Count numbers with difference between number and its digit sum greater than specific value in C++

We are provided two numbers N which defines a range [1,N] and D which is a difference.The goal is to find all the numbers in the range [1,N] such that the [ number - (sum of its digits) ] > D. We will do this by traversing numbers from 1 to N and for each number we will calculate its digit sum using a while loop. Check if the number and calculated digit sum has a difference more than D.Let’s understand with examples.Input <pre class="result notranslate">N=15 D=5</pre>Output <pre class="result notranslate">Numbers such that difference b/w no. and its digit sum greater than value D: 6</pre>Explanation <pre class="result notranslate">Numbers 10, 11, 12, 13, 14, 15 satisfy the condition. ( 10-1, 11-2, 12-3, 13-4, 14-5, 15-6 ) all differences are 9 which is greater than 5.</pre>Input <pre class="result notranslate">N=20 D=10</pre>Output <pre class="result notranslate">Only 20 satisfies the condition. 20-2=18 > 10.</pre>Explanation <pre class="result notranslate">This is list of numbers divisible by all non-zero digits : 100 101 102 104 105 110 111 112 115 120 122 124 126 128 132 135 140 144 150 155 162 168 175 184 200</pre><h2>Approach used in the below program is as follows</h2><ul class="list"><li>We take integers N and D.</li><li>Function digitSum(int n, int d) takes variables N, D and returns the count of numbers with (num-digitsum) >d.</li><li>Take the initial variable count as 0 for such numbers.</li><li>Take variable digsum as 0</li><li>Traverse range of numbers using for loop. i=1 to i=n</li><li>Now for each number num=i, using while loop check if number is >0.</li><li>calculate digsum+=num%10. Reduce num=num/10 to add the next digit.</li><li>At the end of the while, check if ( i - digsum > d ). If true increment count.</li><li>At the end of all loops count will have a total number which satisfies the condition.</li><li>Return the count as result.</li></ul><h2>Example</h2><a class="demo" href="http://tpcg.io/OkkygSu2" rel="nofollow" target="_blank"> Live Demo</a><pre class="prettyprint notranslate">#include <bits/stdc++.h> using namespace std; int digitSum(int n, int d){    int count = 0;    int digsum = 0;    for (int i = 1; i <= n; i++){       int num=i;       digsum=0;       while(num>0){          digsum+=num%10; //sum of digits          num=num/10;       }       if(i-digsum>d) //original number is i {          count++;          //cout<<i<<" ";       }    }    return count; } int main(){    int N = 20;    int D = 8;    cout <<"Numbers such that difference between number and its digit sum greater than specific value: "<<digitSum(N,D);    return 0; }</pre><h2>Output</h2>If we run the above code it will generate the following output −<pre class="result notranslate">Numbers such that difference between number and its digit sum greater than specific value: 11</pre>

Sunidhi Bansal

Updated on 31-Oct-2020 04:53:26

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