HDU 5311:Hidden String【字符串】

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1738    Accepted Submission(s): 612


[align=left]Problem Description[/align]
Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string
s
of length n.
He wants to find three nonoverlapping substrings s[l1..r1],
s[l2..r2],
s[l3..r3]
that:

1. 1≤l1≤r1<l2≤r2<l3≤r3≤n

2. The concatenation of s[l1..r1],
s[l2..r2],
s[l3..r3]
is "anniversary".
 

[align=left]Input[/align]
There are multiple test cases. The first line of input contains an integer
T
(1≤T≤100),
indicating the number of test cases. For each test case:

There's a line containing a string s
(1≤|s|≤100)
consisting of lowercase English letters.
 

[align=left]Output[/align]
For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).
 

[align=left]Sample Input[/align]

2
annivddfdersewwefary
nniversarya

 

[align=left]Sample Output[/align]

YES
NO

 暴力，把字符串anniversary分成三段，枚举每段的长度，分析是否有符合条件的。。。
AC-code:

#include<cstdio>
#include<cstring>
int main()
{
int T,i,j,flag,l;
char s[105],str[]="anniversary";
scanf("%d",&T);
getchar();
while(T--)
{
gets(s);
l=strlen(s);
int flag=0;
for(i=0;i<=8;i++)
{
for(j=i+1;j<=9;j++)
{
int k=0;
while(k<l&&strncmp(s+k,str,i+1)!=0)
k++;
if(k==l)
break;
k+=i+1;
while(k<l&&strncmp(s+k,str+i+1,j-i)!=0)
k++;
if(k==l)
break;
k+=j-i;
while(k<l&&strncmp(s+k,str+j+1,10-j)!=0)
k++;
if(k!=l)
{
flag=1;
break;
}
}
if(flag)
break;
}
if(flag)
puts("YES");
else
puts("NO");
}
return 0;
}