【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=;
     if(k>s||k*(n-)<s)
     {
         printf("NO\n");
         return ;
     }
     printf("YES\n");
     while(k)
     {
         ll t=min(n-,s-(k-));
         if(now-t>) now-=t;
          else now+=t;
         printf("%lld ",now);
         s-=t;
         k--;
     }
      return ;
 }