2018

第一次参加cf比赛，水了一发写了三道水题实在不想写了，然后Rating就低了，好难过哈哈哈哈。

cf#521(div.3)

D：Cutting Out

Description

You are given an array

You have to find any array

Cutting out the copy of

For example, if

To cut out the first copy of
To cut out the second copy of

Your task is to find such array

Input

The first line of the input contains two integers

The second line of the input contains exactly

Output

Examples

Input

7 3
1 2 3 2 4 3 1

Output

1 2 3

Input

10 4
1 3 1 3 10 3 7 7 12 3

Output

7 3 1 3

Input

15 2
1 2 1 1 1 2 1 1 2 1 2 1 1 1 1

Output

1 1

Hint

The first example is described in the problem statement.

In the second example the only answer is

In the third example the array

正确解法：

题目的意思是在n个数中找出k个数，使这k个数在n个数中出现的最多。

给定n和k，求这k个数。

这题目一开始就把我弄懵了，原本想到了桶排序，然后按出现的次数大小排序，找出前k个就好。但是这样的话就不会有重复的，也就是说不是题目所要的。于是我就不会写了。

看答案很简单啊，不过还要多想想。

还是桶排序，然后从1-n开始二分，找出有k个数，在n个数中出现的最大的次数。

因为数最多出现n次，最少出现1次嘛，假设最大次数是mid，有这样的数就是 b[i]/mid 的和。（b[i]是桶排序之后的），还是那个看是否符合条件，这也是二分的思想。先假设出一个解，再找是否符合条件，一直二分就找到了临界值。

知道了出现次数，数又从1开始，按这个数出现次数 / 上一阶段求出的次数就可以知道这个数是否存在且重复，找出前k个数就好啦。（因为可能有多个符合条件的序列，只要输出一个就好，所以我们从1开始，找到k个数就结束，不管后面的啦）

代码倒是挺简单的，我都说这么多，应该就好写了，但是中间二分找次数的思想，以及找是否存在或重复的方法都需要理解并记住鸭

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<string>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 int a[300000], b[300000], n, k,c[300000],cnt=0;
 9 bool can(int aa)
10 {
11     int ans = 0;
12     for (int i = 1; i <= 200000; i++)
13         ans += b[i] / aa;
14     if (ans >= k)    return 1;
15     else return 0;
16 }
17 int main()
18 {
19     cin >> n >> k;
20     for (int i = 1; i <= n; i++)
21     {
22         cin >> a[i];
23         b[a[i]]++;
24     }
25     int l = 1, r = n;
26     while (l+1< r)
27     {
28         int mid = (l + r) / 2;
29         if (can(mid))    l = mid;
30         else r = mid-1;
31     }
32     for (int i = 1; i <= 200000; i++)
33         for (int j = 1; j <= b[i] / l; j++)
34             if (cnt < k)
35                 c[++cnt] = i;
36     for (int i = 1; i < cnt; i++)
37         cout << c[i] << " ";
38     cout << c[cnt] << endl;
39     return 0;
40 }

View Code

E：Thematic Contests

Description

Polycarp has prepared

Polycarp has to host several thematic contests. All problems in each contest should have the same topic, and all contests should have pairwise distinct topics. He may not use all the problems. It is possible that there are no contests for some topics.

Polycarp wants to host competitions on consecutive days, one contest per day. Polycarp wants to host a set of contests in such a way that:

number of problems in each contest is exactly twice as much as in the previous contest (one day ago), the first contest can contain arbitrary number of problems;
the total number of problems in all the contests should be maximized.

Your task is to calculate the maximum number of problems in the set of thematic contests. Note, that you should not maximize the number of contests.

Input

The first line of the input contains one integer

The second line of the input contains

Output

Print one integer — the maximum number of problems in the set of thematic contests.

