## Counting carry operations

count carry in c
given two numbers find the number of carries while adding them geeks
given two numbers. add the numbers and find the count of carries in their addition.
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Can anybody tell me why my program keeps getting wrong answer? It must count the number of carry operations in a sum. I tried every testcase came to my mind. I didn't get wrong output.

Problem Description:

Children are taught to add multi-digit numbers from right-to-left one digit at a time. Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number of carry operations for each of a set of addition problems so that educators may assess their difficulty.

Input

Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0.

Output

For each line of input except the last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below.

Sample Input

```123 456
555 555
123 594
0 0
```

Sample Output

```No carry operation.
3 carry operations.
1 carry operation.
```

Here's my current code:

```#include<stdio.h>
int main()
{
unsigned long long int a,b,m,n,rem_m,rem_n,judge=0,sum,count;
while((scanf("%llu%llu",&m,&n))==2)
{
if(m==0 && n==0)
{
break;
}
count=0;
while(m!=0 && n!=0)
{
rem_m=m%10;
rem_n=n%10;
if(judge==1)
{
rem_m++;
}
sum = rem_m+rem_n;
judge=0;
if(sum>=10)
{
count++;
judge++;
}
m=m/10;
n=n/10;
}
if(count==0)
{
printf("No carry operation.\n");
}
else
{
printf("%llu carry operations.\n",count);
}
}
return 0;
}
```

count the number of carry operations in a sum

Asserting a,b are >= 0:

##### Terse solution

For fun :)

"ds" stands for digit sum.

```int ds(int n){return n == 0 ? 0 : n%10 + ds(n/10);}
int numberOfCarryOperations(int a,int b){return (ds(a) + ds(b) - ds(a+b)) / 9;}
```

Here is a more readable variation.

```int digitSum(int n)
{
int sum;
for (sum=0; n > 0; sum+=n%10,n/=10);
return sum;
}

int numberOfCarryOperations(int a,int b){
// a, b >= 0
return (digitSum(a) + digitSum(b) - digitSum(a+b)) / 9;
}
```

You can proof mathematically: every time you have a carry, the digitSum decreases by 9.

9, because we are in number system 10, so we "lose 10" on one digit if we have carry, and we gain +1 as the carry.

##### Pythonic version

I do not know how to do this in C, but in python it is easy to write a better digitSum function. In python we can easily create the list of digits from a number, and then just use sum() on it to get digitSum of the given number.

Here is a terse python one-liner solution:

```def numberOfCarryOperations(a, b):
# f is the digitSum function
f=lambda n:sum(map(int,str(n)));return(f(a)+f(b)-f(a+b))/9
```

Count the number of carry operations required to add two numbers , Initialise the carry variable and count variable to 0. Now, check from the last index of the strings till both the strings come to an end(one string may be smaller than  Count the number of carry operations required to add two numbers Count the number of carry operations required to add two numbers Given two numbers, the task is to find the number of carry operations required when two numbers are added as below. 1234

The loop condition is wrong. You want `while(m!=0 || n!=0)` (i.e. while at least one of them is not zero) instead of `while(m!=0 && n!=0)`, otherwise the answer will be wrong for things like `999 9`, it will incorrectly stop after one iteration and report 1 carry operation whereas the correct answer should be 3. Think of it like this: you only want to stop when both of them are 0, so the loop must continue as long as at least one of the numbers is not 0.

Also, you forgot to clean up `judge` after printing output. You need to clear it before reading input again, or you could mistakenly have `judge == 1` from a previous computation that ended with a carry (the name choice for this variable seems odd to me, you should rename it to something more meaningful like `carry`, but it's not the main issue here).

`a` and `b` are unused (you should enable compiler warnings).

The sample output shows the word operation (as in, singular) when the count is 1; your program always writes operations (plural). If you're submitting this to an automatic judge, the code will not pass because the output does not match exactly the expected output. To fix that small little detail, replace this:

```else
{
printf("%llu carry operations.\n",count);
}
```

With:

```else
{
printf("%llu carry operation%s.\n",count, count > 1 ? "s" : "");
}
```

