Hidden StringTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 429Accepted Submission(s): 161
Problem Description
Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a stringof length. He wants to find three nonoverlapping substringsthat:1.2. The concatenation ofis "anniversary".
Input
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 stringconsisting of lowercase English letters.
Output
For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).
Sample Input
2annivddfdersewwefarynniversarya
Sample Output
YESNO
Source
问题描述
今天是BestCoder一周年纪念日. 比赛管理员Soda有一个长度为满足下列条件: 1. 2. 依次连接之后得到字符串"anniversary".
输入描述
输入有多组数据. 第一行有一个整数, 表示测试数据组数. 然后对于每组数据:一行包含一个仅含小写字母的字符串.
输出描述
对于每组数据, 如果Soda可以找到这样三个子串, 输出"YES", 否则输出"NO".
输入样例
2annivddfdersewwefarynniversarya
输出样例
YESNO
解题思路：枚举三个子字符串，查找。
代码如下：
#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<ctime>#include<iostream>#include<algorithm>#include<string>#include<vector>#include<deque>#include<list>#include<set>#include<map>#include<stack>#include<queue>#include<numeric>#include<iomanip>#include<bitset>#include<sstream>#include<fstream>#include<limits.h>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-4)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1e9+7using namespace std;typedef long long ll;typedef unsigned long long ULL;int main(){int i,j,k,n,flag,t;char s[200],S[]="anniversary";scanf("%d",&t);while(t–){scanf("%s",s);n=strlen(s);flag=0;for(i=0;i<=8;i++){for(j=i+1;j<=9;j++){k=0;while(k<n&&strncmp(S,s+k,i+1)!=0)k++;if(k==n)continue;k+=i+1;while(k<n&&strncmp(S+i+1,s+k,j-i)!=0)k++;if(k==n)continue;k+=j-i;while(k<n&&strncmp(S+j+1,s+k,10-j)!=0)k++;if(k!=n){flag=1;break;}}}if(flag)printf("YES\n");elseprintf("NO\n");}}
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