Examples

Input

18
2 1 2 10 2 10 10 2 2 1 10 10 10 10 1 1 10 10

Output

Input

10
6 6 6 3 6 1000000000 3 3 6 6

Output

Input

3
1337 1337 1337

Output

Hint

In the first example the optimal sequence of contests is:

In the second example the optimal sequence of contests is:

In the third example you can take all the problems with the topic

正确解法：

首先是桶排序，题目的意思是一样的数的个数能形成以2为d 的等比数列的和的最大值。

其实跟这个数没关系，但跟这个数出现的次数有关，就成了桶排序，可是十亿的数组没办法开数组，全局变量也不可以。

所有人都用map啊，可我不会map啊。

统计过数出现的次数后，从小到大排序。然后从1开始遍历，到2*105 总共有2*105个数，万一数都一样呢。

这题目所说的是 1.2.6 中可以成为1.2.4 ans=7.　　1.3.6 可以成为1.2.4 也可以 3.6 ans=9，求最大的。

从1开始遍历，二分找出第一个大于1的下标。然后答案加上1，1变成2，二分找出第一个大于2的下标，直到走到数据的最大值。

要注意的是，数据的n+1是0，那么这个数不存在，lower—bound（）就会返回n+1.

啊啊啊我研究的答案是用一个数来存上一个数的下标+1，这个数的下标=max（下标，上个数的下标+1）

刚开始以为所有的数都小于这个数时，二分lower 就会返回n+1，直接就break了，还要存上一个数的下标有什么用呢。

后来提交错了，才发现这是害怕这个数即大于1，也大于2.那么这个数就加了两次，我们是不允许这样存在的。

于是这个数的下标必须大于等于上个数的下标+1.

啊说的乱七八糟自己都不知道自己在说什么。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<string>
 6 #include<cstring>
 7 #include<map>//map头文件
 8 using namespace std;
 9 int cc[300010], cnt = 0;
10 int solver(int f)
11 {
12     int ans = 0,ps=0;//ps存上一个数的下标，所以初始为0.
13     while (true)
14     {
15         int id = lower_bound(cc + 1, cc + cnt + 1, f)-cc;//注意二分函数的用法，最后要减去数组
16         id = max(id,ps+1);
17         if (id > cnt)    break;
18         ans += f;
19         f = f * 2;
20         ps = id;
21     }
22     return ans;
23 }
24 int main()
25 {
26     int a, n,aaa=0;
27     map<int, int>M;//map申请一个数组
28     cin >> n;
29     for (int i = 1; i <= n; i++)
30     {
31         cin >> a;
32         M[a]++;
33     }
34     map<int, int>::iterator it;//不知道有什么用，map的基本用法吧
35     for (it = M.begin(); it != M.end(); it++)
36         cc[++cnt] = it->second;//第一个数first是下标，第二个数second是这个数
37     sort(cc + 1, cc + cnt + 1);
38     for (int i = 1; i <= 200000; i++)
39         aaa = max(aaa,solver(i));
40     cout << aaa << endl;
41     return 0;
42 }

View Code

F1:Pictures with Kittens

Description

Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of

Vova wants to repost exactly

each segment of the news feed of at least
the sum of beauty values of reposted pictures is maximum possible.

For example, if

Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.

Input

The first line of the input contains three integers

The second line of the input contains

Output

Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.

Otherwise print one integer — the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.

Examples

Input

5 2 3
5 1 3 10 1

Output

Input

6 1 5
10 30 30 70 10 10

Output

-1

Input

4 3 1
1 100 1 1

Output

100

正确解法：

DP。d[i][j] d[i][j]d[i][j]表示前i ii个数中，在选了ai

的情况下，总共选择了j jj个数的最大值。

则有d[i][j]=max(d[y][j−1]+a[i]),(i−k≤y) d[i][j]=max(d[y][j-1]+a[i]),(i-k\le y)d[i][j]=max(d[y][j−1]+a[i]),(i−k≤y)

最后的答案=max(d[i][x])，(n−k+1≤i) =max(d[i][x])，(n-k+1\le i)=max(d[i][x])，(n−k+1≤i)

其中可以发现转移的部分：d[i][j]=max(d[y][j−1]+a[i]),(i−k≤y) d[i][j]=max(d[y][j-1]+a[i]),(i-k\le y)d[i][j]=max(d[y][j−1]+a[i]),(i−k≤y)

