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≤n2. The concatenation of s[l1..r1], s[l2..r2], s[l3..r3] is "anniversary".

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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.

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

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Sample Input 2 annivddfdersewwefary nniversarya

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Sample Output YES NO

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Source BestCoder 1st Anniversary ($)

?題意：從s串中找出3段連續(xù)的字串組成“anniversary”

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復(fù)習(xí)了下find，substr的用法，老是忘記

s.substr（i，j）表示從s串的i位置開始，長度為j的子串。

s.find（p，i），p為字串，表示從s串的i位置開始，尋找有沒有等于p的子串，如果有返回s的首地址，否則返回-1

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這題枚舉“anniversary”的3個(gè)子串，在給出的s串中尋找就可以了！

1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<queue> 6 #include<cmath> 7 #include<stdlib.h> 8 #include<map> 9 using namespace std; 10 string s; 11 string goal="anniversary"; 12 bool solve(){ 13 for(int i=1;i<=9;i++){ 14 int ans1=s.find(goal.substr(0,i),0); 15 if(ans1<0) continue; 16 for(int j=1;j+i<=10;j++){ 17 int ans2=s.find(goal.substr(i,j),ans1+i); 18 if(ans2<0) continue; 19 int k=11-i-j; 20 int ans3=s.find(goal.substr(i+j,k),ans2+j); 21 if(ans3<0) continue; 22 return true; 23 } 24 } 25 return false; 26 } 27 int main() 28 { 29 int t; 30 scanf("%d",&t); 31 while(t--){ 32 cin>>s; 33 if(solve()){ 34 printf("YES\n"); 35 } 36 else{ 37 printf("NO\n"); 38 } 39 } 40 return 0; 41 } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/UniqueColor/p/4799024.html

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