【CF1015D】Walking Between Houses（构造，贪心）

题意：从1开始走，最多走到n，走k步，总长度为n，不能停留在原地，不能走出1-n，问是否有一组方案，若有则输出

n<=1e9，k<=2e5，s<=1e18

思路：无解的情况分为两种：总长度太大，步数太多

每次贪心从1走到n或从n走到1，但要注意给剩下的步数留出空间

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<int> VI;
#define fi first
#define se second 
#define MP make_pair
#define N  100010
#define M  200
#define MOD 998244353
#define eps 1e-8 
#define pi acos(-1)



int main()
{
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    ll n,k,s;
    scanf("%lld%lld%lld",&n,&k,&s);
    ll now=1;
    if(k>s||k*(n-1)<s)
    {
        printf("NO
");
        return 0;
    }
    printf("YES
");
    while(k)
    {
        ll t=min(n-1,s-(k-1));
        if(now-t>0) now-=t;
         else now+=t;
        printf("%lld ",now);
        s-=t;
        k--;
    }
     return 0;
}