For odd values of $$$n$$$ we can re-express string $$$s$$$ as

Note that $$$s_1+s_2+...+s_{\lfloor \frac{n}{2} \rfloor}=s_{\lceil \frac{n}{2} \rceil+1}+....+s_n$$$ , so we can assume the sum as $$$f$$$.

And let's denote the median digit by $$$h$$$ , thus the sum is $$$2f+h$$$ , it can be shown that $$$h$$$ can be possibly any digit , thus we can construct all sums which lies between minimum and maximum inclusive .

Now Let's do the same for even values of $$$n$$$ we can re-express $$$s$$$ as

Using the symmetric property we can see $$$s_{\lfloor \frac{n}{2} \rfloor}=s_{\lfloor \frac{n}{2} \rfloor+1}$$$ , if we denote one digit by $$$h$$$ thus the sum of medians is $$$2h$$$ and if we denote the left and right segments by $$$f$$$ as we did for odd $$$n$$$ then we have $$$m=2f+2h=2(f+h)$$$ thus for even $$$n$$$ we have an additional condition that sum must be even

Thus Total Complexity for each test case is $$$\mathcal{O}(1)$$$ time .

for _ in range(int(input())): 
  n,m= map(int, input().split())
  if(m >= n and m <= 9 * n and (n % 2 != 0 or m % 2 == 0)):
    print("YES")
  else:
    print("NO")

If $$$x<z$$$, the answer is obviously $$$-1$$$. We only consider the case of $$$x \ge z$$$.

Maximum number of interviews per day can happen is the number of distinct interviews can happen , which is the number of interviewers times the capacity of each interviewer integer divided by the needed interviewers in one interview , Formally

The construction is simple:

consider a sequence of length $$$x \cdot y$$$ — $$$[1,2,\ldots,x,1,2,\ldots,x,\lodts]$$$ , and simply output consecutive segments of length $$$z$$$ in order.

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <random>
#include <chrono>
 
using namespace std;
 
#define _int64 long long
#define mo 998244353


int main()
{
	int i,j,k,n,l,t,m,x,y,o,b1,z,ans,now;
	scanf("%d",&t);
	for (l=0;l<t;l++)
	{
		scanf("%d%d%d",&x,&y,&z);
		if (z>x)
		{
			printf("0\n");
			continue;
		}
		ans=x*y/z;
		printf("%d\n",ans);
		now=1;
		for (i=0;i<ans;i++)
		{
			for (j=0;j<z;j++)
			{
				printf("%d ",now);
				now++;
				if (now>x) now=1;
			}
			printf("\n");
		}
	}
}

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