构造一个长度为$n$的序列$a$，满足$a_i\ge 0$，$\sum _{i=1}^n{a}_i=s$，对于任意的$i\in [1,n-1]$，都有$a_i-a_{i+1}=kor{a}_{i+1}-ai=1$，其中$n,s,k$是给定的数。

假定所有的$a_i$都满足$a_{i+1}-a_i=1$，则对于一个给定的$a_1$，有$\sum _{i=1}^n{a}_i=n\times a_1+\frac{n(n-1)}{2}$，考虑$a_i-a_{i+1}=k$，$\sum _{i=1}^n{a}_i=n\times a_1+\frac{n(n-1)}{2}-x(k+1)$。

考虑对所有数$(\mathrm{m}\mathrm{o}\mathrm{d}k+1)$下找到一个解，满足$\sum _{i=1}^n{a}_i(\mathrm{m}\mathrm{o}\mathrm{d}k+1)\equiv s(\mathrm{m}\mathrm{o}\mathrm{d}k+1)$。

考虑同余下的一个最小解，$a_1,a_1+1,\dots ,k,0,1,2,\dots ,k,0,1,2,\dots$，如果满足$sum\le sandsum(\mathrm{m}\mathrm{o}\mathrm{d}k+1)\equiv s(\mathrm{m}\mathrm{o}\mathrm{d}k+1)$，则说明找到一组解。

获得最小解后，凑得答案，只需每次对最小得加$k+1$，这样可以保证变动之后这个数比其前方的大$1$，后面的比其小$k$，满足要求。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;int n, k, flag;long long s, a[N];vector<int> pos[N];inline int add(int x, int y) {return x + y < k + 1 ? x + y : x + y - k - 1;
}void solve() {for (int i = 1; i <= n; i++) {a[i] = add(a[i - 1], i != 1);pos[a[i]].push_back(i);}int x = a[1], len = min(k - x + 1, n), last = n - len;long long sum = 1ll * (x + x + len - 1) * len / 2;sum += 1ll * (last / (k + 1)) * (0 + k) * (k + 1) / 2;last %= k + 1;sum += 1ll * (0 + 0 + last - 1) * last / 2;s -= sum;sum = s / (k + 1);long long tot = sum / n, more = sum % n;for (int i = 1; i <= n; i++) {a[i] += 1ll * (k + 1) * tot;}for (int i = 0; i <= k; i++) {for (auto it : pos[i]) {if (more) {a[it] += k + 1;more--;}}}for (int i = 1; i <= n; i++) {printf("%lld%c", a[i], i == n ? '\n' : ' ');}
}bool judge(int x) {int len = min(k - x + 1, n), last = n - len;long long sum = 1ll * (x + x + len - 1) * len / 2;sum += 1ll * (last / (k + 1)) * (0 + k) * (k + 1) / 2;last %= k + 1;sum += 1ll * (0 + 0 + last - 1) * last / 2;return sum <= s && s % (k + 1) == sum % (k + 1);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);cin >> n >> k >> s;for (int i = 0; i <= k; i++) {if (judge(i)) {flag = 1;a[0] = i;solve();break;}}if (!flag) {puts("-1");}return 0;
}