String[MAXSIZE];int Front,Middle,Back,Length,Index_Z,Index_J,Count_Z,Count_J;int main(){    while(scanf("%s",String) != EOF)    {        Length = strlen(String);        Index_Z = Index_J = -1;        Count_Z = Count_J = Front = Middle = Back = 0;        for(int i = 0; i < Length; i++)        {            if(String[i] == 'Z')                ++Count_Z;            if(String[i] == 'J')                ++Count_J;            if((String[i] == 'Z') && (Index_Z == -1))                Index_Z = i;            if((String[i] == 'J') && (Index_J == -1))                Index_J = i;            if((String[i] == 'O') && (Index_Z == -1) && (Index_J == -1))                ++Front;            if((String[i] == 'O') && (Index_Z != -1) && (Index_J == -1))                ++Middle;            if((String[i] == 'O') && (Index_J != -1))                ++Back;        }        if((Count_Z == 1) && (Count_J == 1) && (Middle != 0) && (Middle == (Back - Front)))            printf("Accepted\n");        else            printf("Wrong Answer\n");        //printf("Count_Z = %d,Count_J = %d,Front = %d,Middle = %d,Back = %d\n",Count_Z,Count_J,Front,Middle,Back);    }    return 0;}

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