首页 > 代码库 > HDU 5311 Hidden String (优美的暴力)
HDU 5311 Hidden String (优美的暴力)
Hidden String
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 52 Accepted Submission(s): 25
Problem Description
Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string
of length .
He wants to find three nonoverlapping substrings ,,
that:
1.
2. The concatenation of ,, is "anniversary".
1.
2. The concatenation of ,, is "anniversary".
There are multiple test cases. The first line of input contains an integer,
indicating the number of test cases. For each test case:
There‘s a line containing a string consisting of lowercase English letters.
There‘s a line containing a string consisting of lowercase English letters.
For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).
2 annivddfdersewwefary nniversarya
YES NO
BestCoder 1st Anniversary ($)
题目链接：http://acm.hdu.edu.cn/showproblem.php?
pid=5311
题目大意：问在一个字符串里能不能找到连续三个区间拼成anniversary
题目分析：暴力，枚举每段的长度
#include <cstdio>
#include <cstring>
char s[200], con[] = "anniversary";
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%s", s);
int len = strlen(s);
bool flag = false;
for(int i = 0; i <= 8; i++)
{
for(int j = i + 1; j <= 9; j++)
{
int k = 0;
while(k < len && strncmp(con, s + k, i + 1) != 0)
k ++;
if(k == len)
break;
k += i + 1;
while(k < len && strncmp(con + i + 1, s + k, j - i) != 0)
k ++;
if(k == len)
break;
k += j - i;
while(k < len && strncmp(con + j + 1, s + k, 10 - j) != 0)
k ++;
if(k != len)
{
flag = true;
break;
}
}
}
if(flag)
puts("YES");
else
puts("NO");
}
}HDU 5311 Hidden String (优美的暴力)
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。