Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1950    Accepted Submission(s): 724 Problem Description 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".   Input 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.   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 2 annivddfdersewwefary nniversarya   Sample Output YES NO   题意：在给出的字符串里能否找到三段不重叠的子串，使它们组成单词 "anniversary"，能输出YES 不能输出NO 思路：数据到100 ，暴力枚举，使用string类方便些，每次枚举单词"anniversary"的三个子串，然后在给出的串里看能否找到

code

#include <iostream> #include<cstdio> #include<string> using namespace std; int main() { string s,s1,s2,s3,str="anniversary"; int t; scanf("%d",&t); while(t--) { cin>>s; int len_s=s.length(); int len_str=str.length(); int flag=0; for(int i=1;i<=len_str-2;i++) { for(int j=1;j<=len_str-i-1;j++) { s1=str.substr(0,i); s2=str.substr(i,j); s3=str.substr(i+j,len_str-i-j); int st1=s.find(s1,0); if(st1==-1) continue; int st2=s.find(s2,st1+i); if(st2==-1) continue; int st3=s.find(s3,st2+j); if(st3==-1) continue; flag=1; break; } if(flag) break; } if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }