ZOJ问题
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3631 Accepted Submission(s): 1092
Problem 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;
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
Source
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对这道题彻底的无语了,虽然是汉语,但是题的意思太难懂了注意,主要是第三种情况,这里贴一下代码吧
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
char s[1010];
int i,j,a,b,c,ans,flag;
while(scanf("%s",s)!=EOF)
{
flag=ans=c=0;
int len=strlen(s);
for(i=0;i<len;i++)
{
if(s[i]=='z')
{
ans++;
a=i;
}
else if(s[i]=='j')
{
flag++;
b=i;
}
else
c++;
}
if(flag!=1||ans!=1||b-a<2||c==0)
printf("Wrong Answer\n");
else
{
if(a*(b-a-1)==len-b-1)
{
printf("Accepted\n");
}
else
printf("Wrong Answer\n");
}
}
return 0;
}