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63 lines
1.5 KiB
63 lines
1.5 KiB
1 year ago
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#include <bits/stdc++.h>
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using namespace std;
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/*
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14
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3 2 2 3 2 2 3 2 2 2 3 3 3 3
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*/
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const int N = 110;
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int n;
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int a[N];
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int res; // 执行次数
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int st[10]; // 哪些数字使用过了
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int b[10];
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// 结构体
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struct Node {
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int c;
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int count;
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int st, ed;
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bool const operator<(const &Node t) {
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if (count != t.count) return count > t.count;
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if (st < t.st && ed > t.ed) return false;
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return true;
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}
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};
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// 找出未处理的数字中符合条件的那个
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Node find() {
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memset(b, 0, sizeof b);
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vector<Node> q;
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for (int i = 1; i <= n; i++)
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if (!st[a[i]]) b[a[i]]++; // 个数
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for (int i = 1; i <= 9; i++) {
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int start, end;
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for (int j = 1; j <= n; j++)
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if (a[j] == i) {
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start = j;
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break;
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}
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for (int j = n; j >= 1; j--)
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if (a[j] == i) {
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end = j;
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break;
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}
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q.push_back({i, b[i], start, end});
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}
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sort(q.begin(), q.end());
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return q[0];
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}
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int main() {
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> a[i]; // 密码数组
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for (res = 0;; res++) {
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// 找出没有记录过的数字，并且，它的数量最多，如果存在数量一样多的多个数字，就比较谁的区间大，大的优先，如果无法比较谁的区间大，就数小的优先
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Node x = find();
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if (x.count == 0) break;
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}
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cout << res << endl;
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return 0;
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}
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