Here's the fixed version:

```#include <stdio.h>

int main(void)
{
unsigned long long int m,n,rem_m,rem_n,judge=0,sum,count;
while((scanf("%llu%llu",&m,&n))==2)
{
if(m==0 && n==0)
{
break;
}
count=0;

/* We want || here, not && */
while(m!=0 || n!=0)
{
rem_m=m%10;
rem_n=n%10;
if(judge==1)
{
rem_m++;
}
sum = rem_m+rem_n;
judge=0;
if(sum>=10)
{
count++;
judge++;
}
m=m/10;
n=n/10;
}

/* Clean up for next iteration */
judge = 0;

if(count==0)
{
printf("No carry operation.\n");
}
else
{
printf("%llu carry operations.\n",count);
}
}
return 0;
}
```

Python: Count the number of carry operations for each of a set of , Python Exercises, Practice and Solution: Write a Python program to count the number of carry operations for each of a set of addition problems. Many find the "carry" operation - in which 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number of carry operations for each of addition problem so that educators may assess their difficulty. For the input first line contains n number of records which is less then 1000.

A ruby solution would be:

```def count_carry_operations x, y
return 0 if x == 0 && y == 0

count = 0
carry = 0

while true
return count if x == 0 && y == 0

while x != 0 || y != 0
xr = x % 10
yr = y % 10

xr += 1 if carry == 1

sum = xr + yr
carry = 0

if sum >= 10
count += 1
carry += 1
end

x /= 10
y /= 10
end

carry = 0
end

count
end
```

Carry count, Many of us found the carry operation - in which a 1 is carried from one is to count the number of carry operations required for adding two multi  Write a Python program to count the number of carry operations for each of a set of addition problems. According to Wikipedia " In elementary arithmetic, a carry is a digit that is transferred from one column of digits to another column of more significant digits.

A java solution would be:

```public class Main {

public  static int carry_count=0,carry_number=0;
public static void main(String[] args) {
System.out.println(Carry(99511,512));
}
private  static  int Carry(int num1,int num2){
if(num1/10==0 || num2/10==0){
int sum=num1%10+num2%10+carry_number;
if(sum>=10){
carry_number=1;
carry_count++;
return Carry(num1/10,num2/10);
}else{
return carry_count;}
}else {
int sum=num1%10+num2%10+carry_number;
if(sum>=10){
carry_number=1;
carry_count++;
}else {
carry_number=0;
}
return Carry(num1/10,num2/10);
}
}
}
```

Counts the number of carry operation in an addition of two numbers , function numberOfCarryOperations(x,y) {. var xs = x.toString();. var ys = y.toString(​);. var cary = 0;. var operations=0;. var xCurrent;. var yCurrent;. Many of us found the carry operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge at that point of time. (You may find it challenging yet again if you can't solve this D:) Your task is to count the number of carry operations required for adding two multi digit numbers.

Count Carry Problem | Practice Problems, Many find the "carry" operation - in which 1 is carried from one digit position to be Your job is to count the number of carry operations for each of addition  Given two numbers, the task is to find the number of carry operations required when two numbers are added as below.. 1234 + 5678-----6912-----4+8 = 2 and carry 1 carry+3+7 = carry 1 carry+2+6 = 9, carry 0 carry+1+5 = 6. We need two carry operations in this example. Input: First line of input consists of an integer T, denoting the no of test cases.

Carry (arithmetic), When used in subtraction the operation is called a borrow. Carrying is emphasized in traditional mathematics, while curricula based on reform mathematics do  The sequence of counting numbers in Peano arithmetic being defined by applying a successor function; the Church encoding makes that successor function a way of encoding numbers as functions themselves using a recursive structure.

Carry Counting, Carry Counting. Given two positive integers, compute how many carry operations are involved in the standard method for adding them in base 2. Step 6: CARRY treatment: Recommended rugged method (that works in case of any combination of values of the two digits involved in subtraction at the current place): If there is a BORROW of -1 at this ten's digit place where presently subtraction process is going on, we first add 1 to the ten's digit of the bottom number and then carry out the subtraction of the two ten's digits, the bottom one from the top.

• when compiling, always enable all warnings (for gcc, at least: `-Wall -Wextra -pedantic`) then fix the warnings. for instance: for the posted code, the compiler will warn about unused variable 'a' and unused variable 'b'. Strongly suggest placing each variable declaration on a separate line, for readability by us humans and for ease of documentation.
• the variable `count` can never be greater than 10 so why make it a 'llu' variable?
• the variable `judge` can never be other than 1 or 0, so why make it a 'llu' variable? Also, the name `judge` is meaningless in the current context. Suggest something meaningful like: `carry`