#include <iostream>
#include <stdio.h>
#include <iomanip>
#include <math.h>
using namespace std;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int main()
{
	int n;
	cin >> n; 
	n = n * 256 * 1024;
	cout << n;
	return 0;
}

#include<iostream> using namespace std; int main(){ char a; cin>>a; cout<<" "<<a<<endl; cout<<" "<<a<<a<<a<<endl; cout<<a<<a<<a<<a<<a<<endl; return 0; }

python的好处：可以一步到位

print(int(int(input())*1024*1024/4))

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long rzr,m;
    cin>>rzr;
    cout<<rzr*1024*256;
}

样例告诉我们了，$1$ MB 答案就是 $262144$，所以我们只需要输出 $262144\times n$ 即可。

python一行解决：

print(int(input())*262144)

是不是很简单（

样例已经告诉我们了【。。。】

#include <bits/stdc++.h>
using namespace std;
int main() 
{
    int a; 
    cin >> a;
    cout << a * 262144;
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main() 
{
    int n; 
    cin >> n;
    cout << n * 262144;
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main(void)
{
    int m, b;

    cin >> m;

    b = m * 1024 * 1024;

    cout << b / 4;

    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main(void)
{
    int m, b;

    cin >> m;

    b = m * 1024 * 1024;

    cout << b / 4;

    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main(void){
    int n;
    cin>>n;
    cout<<n*1024*1024/4;
    return 0;
}

#include
using namespace std;
int main(){
    int m,b;
	cin>>m;
	b=m*1024*1024/4;
    cout<<b;
}

#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
    int n;
	cin>>n;
	int a=n*1024*1024/4;
	cout<<a<<endl;	
}

#include<bits/stdc++.h>
using namespace std;
int main(){
	int n;
	cin>>n;
	cout<<(n*1024*1024)/4;
	return 0;
}