均分纸牌类型的问题和延申

经典题：均分纸牌p1031

线性传递，相邻之间可传（首尾不相邻），一次操作可传递若干张牌，求最小操作次数。

对每个a[i]，先减去平均值，目标是将ai序列全部变成0，依次向右传递，推导答案易得,

g[i]为前缀和。

#include <bits/stdc++.h>
using namespace std;
int n, a[105], sum;
int main() {
	cin >> n;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i];
		sum += a[i];
	}
	sum /= n;
	int ans = 0;
	for (int i = 1; i <= n; i ++) {
		a[i] -= sum;//减去平均数 
		if (a[i]) {
			ans ++;
			a[i + 1] += a[i];
		}
	}
	cout << ans;
}

环形均分纸牌，p2512糖果传递。

枚举位置k，从k断开转化成问题1，重新统计前缀和，观察最后的表达式，即为货仓选址问题，找中位数即可。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
long long n, a[N], s[N], av, ans;
int main() {
	cin >> n;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i];
		av += a[i];
	} 
	av /= n;
	for (int i = 1; i <= n; i ++) {
		a[i] -= av;
		s[i] = s[i - 1] + a[i];
	}
	sort(s + 1, s + n + 1);
	int k = (n + 1) / 2;
	for (int i = 1; i <= n; i ++)
		ans += abs(s[i] - s[k]);
	cout << ans;
}

七夕祭

行和列互不影响，做两次环状均分纸牌即可。（注意开longlong）

#include <bits/stdc++.h>//环形均分纸牌 (注意开longlong)
#define int long long
using namespace std;
const int N = 1e5 + 10;
const int M = 0x3f3f3f3f;
int n, m, t, r[N], c[N];
int ans1 = M, ans2 = M; 
signed main() {
	cin >> n >> m >> t;
	for (int i = 1; i <= t; i ++) {
		int x, y;
		cin >> x >> y;
		r[x] ++, c[y] ++;
	}
	if (!(t % n)) {
		ans1 = 0;
		for (int i = 1; i <= n; i ++) {
			r[i] -= (t / n);
			r[i] += r[i - 1];//前缀和 
		}
		sort(r + 1, r + n + 1);
		int k = (n + 1) / 2;
		for (int i = 1; i <= n; i ++)
			ans1 += abs(r[i] - r[k]);
	}
	if (!(t % m)) {
		ans2 = 0;
		for (int i = 1; i <= m; i ++) {
			c[i] -= (t / m);
			c[i] += c[i - 1];
		}
		sort(c + 1, c + m + 1);
		int k = (m + 1) / 2;
		for (int i = 1; i <= m; i ++)
			ans2 += abs(c[i] - c[k]);
	}
	if (ans1 != M && ans2 != M) printf("both %lld", ans1 + ans2);
	if (ans1 != M && ans2 == M) printf("row %lld", ans1);
	if (ans1 == M && ans2 != M) printf("column %lld", ans2);
	if (ans1 == M && ans2 == M) printf("impossible");
	return 0;
}