std::cout 如何格式化 output？

Question

How does std::cout work? The following code doesn't pass certain test cases for a question on HackerEarth.com:

double n,a,b;
while(t--){
    cin>>n>>a>>b;
    long long x = round(b*n*1.0/(a+b));
    cout<<((a*x*x) + b*(n-x)*(n-x))<<endl;
}

while, the following one passes all of them:

double n,a,b;
while(t--){
    cin>>n>>a>>b;
    long long x = round(b*n*1.0/(a+b));
    long long ans = (a*x*x) + b*(n-x)*(n-x);
    cout<<ans<<endl;
}

Why does it happen that when I store the calculated value in a variable, then only the test cases are passed? Does the value change if directly printed to console? I am a newbie to C++.

Answer 1

The output format "chosen" by cout (or any std::ostream ) depends on the type which is being outputted (specifically, it depends on the implementation of operator<< for that type).

Let's look at what types are at play in both cases.

In your first example, you are outputting a double , since a , b , and n are doubles. Even though x is a long long , the expression is a double due implicit conversions .

In the second example, you are outputting a long long , since that is the type of ans . Note that your calculation of ans may be truncated, since it is being computed as a double (for the reasons explained above) but stored in a long long .

Without knowing the details of the test cases you are talking about, one difference in the output is that doubles will likely be output in decimal notation (eg 1.23 or 1.0 ) whereas a long long (or any integral type) will be output as a whole number (note: there are ways to change this behavior, which I'm omitting here for simplicity).