CODEWITHSUNDEEP
c++
Ch3steR
Ch3steR

Reputation: 20669

Code produce two different upon removing if statement

While I was solving a question from codeforces(link to the question: https://codeforces.com/contest/1466/problem/A) I encountered weird output from the below code

#include<iostream>
#include<vector>
#include<set>

using namespace std;

void solve(){
    int64_t n;
    cin>>n;
    if (n<2){
            cout<<0<<endl;
            return;
    }
    vector<int64_t> vec(n, 0);
    for(auto &i: vec){
            cin>>i;
    }
    set<int64_t> s;
    for(int64_t i=0; i<n-1; ++i){
        for(int64_t j=i+1; j<n; ++j){
            s.insert(vec[j]-vec[i]);
        }
    }
    cout<<s.size()<<endl;
}

int main(){
    int64_t t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

Input:

8
4
1 2 4 5
3
1 3 5
3
2 6 8
2
1 2
1
50
5
3 4 5 6 8
3
1 25 26
6
1 2 4 8 16 32

Ouptut given by my code:

4
2
3
1
0
53
1
1

When I removed if block it passed all the pretests/testcases.

    if (n<2){
            cout<<0<<endl;
            return;
    }

Expected output:

4
2
3
1
0
5
3
15

After removing the if block I get the same output as above. I'm not able to understand how if block is affecting the size of set in the next iteration.

Compiled with g++ same output with -std=c++11 -std=c++14 and -std=c++17

Upvotes: 0

Views: 60

Answers (1)

scohe001
scohe001

Reputation: 15446

if (n<2){
    cout<<0<<endl;
    return;
}

Since you're returning here, it means that if n is 1, you never read that one value out of std::cin.

If you remove that if, then that one value will get read in here:

vector<int64_t> vec(n, 0);
for(auto &i: vec){
        cin>>i;
}

So if you want the if to work as expected, it should be:

if (n<2){
    cout<<0<<endl;
    if(n == 1) { std::cin >> n; } // remove 1 item from stream
    return;
}

Upvotes: 2

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