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@ -0,0 +1,36 @@
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 120;
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int a[N], b[N];
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const int INF = 0x3f3f3f3f;
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/*
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3
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2 6 7
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*/
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int n, sum, avg;
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int main() {
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cin >> n;
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for (int i = 0; i < n; i++) {
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cin >> a[i];
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sum += a[i];
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a[i + n] = a[i]; // 破环成链, 0 1 2-->a[3]=a[0],a[4]=a[1],a[5]=a[2]
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}
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avg = sum / n; // 平均数
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int mi = INF; // 最少次数,给初始值最大
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for (int i = 0; i < n; i++) { // 按每个石子堆为起点
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for (int j = 0; j < n; j++) b[j] = a[i + j]; // 从原始的破环成链数组中拷贝出来现在的临时数组
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int cnt = 0;
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for (int j = 0; j < n - 1; j++) {
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cnt += abs(b[j] - avg); // 需要调整几个石子到下一个石子堆
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b[j + 1] += b[j] - avg; // 下一个石子堆接着这些石子
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b[j] = avg; // j调整为平均值,其实这句没有发挥作用,可省略
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}
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// 以第i个石子为起点的讨论结束,判断是不是我们需要的最优解
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mi = min(mi, cnt);
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}
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cout << mi << endl;
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return 0;
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}
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