Description
对给定的字符串(只包含'z','o','j'三种字符),判断他是否能AC。
是否AC的规则如下:
1. zoj能AC;
2. 若字符串形式为xzojx,则也能AC,其中x可以是N个'o' 或者为空;
3. 若azbjc 能AC,则azbojac也能AC,其中a,b,c为N个'o'或者为空;
Input
输入包含多组测试用例,每行有一个只包含'z','o','j'三种字符的字符串,字符串长度小于等于1000;
Output
对于给定的字符串,如果能AC则请输出字符串“Accepted”,否则请输出“Wrong Answer”。
Sample Input
zoj ozojo ozoojoo oozoojoooo zooj ozojo oooozojo zojoooo
Sample Output
Accepted Accepted Accepted Accepted Accepted Accepted Wrong Answer Wrong Answer
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char s[1005];
int main()
{
int len;
int i;
int a, b, c;
while (gets(s)){
len = strlen(s);
a = b = c = 0;
i = 0;
while (i < len){
if (s[i] == 'z')
a = i;
else if (s[i] == 'j'){
b = i;
break;
}
i++;
}
// 原式的 a * b == c 且 b > 0
if (a*(b-a-1)==(len-b-1) && (b-a-1)>0)
printf("Accepted\n");
else
printf("Wrong Answer\n");
}
return 0;
}