10035 – Primary Arithmetic Solution

This is an easy problem from uva judge https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=976

• I read the input until I find a 0 0.
• Set a carry variable with 0 at the beginning.
• I divide both number with 10 and I get the module 10 until both numbers are 0.
• In each division I get the last digit of the numbers dividing by 10. I add both last digits and a carry variable. If the value is greater than 9 I set the carry variable with a value different than 0, otherwise I set carry to 0.
• if the carry is greater than 0 I increase the the answer by 1 .
• Print the answer.
• Accepted!

Hints.

• Check all the carries.

Critical Input

999 1
1 999
1 989
989 1
5 5

Critical Output

3 carry operations.
3 carry operations.
1 carry operation.
1 carry operation.
1 carry operation.

Code

/*
*Author: Fabian Calsina
*Date: 30/06/2016
*/

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int carry;
    int a,b, ans;

    while(scanf("%d %d", &a, &b))
    {
        if(!a && !b) break;
        ans = 0;


        carry = 0;
        while(1)
        {
            int tmpA = a%10;
            int tmpB = b%10;

            if(tmpA + tmpB + carry >= 10 )
            {
                carry = (tmpA + tmpB + carry)/10;
            }
            else
                carry = 0;

            if(carry) ans++;
            a/=10;
            b/=10;

            //if(carry == 1)
              //  ans++;
            if(a == 0 && b == 0)
                break;
        }

        if(ans == 0)
            printf("No carry operation.\n");
        else if(ans == 1)
            printf("1 carry operation.\n");
        else
            printf("%d carry operations.\n", ans);

    }
return 0;
}

10055 – Hashmat the Brave Warrior Solution

This is an easy problem from uva judge https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=996

How I did it

• I read the input until I find EOF.
• Print the absolute value of the substraction.
• Accepted!

Hints.

• Don´t use int.

Code

/*
*Author: Fabian Calsina
*Date: 30/06/2016
*/

#include <bits/stdc++.h>

using namespace std;

int main()
{
    long a, b;

    while(scanf("%ld %ld", &a, &b) != EOF)

        printf("%ld\n", abs(a-b));

return 0;
}