CODEWITHSUNDEEP
c++cstringalgorithmloops
sgupta
sgupta

Reputation: 1272

Calculating the sum of digits times the length of string

Last week i had a project from my teacher asking me to develop a program which takes in a string (stream of integers to be precise) and calculates the sum of the numbers in the string for each number in the string ie.

if input is 31456

I did submit a working project but it is full of spaghetti code (nested loops which ends if string length is less than the number of loops) which is not a clean approach. I wondered and studied a quite a lot over this situation in vain. I have not found any way of doing this without nested for loops in C (or maybe i gave up too fast ?)

Again, i am not asking you guys for an answer to my problem but wanted to know if there is a way of doing this without the nested loops (which will have problem if length of input > number of nested loops).

Upvotes: 1

Views: 721

Answers (7)

Elementary
Elementary

Reputation: 1453

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


static void printResults(int* results,int size);
int main( int argc, char** argv ){
    char nums[] = "314156";/* "31456" ? */
    int i,tmp;
    char tmpc[2];
    int length=strlen(nums);
    #define LENGTH length
    int results[LENGTH];
    results[0]=0;
    tmpc[0]=nums[0];
    sscanf(tmpc, "%d", &tmp);
    results[1]=tmp;
    int result=results[1];
    for(i=2 ; i< length ; ++i){
            tmpc[0]=nums[i];
            sscanf(tmpc, "%d", &tmp);
            result+=tmp;
            results[i]=result;
    }
    printResults(results,length);
    return 0;
}

static void printResults(int* results,int size){
    int i;
    for(i=0;i<size;i++){
       printf("%d time loop sum is %d\n", i+1,results[i]);
    }

}

This is a really basic and understandable code in C for beginner and the ouput is:

output

Upvotes: 0

Arslan
Arslan

Reputation: 1

#include<iostream>
#include<cctype>
using namespace std;
void main()
{
const int SIZE=10;
char myArray[SIZE];
int length=0,sum=0;
cout<<"enter Array of digit from 1-9\n\n ";
cin>>myArray;
cout<<myArray<<endl;
for(int i=0;i<myArray[i];i++)
{
    if(!isspace(myArray[i]))
    //if(myArray[i]!=NULL)

    length++;
}
    cout<<length<<endl;


for(int i=0;i<length;i++)
{
    sum+=myArray[i]-'0';

}



    cout<<sum;
    }

Upvotes: -1

Blastfurnace
Blastfurnace

Reputation: 18652

There's a C++ Standard Library function called partial_sum() that performs the series of sums you describe in a single pass over the input.

int sums[] = { 0, 3, 1, 4, 1, 5, 6 };

std::partial_sum(sums, sums + 7, sums);

// The results are left in sums[0]..sums[6]

Upvotes: 0

BLUEPIXY
BLUEPIXY

Reputation: 40145

#include <stdio.h>

int sum(const char ch){
    static int sum = 0;
    int retValue = sum;

    sum += ch -'0';

    return retValue;
}

int main(){
    char nums[] = "314156";/* "31456" ? */
    int size = sizeof(nums)/sizeof(char);
    int i;

    for(i=0 ; i< size ; ++i){
        printf("%d time loop sum is %d\n", i+1, sum(nums[i]));
    }

    return 0;
}

DEMO

1 time loop sum is 0
2 time loop sum is 3
3 time loop sum is 4
4 time loop sum is 8
5 time loop sum is 9
6 time loop sum is 14
7 time loop sum is 20

Upvotes: 0

Shahbaz
Shahbaz

Reputation: 47503

A pseudo-code:

array of sums
sums[0] = 0;              // That first time that you want to get 0!!
for i = 1 to length of str
    sums[i] <- sums[i-1] + str[i]

Upvotes: 0

NPE
NPE

Reputation: 500267

Here is my advice: stop thinking in "loops" and start thinking in "steps". If the input string has n characters, you have n+1 steps.

Now, ponder the following three questions:

  1. Do you know the solution to the first step (no digits)?
  2. Given the solution to step k, how can you compute the solution to step k+1?
  3. How can you combine questions 1 and 2 to solve your entire problem with just a single loop?

Since this is homework, I'll let you take it from here.

Upvotes: 3

Sandy
Sandy

Reputation: 11687

If i am getting you right

double temp = 0;
for(int i = 0; i < str.Length; i++)  // str is your complete number
{
   temp += Convert.ToDouble(str[i]);
}

Hope it helps.

Upvotes: 0

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