题目描述
You are given string S and T consisting of lowercase English letters.
Determine if S equals T after rotation.
That is, determine if S equals T after the following operation is performed some number of times:
Operation: Let S=S1S2…S|S|. Change S to S|S|S1S2…S|S|−1.
Here, |X| denotes the length of the string X.
Constraints
·2≤|S|≤100
·|S|=|T|
·S and T consist of lowercase English letters.
输入
Input is given from Standard Input in the following format: S T
输出
If S equals T after rotation, print Yes; if it does not, print No.
样例输入
kyoto tokyo
样例输出
Yes
提示
In the first operation, kyoto becomes okyot.
In the second operation, okyot becomes tokyo.
题意很简单，不过醉翁之意不在酒，想要记录下一种好玩的做法，很好理解。
有关substr函数的应用
附上大佬的链接：
https://blog.csdn.net/Velly_zheng/article/details/89712735?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522158423930019724848339647%2522%252C%2522scm%2522%253A%252220140713.130056874…%2522%257D&request_id=158423930019724848339647&biz_id=0&utm_source=distribute.pc_search_result.none-task
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math") #pragma GCC optimize("Ofast") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #pragma comment(linker, "/stack:200000000") #pragma GCC optimize (2) #pragma G++ optimize (2) #include <bits/stdc++.h> #include <algorithm> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> using namespace std; #define wuyt main typedef long long ll; #define HEAP(...) priority_queue<__VA_ARGS__ > #define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > template<class T> inline T min(T &x,const T &y){return x>y?y:x;} template<class T> inline T max(T &x,const T &y){return x<y?y:x;} //#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++) //char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf; ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar(); if(c == '-')Nig = -1,c = getchar(); while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar(); return Nig*x;} #define read read() const ll inf = 1e15; const int maxn = 2e5 + 7; const int mod = 1e9 + 7; #define start int wuyt() #define end return 0 start{ /** int num[4]; cin>>num[1]>>num[2]>>num[3]; sort(num+1,num+4); cout<<num[2]-num[1]+num[3]-num[2];**/ string s,t; cin>>s>>t; int ans=0; for(int i=0;i<s.size();i++) { if(s==t) { ans=1; break; } s=s.back()+s.substr(0,s.size()-1); } if(ans==1) printf("Yes\n"); else printf("No\n"); end; }
