Algorithm

D. Walking Between Houses
time limit per test
2 seconds
memory limit per test
256 megabytes
problem link- https://codeforces.com/contest/1015/problem/D
input
standard input
output
standard output

There are n� houses in a row. They are numbered from 11 to n� in order from left to right. Initially you are in the house 11.

You have to perform k� moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x� to the house y�, the total distance you walked increases by |x−y||�−�| units of distance, where |a||�| is the absolute value of a�. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).

Your goal is to walk exactly s� units of distance in total.

If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k� moves.

Input

The first line of the input contains three integers n�, k�, s� (2≤n≤1092≤�≤109, 1≤k≤2⋅1051≤�≤2⋅105, 1≤s≤10181≤�≤1018) — the number of houses, the number of moves and the total distance you want to walk.

Output

If you cannot perform k� moves with total walking distance equal to s�, print "NO".

Otherwise print "YES" on the first line and then print exactly k� integers hiℎ� (1≤hi≤n1≤ℎ�≤�) on the second line, where hiℎ� is the house you visit on the i�-th move.

For each j� from 11 to k−1�−1 the following condition should be satisfied: hj≠hj+1ℎ�≠ℎ�+1. Also h1≠1ℎ1≠1 should be satisfied.

Examples
input
Copy
10 2 15
output
Copy
YES
10 4
input
Copy
10 9 45
output
Copy
YES
10 1 10 1 2 1 2 1 6
input
Copy
10 9 81
output
Copy
YES
10 1 10 1 10 1 10 1 10
input
Copy
10 9 82
output
Copy
NO

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

long long n, k, s;
vector<int> sol;

int main() {
  scanf("%lld %lld %lld", &n, &k, &s);

  if(s < k || s > k * (n - 1)) {
    puts("NO");
    return 0;
  }

  sol.push_back(1);

  bool ok = true;
  long long cur = 1, nxt[] = {1, n};
  while(s >= abs(cur - nxt[ok])) {
    s -= abs(cur - nxt[ok]);
    cur = nxt[ok];
    ok = !ok;
    sol.push_back(cur);
    --k;
  }

  int step;
  if(s == k) {
    if(!ok) {
      step = -1;
      for(int i = cur - 1; k != 0; --k, i += step) {
        if(i == 0)
          i = 2, step = 1;
        if(i == n + 1)
          i = n - 1, step = -1;
        sol.push_back(i);
      }
    } else {
      step = 1;
      for(int i = cur + 1; k != 0; --k, i += step) {
        if(i == 0)
          i = 2, step = 1;
        if(i == n + 1)
          i = n - 1, step = -1;
        sol.push_back(i);
      }
    }
  } else if(s > k) {
    long long need = s - k + 1;
    s -= need;
    --k;
    if(ok) {
      step = 1;
      sol.push_back(sol.back() + need);
      for(int i = sol.back() + 1; k != 0; --k, i += step) {
        if(i == 0)
          i = 2, step = 1;
        if(i == n + 1)
          i = n - 1, step = -1;
        sol.push_back(i);
      }
    } else {
      step = -1;
      sol.push_back(sol.back() - need);
      for(int i = sol.back() - 1; k != 0; --k, i += step) {
        if(i == 0)
          i = 2, step = 1;
        if(i == n + 1)
          i = n - 1, step = -1;
        sol.push_back(i);
      }
    }
  } else {
    while(s < k) {
      sol.pop_back();
      s += abs(1 - n);
      ++k;
      ok = !ok;
    }
    long long need = s - k + 1;
    s -= need;
    --k;
    if(ok) {
      step = 1;
      sol.push_back(sol.back() + need);
      for(int i = sol.back() + 1; k != 0; --k, i += step) {
        if(i == 0)
          i = 2, step = 1;
        if(i == n + 1)
          i = n - 1, step = -1;
        sol.push_back(i);
      }
    } else {
      step = -1;
      sol.push_back(sol.back() - need);
      for(int i = sol.back() - 1; k != 0; --k, i += step) {
        if(i == 0)
          i = 2, step = 1;
        if(i == n + 1)
          i = n - 1, step = -1;
        sol.push_back(i);
      }
    }
  }

  puts("YES");
  for(int i = 1; i < sol.size(); ++i)
    printf("%d ", sol[i]);
  puts("");

  return 0;
}
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Input

x
–
+
cmd
10 2 15

Output

x
–
+
cmd
YES
10 4
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Demonstration

Codeforces Solution-D. Walking Between Houses-Solution in C, C++, Java, Python,Walking Between Houses,Codeforces Solution

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