Codeforces Round #501 (Div. 3) D. Walking Between Houses (思维,构造)

• 题意:一共有\(n\)个房子,你需要访问\(k\)次,每次访问的距离是\(|x-y|\),每次都不能停留,问是否能使访问的总距离为\(s\),若能,输出\(YES\)和每次访问的房屋,反正输出\(NO\).

• 题解:最优解一定是让每次访问的距离为\(s/k\),然后将余数\(s\ mod\ k\)平均分配到前s%k的房屋中,之后每次访问\(s/k\)即可,也就是构造一个类似于\(x\ 1\ x\ 1\ x\ 1\)这样的一个序列,但因为有余数,所以要修改.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  
  ll n,k,s;
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
      cin>>n>>k>>s;
      if(k*(n-1)<s || k>s) puts("NO");
      else{
          puts("YES");
          ll num=s/k;
          ll rest=s%k;
          ll tmp=1;
          for(ll i=1;i<=rest;++i){
              if(tmp==1){
                  tmp=num+2;
                  printf("%lld ",tmp);
              }
              else{
                  tmp=1;
                  printf("1 ");
              }
          }
          for(ll i=rest+1;i<=k;++i){
              if(tmp<=num) tmp+=num;
              else tmp-=num;
              printf("%lld ",tmp);
          }
  
      }
  
      return 0;
  }