ABC145 B - Echo

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# Problem

Given a string SS, report if SS is twice repetitions of a string or not.

# Explanation

If the length of SS is odd, report No.
Otherwise, split SS in the middle into S1S1 and S2S2.
If S1S1 equals to S2S2, report Yes.

# Time complexity

Half of the length of SS.

O(N/2)O(N / 2)

# Solution

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;
Int N;
string S;

void input() {
  cin >> N >> S;

void solve() {
  if (N % 2 == 1) {
    cout << "No" << endl;

  if (S.substr(0, N/2) == S.substr(N/2)) cout << "Yes" << endl;
  else cout << "No" << endl;

int main(void) {
  return 0;

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