Problem in using printf() function instead of cout

Question

I was trying to solve the record breaking day problem from kickstart .

I have two questions:

1. Why the printf function is not working but the cout does the job (which means there is no problem with the logic)
2. Why the code is not passing the test case 1 given in the kickstart.

#include <bits/stdc++.h>
using namespace std;
long long solve(){
    long long n;
    cin >>n;
    long long a[n];
    for(long long j=0;j<n;j++)
            cin>>a[j];
    long long k=0,ans=0;
    for(long long j=0;j<n-1;j++){
        if(a[j]>k){
            k=a[j];
            if(a[j]>a[j+1]){
                ans++;
            }
        }
    }
    if(a[n-1]>k)
        ans++;
    return ans;
}
int main(){
    long long t;
    cin >> t;
    for(long long i=1;i<=t;i++){
        printf("Case #%d: %d\n",i,solve());//instead of printf() this works: cout<<"Case #"<<i<<": "<<solve()<<endl;
    }
    return 0;
}

sample test case from kickstart:

4
8
1 2 0 7 2 0 2 0
6
4 8 15 16 23 42
9
3 1 4 1 5 9 2 6 5
6
9 9 9 9 9 9

output with printf():

Case #1: 0
Case #2: 0
Case #3: 0
Case #4: 0

expected output(output that i get with cout):

Case #1: 2
Case #2: 1
Case #3: 3
Case #4: 0

Answer 1

The printf format code %d expects an int not a long long .

The max value for an int is 2,147,483,647 and is probably big enough for your usage.

---

The reason that cout is working is because std::ostream::operator<< is overloaded to understand long long values.

Answer 2

You're using the wrong format specifier to printf .

The %d format specifier expects an int but you're passing in a long long . Using the wrong format specifier triggers undefined behavior .

To print a long long , you need to use %lld .

Using the << operator on cout has overloads for various parameter types so it knows the type based on the chosen overload.