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# Alok-nath and His Sanskars CodeChef Solution

## Alok-nath and His Sanskars CodeChef Solution in C++14

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

typedef long long int ll;
typedef vector<long long int> vll;
typedef vector<int> vi;
typedef pair<long long int, long long int> pll;
typedef pair<int, int> pi;

// constraints n<100 W<1000
ll t[100][1002];
// int minDiff = INT_MAX;

bool backtrack(vll& nums, vector<bool>& used, ll tsum, ll currsum, int k, int index) {
if (k == 1) {
return true;
}
if (currsum == tsum) {
return backtrack(nums, used, tsum, 0, k - 1, 0);
}

for (int i = index; i < nums.size(); i++) {
if (used[i] || (currsum + nums[i]) > tsum) {
continue;
}
used[i] = true;
if (backtrack(nums, used, tsum, currsum + nums[i], k, i + 1)) {
return true;
}
used[i] = false;
if (currsum == 0) {
break;
}
}
return false;
}

bool canPartitionKSubsets(vll& nums, int k) {
ll sum = 0;
int n = nums.size();
for (int i = 0; i < n; i++) {
sum += nums[i];
}
if (k>n || sum % k != 0) {
return false;
}
ll tsum = sum / k, currsum = 0;

vector<bool> used(n, false);
sort(nums.begin(), nums.end(), greater<int>());
if (nums[0] > tsum) {
return false;
}
return backtrack(nums, used, tsum, currsum, k, 0);
}

void solve() {
ll n, k;
memset(t, -1, sizeof(t));
cin >> n >> k;
vll arr(n);
ll sum = 0;
for (ll i = 0; i < n; i++) {
cin >> arr[i];
}
if(canPartitionKSubsets(arr, k)){
cout<<"yes";
}
else cout<<"no";

cout<<"\n";
}

int main() {
ios::sync_with_stdio(0);
cin.tie(0);
ll t;
cin >> t;
while (t--) {
solve();
}
return 0;
}

## Alok-nath and His Sanskars CodeChef Solution in PYTH 3

def hasSubsetSum(A,V):
if V==0:
return (True,[])
if len(A)==0:
return (False,[])
if V<A[len(A)-1]:
return hasSubsetSum(A[0:len(A)-1],V)
else:
L=[(len(A)-1)]
TMP=hasSubsetSum(A[0:len(A)-1],V-A[len(A)-1])
if TMP[0]:
TMP[1].append((len(A)-1))
return (True, TMP[1])
else:
return hasSubsetSum(A[0:len(A)-1],V)
#assume sum of A is divisible by K and equals SSUM
def hasPartition(A,SSUM,K):
if K==0:
return True
TMP=hasSubsetSum(A,SSUM)
if TMP[0]:
B=[A[j] for j in range(len(A)) if not(j in TMP[1])]
return hasPartition(B,SSUM,K-1)
else:
return False

T=int(input())
for _ in range(T):
N,K=map(int,input().split())
A=list(map(int,input().split()))
if K>N:
print('no')
continue
SUM=0
for j in range(N):
SUM=SUM+A[j]
if SUM%K !=0:
print("no")
continue
SSUM=SUM//K
#Y=[[False]*N]*SSUM
if(hasPartition(A,SSUM,K)):
print('yes')
else:
print('no')

## Alok-nath and His Sanskars CodeChef Solution in C

#include<stdio.h>
int isSubsetSum(long long *a,int n,int m)
{
if(m==0)
return 1;
if(n==0)
return 0;
if(a[n-1]>m||a[n-1]<0)
return isSubsetSum(a,n-1,m);
if(isSubsetSum(a,n-1,m-a[n-1]))
{
a[n-1]=-1;
return 1;
}
return isSubsetSum(a,n-1,m);
}
int main()
{
int i,t,n,m,cnt,flag;
long long sum,a[22],x;
scanf("%d",&t);
while(t--)
{
sum=0;cnt=0;flag=0;
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
{
scanf("%lld",&a[i]);
sum+=a[i];
}
if(n<m)
flag=0;
else if(sum%m==0)
{
x=sum/m;

for(i=0;i<m;i++)
{
if(isSubsetSum(a,n,x))
{
cnt++;
}
else
{
break;
}
}
if(cnt==m)
flag=1;
else
flag=0;
}
else
{
flag=0;
}
if(flag)
printf("yes\n");
else
printf("no\n");
}
return 0;
}
/* 1541256279063 */

## Alok-nath and His Sanskars CodeChef Solution in JAVA

import java.util.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner scanner = new Scanner(System.in);
int testcases = scanner.nextInt();

for(int t =0; t <testcases; t++){
Long sum = 0L;
int n = scanner.nextInt();
int k = scanner.nextInt();

//sankars
Long[] arr = new Long[n];
for(int i = 0; i<n; i++){
arr[i] = scanner.nextLong();
sum += arr[i];
}

//corners
if(k>n){
System.out.println("no");
continue;
}
if(sum%k!=0){
System.out.println("no");
continue;
}

//k shares
Long share = sum / Long.parseLong(k + "");
int ans = 1;

for(int i=0; i<k; i++){
if(!cal(arr, n-1, share)){
ans = 0;
break;
}
}

System.out.println((ans == 1) ? "yes" : "no");
}

scanner.close();
}

public static boolean cal(Long[] arr, int i, Long sum){
if(sum == 0) return true;
if(i < 0) return false;

