Codechef Minimize Digit Sum Solution

Minimize Digit Sum Solution September Long Challenge 2021

Let f(n,B)f(n,B) be the sum of digits of the integer nn when written in base BB.

Given QQ queries, each consisting of three integers n,ln,l and rr. Find the value of BB corresponding to which f(n,B)f(n,B) is minimum for all l≤B≤rl≤B≤r. If there are multiple such values, you can print any of them.

Also See: September Long Challenge 2021 Solutions

Input Format

  • The first line contains in single integer QQ, the number of queries
  • Each of the next Q lines contain three space separated integers n,ln,l and rr respectively.

Output Format

  • For each query (n l r), print the value of base BB which lies within [l,r][l,r] such that f(n,B)f(n,B) is minimum.

Constraints

  • 1≤Q≤1031≤Q≤103
  • 2≤n≤1092≤n≤109
  • 2≤l≤r≤1092≤l≤r≤109

Subtasks

Subtask #1 (50 points): original constraints

This problem is worth a total of 50 points and is meant to be complementary to the problem “MNDIGSM2” (also worth 50 points) which is very similar to this problem, but has slightly different constraints.

Sample Input 1 

3
216 2 7
256 2 4
31 3 5

Sample Output 1 

6
2
5

Explanation

Test case 11: We have f(216,2)=f(216,3)=4f(216,2)=f(216,3)=4, f(216,4)=6f(216,4)=6, f(216,5)=8f(216,5)=8, f(216,6)=1f(216,6)=1 and finally f(216,7)=12f(216,7)=12. Clearly the minimum is obtained when B=6B=6.

Test case 22: Note that f(256,2)=f(256,4)f(256,2)=f(256,4) = 22, therefore both the answers 22 and 44 will be considered correct.

Test case 33: f(31,3)=f(31,5)=3f(31,3)=f(31,5)=3 and f(31,4)=7f(31,4)=7, therefore both the answers 33 and 55 will be considered correct.

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Solution

Program C:

#include <limits.h>
#include <stdio.h>

int main() {
    int q, n, l, r, inputNum, sum, min, sum2;
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
        scanf("%d%d%d", &n, &l, &r);
        if (n < l) {
            printf("%d\n", l);
            continue;
        }
        if (n >= l && n <= r) {
            printf("%d\n", n);
            continue;
        }
        sum2 = INT_MAX;
        min = 0;
        /* optimize 2 digit case with high base */
        while (l < r && n / r < r) {
            int d1 = n / r;
            int d2 = n % r;
            if (sum2 > d1 + d2) {
                sum2 = d1 + d2;
                min = r;
            }
            /* skip all lower bases with same d1 */
            r = n / (d1 + 1);
        }
        while (l <= r) {
            inputNum = n;
            sum = 0;
            for (;;) {
                if (inputNum < l) {
                    sum += inputNum;
                    if (sum < sum2) {
                        min = l;
                        sum2 = sum;
                    }
                    break;
                }
                sum += inputNum % l;
                inputNum /= l;
                if (sum >= sum2)
                    break;
            }
            l++;
        }
        printf("%d\n", min);
    }
    return 0;
}

Program C++:

#include <bits/stdc++.h>
using namespace std;

class Test{
    public:
    int solve(){
    int q, n, l, r, enterInt, sum, min, sum2;
    cin >> q;
    for (int i = 0; i < q; i++) {
        cin >> n >> l >> r;
        
        if (n >= l && n <= r) {
            cout << n << "\n";
            continue;
        }
        
        if (n < l) {
            cout << l << "\n";
            continue;
        }
    
        sum2 = INT_MAX;
        min = 0;
       
        while (l < r && n / r < r) {
            int temp1 = n / r, temp2 = n % r;
            if (sum2 > temp1 + temp2) {
                sum2 = temp1 + temp2;
                min = r;
            }
            
            r = n / (temp1 + 1);
        }
        while (l <= r) {
            enterInt = n;
            sum = 0;
            for (;;) {
                if (enterInt < l) {
                    sum += enterInt;
                    if (sum < sum2) {
                        min = l;
                        sum2 = sum;
                    }
                    break;
                }
                sum += enterInt % l;
                enterInt /= l;
                if (sum >= sum2)
                    break;
            }
            l++;
        }
       cout << min << "\n";
    }
    }
    
};

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    
    Test tt;
    
    tt.solve();
    return 0;
}

Program Java:

import java.util.*;
import java.lang.*;
import java.io.*;

class Codechef
{
    public static int func(int n,int b)
    {
        int sum=0;
        for(int i=n;i!=0;i/=b)
        {
            sum+=i%b;
        }
        return sum;
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(br.readLine());
		while(t-->0)
		{
		    String [] s=br.readLine().split(" ");
		    int n=Integer.parseInt(s[0]);
		    int l=Integer.parseInt(s[1]);
		    int r=Integer.parseInt(s[2]);
		    int min=999999999,ind=-1;
		    if(r>n && l>=n)
		    {
		        System.out.println(l);
		        continue;
		    }
		    if(r>=n)
		    {
		        System.out.println(n);
		        continue;
		    }
		    while(l<r && n/r < r)
		    {
		        int t1=n/r;
		        int t2=n%r;
		        if(min>(t1+t2))
		        {
		        min=t1+t2;
		        ind=r;
		        }
		        r=n/(t1+1);
		    }
		    while(l<=r)
		    {
		        int e=n;
		        int sum=0;
		        for(;;)
		        {
		            if(e<l){
		                sum+=e;
		                if(sum<min)
		                {
		                    min=sum;
		                    ind=l;
		                }
		                break;
		            }
		            sum+=e%l;
		            e/=l;
		            if(sum>=min)
		            break;
		        }
		        l++;
		    }
		   
		    System.out.println(ind);
		}
	
	}
}

Program Python:

def ff(m,j):
    s=0
    while(m>0):
        s+=m%j
        m=m//j
    return s
t=int(input())
for item in range(t):
    n,l,r=map(int,input().split())
    s=0
    mn=n
    index=0
    if r>n:
        r=n
    for j in range(l,r+1):
        s=ff(n,j)
        if mn > s:
            mn=s
            index=j
    print(index)

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