Bonetrousle

                            Solution in C :

In C++ :





#include <bits/stdc++.h>

#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)

#define LL long long
#define Ldouble long double
#define PI 3.1415926535897932384626

#define PII pair<int,int>
#define PLL pair<LL,LL>
#define mp make_pair
#define fi first
#define se second

using namespace std;

int t, n;
LL ttl, sto, s[100005];

void solve(){
	LL bs = 1LL * n * (n + 1) / 2, tar;
	if(ttl < bs){
		printf("-1\n");
		return;
	}
	FI(i, 1, n) s[i] = i;
	
	tar = ttl - bs;
	
	FD(i, n, 1){
		LL incre = max(0LL, min(tar, sto - (n - i) - s[i]));
		s[i] += incre;
		tar -= incre;
	}
	
	if(tar > 0){
		printf("-1\n");
		return;
	}
	
	FI(i, 1, n) printf("%lld%c", s[i], i == n ? '\n':' ');
	return;
}

int main(){
	scanf("%d", &t);
	while(t--){
		scanf("%lld %lld %d", &ttl, &sto, &n);
		solve();
	}
	return 0;
}









In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int z = 0; z < t; z++) {
            long n = sc.nextLong();
            long k = sc.nextLong();
            int b = sc.nextInt();
            long diff = k-b;
            long[] selected = new long[b];
            long sum = 0;
            for (int i = 0; i < b; i++) {
                selected[i] = i+1;
                sum += i+1;
            }
            if (sum > n) {
                System.out.println(-1);
                continue;
            }
            long next = k+1;
            for (int i = b-1; i >= 0; i--) {
                sum += diff;
                if (sum < n) {
                    selected[i] = --next;
                } else {
                    sum -= diff;
                    selected[i] = n-sum+selected[i];
                    sum = n;
                    break;
                }
            }
            if (sum < n) {
                System.out.println(-1);
                continue;
            }
            StringBuilder print = new StringBuilder();
            for (int i = 0; i < b; i++) {
                if (i > 0)
                    print.append(" ");
                print.append(selected[i]);
            }
            System.out.println(print);
        }
    }
}








In C :





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

int main() {
    int t;
    long long int n, k, b;
    long long int llt, llt2, llt3;
    long long int i;
    int has_printed;
    
    scanf("%d", &t);
    
    while (t--) {
        scanf("%lld %lld %lld", &n, &k, &b);
        
        llt = (b * (b + 1)) >> 1;
        if (llt > n) {
            printf("-1\n");
            continue;
        }
        llt2 = n - llt;
        
        // divide llt2 with b
        if ((llt2 % b) == 0) {
            llt3 = llt2 / b;
            has_printed = 0;
            // if it is within limits, then we are good
            if ((b + llt3) <= k) {
                for (i = llt3 + 1; i <= llt3 + b; i++) {
                    if (has_printed) {
                        printf(" ");
                    }
                    printf("%lld", i);
                    has_printed = 1;
                }
                printf("\n");
            } else {
                printf("-1\n");
            }
        } else {
            llt3 = llt2 / b;
            // we are starting at lower end
            if ((llt3 + b) >= k) {
                printf("-1\n");
            } else {
                llt = llt2 % b;
                has_printed = 0;
                for (i = llt3 + 1; i <= (llt3 + b - llt); i++) {
                    if (has_printed) {
                        printf(" ");
                    }
                    printf("%lld", i);
                    has_printed = 1;
                }
                for (; i <= (llt3 + b); i++) {
                    printf(" ");
                    printf("%lld", i + 1);
                }
                printf("\n");
            }
        }
    }
    
    return 0;
}








In Python3 :





def sum(l, r):
    return r * (r + 1) // 2 - l * (l - 1) // 2

t = int(input())
while t > 0:
    t -= 1
    
    n, k, b = map(int, input().split())

    if sum(1, b) > n or sum(k - b + 1, k) < n: print(-1)
    else:
        a = []
        while n > 0:
            x = min(k, n - sum(1, b - 1))
            a += [x]

            k = x - 1
            b -= 1
            n -= x
        
        print(' '.join(map(str, a)))