1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 signed main(void)
 5 {
 6     int n; cin >> n;
 7     string s; cin >> s;
 8     
 9     int t = 0;
10     
11     for (auto i : s)
12         if (i == ‘A‘)
13             ++t;
14         else
15             --t;
16     
17     if (!t)
18         cout << "Friendship" << endl;
19     else if (t > 0)
20         cout << "Anton" << endl;
21     else
22         cout << "Danik" << endl;
23 }

Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare npotions.

Anton has a special kettle, that can prepare one potions in x seconds. Also, he knows spells of two types that can faster the process of preparing potions.

1. Spells of this type speed up the preparation time of one potion. There are m spells of this type, the i-th of them costs bi manapoints and changes the preparation time of each potion to ai instead of x.
2. Spells of this type immediately prepare some number of potions. There are k such spells, the i-th of them costs di manapoints and instantly create ci potions.

Anton can use no more than one spell of the first type and no more than one spell of the second type, and the total number of manapoints spent should not exceed s. Consider that all spells are used instantly and right before Anton starts to prepare potions.

Anton wants to get to the next level as fast as possible, so he is interested in the minimum number of time he needs to spent in order to prepare at least n potions.

The first line of the input contains three integers n, m, k (1?≤?n?≤?2·109,?1?≤?m,?k?≤?2·105) — the number of potions, Anton has to make, the number of spells of the first type and the number of spells of the second type.

The second line of the input contains two integers x and s (2?≤?x?≤?2·109,?1?≤?s?≤?2·109) — the initial number of seconds required to prepare one potion and the number of manapoints Anton can use.

The third line contains m integers ai (1?≤?ai?<?x) — the number of seconds it will take to prepare one potion if the i-th spell of the first type is used.

The fourth line contains m integers bi (1?≤?bi?≤?2·109) — the number of manapoints to use the i-th spell of the first type.

There are k integers ci (1?≤?ci?≤?n) in the fifth line — the number of potions that will be immediately created if the i-th spell of the second type is used. It‘s guaranteed that ci are not decreasing, i.e. ci?≤?cj if i?<?j.

The sixth line contains k integers di (1?≤?di?≤?2·109) — the number of manapoints required to use the i-th spell of the second type. It‘s guaranteed that di are not decreasing, i.e. di?≤?dj if i?<?j.

Print one integer — the minimum time one has to spent in order to prepare n potions.

需要制造n瓶毒药（也许是饮料，不知道），制造一瓶需要x单位时间。现在有2类魔法，每类最多只能使用一种且一次。

魔法一：有m种，花费bi金钱，使得制造一瓶毒药的时间变成ai。

魔法二：有k种，花费di金钱，使得在0单位时间内制造ci瓶毒药。

先给出最多花费金钱，求最小完成时间。

对于每种魔法，按照花费从小到大排序后，花费大而效果差的一定没用，直接忽略即可。

筛出两个按照花费单调上升的魔法序列，从左向右枚举一种魔法的时候，根据最大金钱的限制，另一种魔法对应的最优决策一定是从右向左，维护两个指针即可。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 200005
 5 #define int long long
 6 
 7 typedef long long LL;
 8 typedef pair<int, int> pr;
 9 
10 int n, m, k, c, s;
11 
12 pr sp1[N]; int t1;
13 pr sp2[N]; int t2;
14 
15 signed main(void)
16 {
17     cin >> n >> m >> k >> c >> s;
18     
19     for (int i = 1; i <= m; ++i)
20         cin >> sp1[i].second;
21         
22     for (int i = 1; i <= m; ++i)
23         cin >> sp1[i].first;
24         
25     for (int i = 1; i <= k; ++i)
26         cin >> sp2[i].second;
27     
28     for (int i = 1; i <= k; ++i)
29         cin >> sp2[i].first;
30         
31     sp1[++m] = make_pair(0, c);
32     sp2[++k] = make_pair(0, 0);
33         
34     sort(sp1 + 1, sp1 + 1 + m);
35     sort(sp2 + 1, sp2 + 1 + k);
36     
37     t1 = t2 = 0;
38     
39     for (int i = 1; i <= m; ++i)if (sp1[i].first <= s)
40         if (!t1 || sp1[i].second < sp1[t1].second)sp1[++t1] = sp1[i];
41             
42     for (int i = 1; i <= k; ++i)if (sp2[i].first <= s)
43         if (!t2 || sp2[i].second > sp2[t2].second)sp2[++t2] = sp2[i];
44         
45     if (sp1[t1].second <= 0)return puts("0"), 0;
46     if (sp2[t2].second >= n)return puts("0"), 0;
47     
48     LL answer = 4e18 + 7;
49     
50     for (int i = 1, j = t1; i <= t2; ++i)
51     {
52         while (j && sp1[j].first + sp2[i].first > s)--j;
53             
54         if (j)answer = min(answer, (LL)sp1[j].second * (n - sp2[i].second));
55     }
56     
57     cout << answer << endl;
58 }

C.cpp

D. Anton and Chess

Output

The only line of the output should contains "YES" (without quotes) if the white king is in check and "NO" (without quotes) otherwise.

 

在无限大的国际象棋棋盘上，给出白子King的位置，以及一些黑子的种类(后，车，象）和位置，求是否in check。注意棋子不能 越“子”强杀。

 

枚举王的8个方向，看最近的棋子是否符合这个方向即可。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 500005
 5 #define inf 2e9 + 7
 6 
 7 int n;
 8 int tx;
 9 int ty;
10 int vis[8];
11 int dis[8];
12 
13 signed main(void)
14 {
15     cin >> n >> tx >> ty;
16     
17     for (int i = 0; i < 8; ++i)
18         dis[i] = inf;
19     
20     for (int i = 1; i <= n; ++i)
21     {
22         char c; int x, y; 
23         cin >> c >> x >> y;
24 
25         if (x == tx && y < ty && ty - y < dis[0])
26             dis[0] = ty - y, vis[0] = (c == ‘Q‘ || c == ‘R‘);
27         if (x == tx && y > ty && y - ty < dis[1])
28             dis[1] = y - ty, vis[1] = (c == ‘Q‘ || c == ‘R‘);
29         if (y == ty && x < tx && tx - x < dis[2])
30             dis[2] = tx - x, vis[2] = (c == ‘Q‘ || c == ‘R‘);
31         if (y == ty && x > tx && x - tx < dis[3])
32             dis[3] = x - tx, vis[3] = (c == ‘Q‘ || c == ‘R‘);
33             
34         if (tx + ty == x + y && x < tx && tx - x < dis[4])
35             dis[4] = tx - x, vis[4] = (c == ‘Q‘ || c == ‘B‘);
36         if (tx + ty == x + y && x > tx && x - tx < dis[5])
37             dis[5] = x - tx, vis[5] = (c == ‘Q‘ || c == ‘B‘);
38         if (tx - ty == x - y && x < tx && tx - x < dis[6])
39             dis[6] = tx - x, vis[6] = (c == ‘Q‘ || c == ‘B‘);
40         if (tx - ty == x - y && x > tx && x - tx < dis[7])
41             dis[7] = x - tx, vis[7] = (c == ‘Q‘ || c == ‘B‘);
42     }
43     
44     int answer = 0;
45     
46     for (int i = 0; i < 8; ++i)
47         answer |= vis[i];
48         
49     puts(answer ? "YES" : "NO");
50 }

D.cpp

 

后记：貌似这次CF的题都很友（水）好（水）。

 

@Author: YouSiki