There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1
.
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≤109, 1≤k≤2⋅105, 1≤s≤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 hi (1≤hi≤n) on the second line, where hi is the house you visit on the i
-th move.
For each j
from 1 to k−1 the following condition should be satisfied: hj≠hj+1. Also h1≠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
题意：现在有n个房子排成一列，编号为1~n，起初你在第1个房子里，现在你要进行k次移动，每次移动一都可以从一个房子i移动到另外一个其他的房子j里（i != j），移动的距离为|j - i|。问你进过k次移动后，移动的总和可以刚好是s吗？若可以则输出YES并依次输出每次到达的房子的编号，否则输出NO。
题解：每次至h少移动一个单位的距离，至多移动n-1个单位长度，所以要满足 k<=s&&k*(n-1)>=s 然后我们要将首先要将k的值与s值尽快相等 所以 rm=（s-k,n-1)
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#define N 0x3f3f3f3f
#define ll long long
using namespace std;
int main()
{
ios::sync_with_stdio(false);
ll n,k,s;
cin>>n>>k>>s;
if(k>s||k*(n-1)<s)
{
cout<<"NO"<<endl;
return 0;
}
cout<<"YES"<<endl;
int st=1;
while(k--)
{
int rm=min(s-k,n-1);
s-=rm;
if(rm+st<=n)
{
st=st+rm;
}
else
st=st-rm;
cout<<st<<" ";
}
cout<<endl;
}