这个问题类型以前没学过，记一下

P1031 [NOIP2002 提高组] 均分纸牌
https://www.luogu.com.cn/problem/P1031

显然最后均分完纸牌后，各个堆的纸牌数量均为平均数avg
考虑一个堆一个堆处理
如果a[i] == avg，则这个堆已经满足 不用处理
否则，则需要将(a[i] - avg)加到a[i + 1]上（表示把牌放到i+1那个堆或者从i+1那个堆拿牌），同时数量增加
这里移动若干张纸牌就算1次 如果移动x张就算x次操作的话，就能发现其实就是求|a[i] - avg|的前缀和

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int n, a[110], d[110], sum, ans, avg;
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  avg = sum / n;
  for (int i = 1; i < n; i++)
  {
    if (a[i] == avg)
      continue;
    a[i + 1] += (a[i] - avg);
    ans++;
  }
  cout << ans;
  return 0;
}

P2512 [HAOI2008] 糖果传递
https://www.luogu.com.cn/problem/P2512
环形均分纸牌问题

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 1e6 + 10;
int n;
LL a[N], c[N], sum, avg;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  avg = sum / n;

  for (int i = 1; i <= n; i++) {
    c[i] = c[i - 1] + a[i] - avg;
  }

  sort(c + 1, c + n + 1);

  LL t = c[(n + 1) / 2], ans = 0;

  for (int i = 1; i <= n; i++)
  {
    ans += abs(t - c[i]);
  }

  cout << ans;

  return 0;
}

P10453 七夕祭
https://www.luogu.com.cn/problem/P10453
读一下发现，行和列处理互不影响
相当于对行和列分别进行均分纸牌

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 1e5 + 10;
int n, m, t, r[N], c[N];
LL d[N], ans, sum, avg;
int main() {
  cin >> n >> m >> t;
  if (t % n != 0 && t % m != 0) {
    cout << "impossible";
    return 0;
  }

  for (int i = 1; i <= t; i++) {
    int x, y;
    cin >> x >> y;
    r[x]++, c[y]++;
  }


  if (t % n == 0) // 对行进行环形均分纸牌
  { 
    sum = 0, avg = 0;
    for (int i = 1; i <= n; i++)
      sum += r[i];
    avg = sum / n;
    for (int i = 1; i <= n; i++)
      d[i] = d[i - 1] + avg - r[i];
    sort(d + 1, d + n + 1);
    for (int i = 1; i <= n; i++)
      ans += abs(d[(n + 1) / 2] - d[i]);
  }

  if (t % m == 0) // 对列进行环形均分纸牌
  {
    sum = 0, avg = 0;
    for (int i = 1; i <= m; i++)
      sum += c[i];
    avg = sum / m;
    for (int i = 1; i <= m; i++)
      d[i] = d[i - 1] + avg - c[i];
    sort(d + 1, d + m + 1);
    for (int i = 1; i <= m; i++)
      ans += abs(d[(m + 1) / 2] - d[i]);
  }

  if (t % m != 0)
    cout << "row ";
  else if (t % n != 0)
    cout << "column ";
  else
    cout << "both ";
  cout << ans;









  return 0;
}