在 HackerRank 练习题中使用 Printf 和 Scanf 的问题

Question

I have been trying to solve this problem in Hackerrank for quite a while now.The issue that i am facing is that I being a beginner on reading the question ended up using cin and cout instead of scanf and printf while taking user input just so as to be able to check as to whether my reasoning was correct or not and although I have solved the problem correctly using the former but am facing difficulty using the latter.

Here is the problem that I have been trying to solve:
https://www.hackerrank.com/challenges/acm-icpc-team/problem

Here is code using cin and cout :

#include<iostream>
#include<stdio.h>
using namespace std;

int main(){
int n,m;
 cin>>n>>m;
int a[n][m]={};
for (int i = 0; i < n; i++)
{
    for (int j = 0; j < m; j++)
    {
    
       cin>>a[i][j];
    }
    
}
int b[n*(n-1)/2][m]={};

int x=0;
for (int i = 0; i < n-1; i++)
{   
    for (int j = i+1; j < n; j++)
    {
       for (int k = 0; k < m; k++)
       {
           b[x][k] = (a[i][k]|a[j][k]);
       }
       x=x+1;
       
    }    
}

int count = 0;
int c[n*(n-1)/2]={};

for (int i = 0; i < n*(n-1)/2; i++)
{
    c[i]=0;
}
for (int i = 0; i < n*(n-1)/2; i++)
{
    for (int j = 0; j < m; j++)
    {
        if (b[i][j]==1)
        {
           count++;
        }
        
    }
    c[i]=count;
    count=0;
    
}

for (int i = 0; i < n; i++)
{
   int temp=c[i];
   int j=i;
   while (j>0 && temp<c[j-1])
   {
       c[j]=c[j-1];
       j=j-1;
   }
   c[j]=temp;
    
}
int sum=0;
for (int i = 0; i <  n*(n-1)/2; i++)
{
    if (c[ n*(n-1)/2 - 1]==c[i])
    {
        sum++;
    }
    
}
cout<<c[ n*(n-1)/2 - 1]<<endl;
cout<<sum;

    return 0;
}

While writing the entire code using "printf" and "scanf", I simply replaced "cin and cout":

#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    int n, m;
    scanf("%d", "%d", &n, &m);
    int a[n][m] = {};
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            scanf("%d", &a[i][j]);
        }
    }
    int b[n * (n - 1) / 2][m] = {};

    int x = 0;
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            for (int k = 0; k < m; k++)
            {
                b[x][k] = (a[i][k] | a[j][k]);
            }
            x = x + 1;
        }
    }

    int count = 0;
    int c[n * (n - 1) / 2] = {};

    for (int i = 0; i < n * (n - 1) / 2; i++)
    {
        c[i] = 0;
    }
    for (int i = 0; i < n * (n - 1) / 2; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (b[i][j] == 1)
            {
                count++;
            }
        }
        c[i] = count;
        count = 0;
    }

    for (int i = 0; i < n; i++)
    {
        int temp = c[i];
        int j = i;
        while (j > 0 && temp < c[j - 1])
        {
            c[j] = c[j - 1];
            j = j - 1;
        }
        c[j] = temp;
    }
    int sum = 0;
    for (int i = 0; i < n * (n - 1) / 2; i++)
    {
        if (c[n * (n - 1) / 2 - 1] == c[i])
        {
            sum++;
        }
    }

    printf("sum");

    return 0;
}

I am also attaching a copy of the output that I am getting using the former code: The output is coming out fine when I am manually giving whitespace characters but when am running the same code using printf and scanf it's not working for me for some reason:

Could someone plz guide me as to where was I going wrong?

Answer 1

scanf("%d", "%d", &n, &m);

should be

scanf("%d %d", &n, &m);

scanf takes one format string as it's first argument and then (depending on the format string) zero or more arguments after that.

printf is similar, so

printf("sum");

should be

printf("%d", sum);

All of this is well documented here and here .

Answer 2

you have 4 problems.

1. scanf("%d", "%d", &n, &m); will not load the integers into n and m. It will load try to load 1 integer into the second argument(int this case "%d") then ignore n and m. To load multiple integers in the same scanf you need to add all of the %d's in the same string:

scanf("%d%d", &n, &m);

2. the second scanf currently has no way of knowing how many digits you want to scan in as the integer and will therefore attempt to load the largest integer possible, because the maximum integer is 2147483647 it has no problem loading 10101 for example as a single integer. This works differently with std::cin as the ">>" operator will read 1 character at a time until it has enough to create a valid integer(1 is a valid integer). You can force scanf to only load a set number of digits by adding the number of digits to load to the format string:

scanf("%1d", &a[i][j]);

3. Your printf function will simply print "sum" not the value of the variable sum. To get it to print the value of the sum variable you need to use the same type of formatting string you use in scanf:

printf("%d", sum);

4. the original code prints out the value of c[ n*(n-1)/2 - 1] the second doesn't. You could either print this out in a second printf:

printf("%d\n", c[ n*(n-1)/2 - 1]); //note use of \n to force a new line on output

or in the same printf:

printf("%d\n%d", c[ n*(n-1)/2 - 1], sum);