Long val = arr[i];
if(val <= sum){
arr[i] = 0L;
boolean isBalanced = cal(arr, i-1, sum-val);
if(isBalanced) return true;

arr[i] = val;
return cal(arr, i-1, sum);
}
else{
return cal(arr, i-1, sum);
}
}
}

## Alok-nath and His Sanskars CodeChef Solution in PYPY 3

import sys
import math

sys.setrecursionlimit(10**6)

def isKPartitionPossibleRec(arr, subsetSum, taken,
subset, K, N, curIdx, limitIdx):
if subsetSum[curIdx] == subset:

""" current index (K - 2) represents (K - 1)
subsets of equal sum last partition will
if (curIdx == K - 2):
return True

# recursive call for next subsetition
return isKPartitionPossibleRec(arr, subsetSum, taken,
subset, K, N, curIdx + 1 , N - 1)

# start from limitIdx and include
# elements into current partition
for i in range(limitIdx, -1, -1):

if (taken[i]):
continue
tmp = subsetSum[curIdx] + arr[i]

# if temp is less than subset, then only
# include the element and call recursively
if (tmp <= subset):

# mark the element and include into
# current partition sum
taken[i] = True
subsetSum[curIdx] += arr[i]
nxt = isKPartitionPossibleRec(arr, subsetSum, taken,
subset, K, N, curIdx, i - 1)

# after recursive call unmark the element and
# remove from subsetition sum
taken[i] = False
subsetSum[curIdx] -= arr[i]
if (nxt):
return True
return False

# Method returns True if arr can be
# partitioned into K subsets with equal sum
def isKPartitionPossible(arr, N, K):

# If K is 1,
# then complete array will be our answer
if (K == 1):
return True

# If total number of partitions are more than N,
# then division is not possible
if (N < K):
return False

# if array sum is not divisible by K then
# we can't divide array into K partitions
sum = 0
for i in range(N):
sum += arr[i]
if (sum % K != 0):
return False

# the sum of each subset should be subset (= sum / K)
subset = sum // K
subsetSum = [0] * K
taken = [0] * N

# Initialize sum of each subset from 0
for i in range(K):
subsetSum[i] = 0

# mark all elements as not taken
for i in range(N):
taken[i] = False

# initialize first subsubset sum as
# last element of array and mark that as taken
subsetSum[0] = arr[N - 1]
taken[N - 1] = True

# call recursive method to check
# K-substitution condition
return isKPartitionPossibleRec(arr, subsetSum, taken,
subset, K, N, 0, N - 1)

try:

T = int(input())
for l in range(T):
n,k = [int(i) for i in input().strip().split(" ")]

path = [int(i) for i in input().strip().split(" ")]
if isKPartitionPossible(path,n,k):
print("yes")
else:
print("no")
except EOFError:
pass

## Alok-nath and His Sanskars CodeChef Solution in PYTH

def donext(cv,V,K,B,p,N):
if D[0] == 0:
for i in range(p,N):
if (B[i] == 0) and (A[i]+cv <= V):
NB = list(B)
NB[i] = 1
if A[i]+cv == V:
if K == 1:
D[0] = 1
else:
k = 0
while NB[k] == 1:
k += 1
# endwhile
NB[k] = 1
donext(A[k],V,K-1,NB,k+1,N)
# endif
else:
donext(A[i]+cv,V,K,NB,i+1,N)
# endif
# endif
# endfor i
# endif
# end fun
t = int(raw_input())
for i in range(t):
st = raw_input().split()
N = int(st[0])
K = int(st[1])
st = raw_input().split()
A = []
tot = 0
for x in st:
n = int(x)
A.append(n)
tot += n
# endfor x
A.sort()
A.reverse()
V = tot/K
if (tot%K > 0) or (A[0] > V) or ((tot == 0) and (N < K)):
r = 0
else:
if K == 1:
r = 1
else:
while (N > 0) and (A[0] == V):
N -= 1
K -= 1
del(A[0])
# endwhile
if K < 2:
r = 1
else:
B = [0 for x in range(N)]
B[0] = 1
K -= 1
D = [0]
donext(A[0],V,K,B,1,N)
r = D[0]
# endif
# endif
# endif
if r == 0:
print 'no'
else:
print 'yes'
# endif
# endfor i

## Alok-nath and His Sanskars CodeChef Solution in C#

using System;
using System.Linq;
class GFG
{
static bool issubsetsum(long[] a, long n, long sum)
{
if (sum == 0)
return true;
if (n == 0)
return false;
if (a[n - 1] > sum || a[n - 1] < 0)
return issubsetsum(a, n - 1, sum);
if (issubsetsum(a, n - 1, sum - a[n - 1]))
{
a[n - 1] = -1;
return true;
}
return issubsetsum(a, n - 1, sum);
}

// Driver Code
public static void Main()
{

for (int f = 0; f < t; f++)
{
var n = int.Parse(input1[0]);
var k = int.Parse(input1[1]);

long[] arr = new long[n];

for (int i = 0; i < input2.Length; i++)
{
if (!string.IsNullOrEmpty(input2[i]))
arr[i] = int.Parse(input2[i]);
}
var ans = canPartitionKSubsets(arr, k, n) == true ? "yes" : "no";
Console.Write(ans);
Console.WriteLine();
}
}

private static bool canPartitionKSubsets(long[] arr, long k, long n)
{
var sum = arr.Sum();

if (sum % k != 0 || n < k)
{
return false;
}

int count = 0;
long p = sum / k;
for (int i = 0; i < k; i++)
{
if (issubsetsum(arr, n, p))
count++;
else break;
}
if (count == k)
return true;
return false;
}
}
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