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## B - Same Name Editorial by en_translator

Since the constraint of $$N$$ is as small as $$1000$$, we may check for every pair of integers $$(i,j)$$ such that $$1 \leq i \lt j \leq N$$ if it satisfies the condition described in the program, that is, check if both $$S_i=S_j$$ and $$T_i=T_j$$ holds.

The complexity is $$O(N(X+Y))$$, where $$X$$ is the sum of lengths of $$S_i$$ and $$Y$$ is the sum of lengths of $$T_i$$.

Sample code (C++)

#include<bits/stdc++.h>
using namespace std;

int main(){
int N; cin >> N;
vector<string> S(N),T(N);
string ans = "No";
for(int i=0; i<N; i++){
cin >> S[i] >> T[i];
for(int j=0; j<i; j++){
if(S[i]==S[j] && T[i]==T[j]) ans = "Yes";
}
}
cout << ans << endl;
}


Sample code (Python)

N = int(input())
S = ['']*N
T = ['']*N
for i in range(N):
S[i],T[i] = map(str,input().split())
ans = 'No'
for i in range(N):
for j in range(i+1,N):
if S[i]==S[j] and T[i]==T[j]:
ans = 'Yes'
print(ans)


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