C Programming Help

Code:

#include<stdio.h>
int main (void)
{
    // declaration (i equate them to 0 as a practice, safety reason)
    int a = b = c = 0;
    int sum1 = sum2 = sum3 = 0;

    // user input
    scanf("%i", &a);
    scanf("%i", &b);
    scanf("%i", &c);

    // computing sum
    sum1 = a+b;
    sum2 = a+c;
    sum3 = b+c;

    // if statements
    if(a != sum3) //if A is not the sum of b & c, continue testing
    {
          if(b != sum2) //if B is not the sum of a & c, continue testing
          {
              if(c != sum1) //if C is not the sum of a & b, there is nothing left to test
              {
                    printf("No.\n"); //nothing to test
              }

              else
              {
                  printf("%i", sum1); //C is the sum of the other 2 numbers a & b! print out sum1.
              }
          }

          else
          {
              printf("%i", sum2); //B is the sum of the other 2 numbers a & c! print out sum2.
          }
    }

    else
    {
          printf("%i", sum3); //A is the sum of the other 2 numbers b & c! print out sum3.
    }
   
    return 0;
}

Code:

#include<stdio.h>
int main (void)
{
    // declaration (i equate them to 0 as a practice, safety reason)
    int a = b = c = 0;
    int sum1 = sum2 = sum3 = 0;

    // user input
    scanf("%i", &a);
    scanf("%i", &b);
    scanf("%i", &c);

    // computing sum
    sum1 = a+b;
    sum2 = a+c;
    sum3 = b+c;

    // if statements
    if(a == sum3) // is a the sum of b and c?
    {
          printf("%i", sum3); // Great! a is the sum of b and c.
    }

    else // damn it a is not the sum of b and c, move on to test the other numbers
    {
          if(b == sum2) // is b the sum of a and c?
          {
              printf("%i", sum2); // Great! b is the sum of a and c.
          }
         
          else // damn it b is not the sum of a and c, move on to test the other numbers
          {
              if(c == sum1) // is c the sum of a and b?
              {
                    printf("%i", sum1); // Great! c is the sum of a and b.
              }

              else
              {
                    printf("No.\n"); // damn it there are no 2 numbers that add up to either of the 3 numbers.
              }
          }
    }
   
    return 0; // exit

}

Code:

if(x >= 100000 && x <= 999999) // x in the range of 100000 to 999999
{
    if(n >= 0 && n <= 1) // n in the range of 0 to 1
    {
          // code here
    }
}

Code:

ans = ((x%10)^n)/(10^(n-1))

wyehuong wrote:sorry i cant c ur code from the screen shot i think u need to crop it esp the right side and show me the impt parts..

i tried to c some of it.. ur if statement the x part is logical but the n part is weird.. maybe i nvr c clearly but how can n be bigger or equal to 1 and be equal to 0 at the same time. 0 is smaller than 1..

hmm.. normally is i have so many parameters in my if statement.. i will try to split it into 2 if statements.. so i can spot the error easier..

Code:

if(x >= 100000 && x <= 999999) // x in the range of 100000 to 999999
{
    if(n >= 0 && n <= 1) // n in the range of 0 to 1
    {
          // code here
    }
}

and line 22.. u using modulus and square? i also rusty liao.. but then i think u shuld put some more brackets.. the compiler might hav executed the equation in a wrong sequence. modulus if i nt wrong is the remainder when u divide two integers.. correct me if im wrong

smth like..

Code:

ans = ((x%10)^n)/(10^(n-1))

Code:

if(x>=100000 && x<= 999999)
{
    if(n==1)
    {
          // code
    }

    else if(n==2)
    {
          // code
    }

    else if(n==3)
    {
          // code
    }

    else if(n==4)
    {
          // code
    }

    else if(n==5)
    {
          // code
    }

    else if(n==6)
    {
          // code
    }

    else // this else is for if n is equal to anything else besides 1 to 6
    {
          printf("invalid\n");
    }

}

else // this else is for x to be other than 100000 to 999999
{
    printf("invalid\n");
}



Code:

// NDigit.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <math.h>
#include <iostream>

using namespace std;

int main (void)
{
   // declaration
   int mDigit = 0;
   int n = 0;
   int mAnswer = 0;

   while(true)
   {
      cout << "Please Enter Digit: ";
      cin >> mDigit;

      cout << "Please Enter Nth Number: ";
      cin >> n;

      /*
      if((mDigit >= 100000 && mDigit <= 999999) && (n >= 1 && n <= 6))
      {
         mAnswer = mDigit % 10 ^ n / 10 ^ (n - 1);

         cout << endl << "The answer is: " << mAnswer << endl << endl;
      }

      else
      {
         cout << "Error" << endl;
      }
      */

      if(mDigit >= 100000 && mDigit <= 999999)
      {
         if(n >= 1 && n <=6)
         {
            switch(n)
            {
               case 1:
                  mAnswer = (mDigit%10)/1;
                  break;
               
               case 2:
                  mAnswer = (mDigit%100)/10;
                  break;
               
               case 3:
                  mAnswer = (mDigit%1000)/100;
                  break;

               case 4:
                  mAnswer = (mDigit%10000)/1000;
                  break;

               case 5:
                  mAnswer = (mDigit%100000)/10000;
                  break;

               case 6:
                  mAnswer = (mDigit%1000000)/100000;
                  break;
            }
            
            cout << endl << "The answer is: " << mAnswer << endl << endl;

         }

         else // if n is not within range
         {
            cout << "Invalid" << endl;
         }
      }

      else // if mDigit is not within range
      {
         cout << "Invalid" << endl;
      }
   }

   return 0;
}



Code:


for(i = 5; i < 10;)
{
   y = ++i;
   cout << "value of y is: " << y << endl;
}


Code:


for(i = 5; i < 10;)
{
   y = i++;
   cout << "value of y is: " << y << endl;
}