D. Walking Between Houses

最新推荐文章于 2021-01-02 22:33:20 发布

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题目链接 D. Walking Between Houses

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

You have to perform kk 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 xx to the house yy , the total distance you walked increases by |x−y||x−y| units of distance, where |a||a| is the absolute value of aa . 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 ss 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 kk moves.

Input

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

Output

If you cannot perform kk moves with total walking distance equal to ss , print "NO".

Otherwise print "YES" on the first line and then print exactly kk integers hihi (1≤hi≤n1≤hi≤n ) on the second line, where hihi is the house you visit on the ii -th move.

For each jj from 11 to k−1k−1 the following condition should be satisfied: hj≠hj+1hj≠hj+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

思路:刚开始队友让我用dfs写，写了一会儿发现不对呀，能直接暴力的呀，结果一交，wa了，TAT

wa代码:没有考虑10 52 449 这种情况

#include<iostream>
#include<cstdio>//
#include<cmath>
using namespace std;
typedef long long int ll;//199979000100000
                         //200000000000000
ll s,y,sum,c,average,yu,n,m;
int main()
{
	while(cin>>n>>m>>c){
		s=(n-1)*m;//2 
		if(s<c||m>c) printf("NO\n");
		else{
			printf("YES\n"); 
			average=c/m;//10 2 15 得到7 
			yu=c-average*(m-1);//15-7==8
			printf("%d",yu+1);
			ll aa=yu+1-average,bb=yu+1;//aa==9-7=2,bb=8+1
			for(ll i=1;i<m;i++){
				if(i&1) cout<<" "<<aa;
				else cout<<" "<<bb;
			}
			printf("\n");
		}
	}
	return 0;	
}

实在是没想出来，参考了大神代码，点击

AC代码:

#include <bits/stdc++.h>
using namespace std;
int main(){
    long long n, k, s;
    cin >> n >> k >> s;
    if(s<k || s>(n-1)*k){
        cout << "NO" << endl;
        return 0;
    }
    cout << "YES" << endl;
    int st=1;
    while(k--){//满足 （k-1 <= s-x） && （x <= n-1）
        int rm=min(s-k, n-1);// 
        s-=rm;
        if(st+rm<=n) st+=rm;
        else st-=rm;
        cout << st << " ";

    }
    cout << endl;
}