srm 554

欢迎点此阅读QvQ

250

---

Description

给高度为h1的棍子r1个，h2的r2个，轮流放，问最多放出多少种不同的高度

Solution

考虑h1是否等于h2，再讨论即可

Code

#include <bits/stdc++.h>using namespace std;#define pb push_back#define mp make_pair#define F first#define S secondtypedef long long LL;typedef pair<int, int> pii;struct TheBrickTowerEasyDivOne {    int find(int r1, int h1, int r2, int h2) {        if (h1 == h2) {            return 2 * min(r1, r2) + (r1 == r2 ? 0 : 1);        }        else {            return 2 * min(r1, r2) + (r1 > r2) + min(r1, r2) + (r2 > r1);        }    }};

500

---

Description

将n个数重新排列，使得相邻两个数最大值之和尽量小，同时字典序最小。

Solution

n个数算n−1的最大值，很显然尽量小的情况是除了最小值所有值都被计算过一次。观察可得，序列应该是先递减再递增的，按字典序排即可。

Code

#include <bits/stdc++.h>using namespace std;#define pb push_back#define mp make_pair#define F first#define S secondtypedef long long LL;typedef pair<int, int> pii;const int N = 50;vector<pii> s;vector<int> ans;bool vis[N];struct TheBrickTowerMediumDivOne {    vector <int> find(vector <int> a) {        int n = a.size();        ans.pb(0), vis[0] = 1;        for (int i = 1; i < n; ++i) {            int p = -1;            for (int j = 0; j < n; ++j) {                if (!vis[j] && a[j] <= a[ans[i - 1]]) {                    p = j;                    break;                }            }            if (~p) ans.pb(p), vis[p] = 1;            else break;        }        for (int i = 0; i < n; ++i)            if (!vis[i])    s.pb(mp(a[i], i));        sort(s.begin(), s.end());        for (int i = 0; i < s.size(); ++i)  ans.pb(s[i].S);        return ans;    }};