是求d[y][j−1] d[y][j-1]d[y][j−1]，y yy在区间[i−k,i−1] [i-k,i-1][i−k,i−1]的最大值，于是可以构造一个单调递减的栈优化DP。

不会DP，先找别人的题解看一下qaq

  1 #include <iostream>
  2 #include <string>
  3 #include <complex>
  4 #include <math.h>
  5 #include <set>
  6 #include <vector>
  7 #include <map>
  8 #include <queue>
  9 #include <stdio.h>
 10 #include <stack>
 11 #include <algorithm>
 12 #include <list>
 13 #include <ctime>
 14 
 15 #include <memory.h>
 16 #include <assert.h>
 17 
 18 #define y0 sdkfaslhagaklsldk
 19 
 20 #define y1 aasdfasdfasdf
 21 #define yn askfhwqriuperikldjk
 22 #define j1 assdgsdgasghsf
 23 #define tm sdfjahlfasfh
 24 #define lr asgasgash
 25 #define norm asdfasdgasdgsd
 26 #define have adsgagshdshfhds
 27 #define ends asdgahhfdsfshdshfd
 28 #define prev asdgSHJsfgsdfhdsh
 29 #define hash asdgasdgasdgdfrywewery
 30 
 31 #define eps 1e-8
 32 #define M_PI 3.141592653589793
 33 #define bsize 512
 34 
 35 #define ldouble long double
 36 using namespace std;
 37 
 38 #define bs 1000000007
 39 
 40 const int N = 300031;
 41 
 42 int n,k,x,ar[5031];
 43 
 44 long long dp[5031][5031];
 45 
 46 int L[5031];
 47 
 48 long long sparse[20][5031];
 49 
 50 long long get(int l,int r){
 51     int span=r-l+1;
 52     span=L[span];
 53     return max(sparse[span][l],sparse[span][r-(1<<span)+1]);
 54 }
 55 
 56 void build_sparse(int lev){
 57     for (int i=1;i<=n;i++){
 58         sparse[0][i]=dp[lev][i];
 59     }
 60     for (int lev=1;lev<15;lev++){
 61         for (int i=1;i+(1<<lev)-1<=n;i++){
 62             sparse[lev][i]=max(sparse[lev-1][i],sparse[lev-1][i+(1<<lev)/2]);
 63         }
 64     }
 65 }
 66 
 67 int main(){
 68 //    freopen("apache.in","r",stdin);
 69 //    freopen("apache.out","w",stdout);
 70 //    freopen("input.txt", "r", stdin);
 71 //    freopen("output.txt", "w", stdout);
 72     ios_base::sync_with_stdio(0);
 73     //cin.tie(0);
 74 
 75     for (int i=2;i<=5000;i++){
 76         L[i]=L[i/2]+1;
 77     }
 78 
 79     cin>>n>>k>>x;
 80     for (int i=1;i<=n;i++){
 81         cin>>ar[i];
 82     }
 83 
 84     for (int lev=1;lev<=x;lev++){
 85         for (int i=1;i<=n;i++){
 86             dp[lev][i]=-1e18;
 87         }
 88     }
 89 
 90     for (int i=1;i<=k;i++){
 91         dp[1][i]=ar[i];
 92     }
 93 
 94     for (int lev=2;lev<=x;lev++){
 95 
 96         build_sparse(lev-1);
 97         for (int i=2;i<=n;i++){
 98             int ps=max(1,i-k);
 99     //        cout<<ps<<" "<<i-1<<" "<<get(ps,i-1)<<endl;
100             dp[lev][i]=get(ps,i-1)+ar[i];
101         }
102     }
103 
104     long long ans=-1e18;
105     for (int i=n-k+1;i<=n;i++){
106         ans=max(ans,dp[x][i]);
107     }
108 
109     if (ans<0)
110         ans=-1;
111 
112     cout<<ans<<endl;
113 
114 //    cin.get(); cin.get();
115     return 0;
116 }

View Code

posted @ 2018-11-17 21:21 kaike 阅读(310) 评论(0) 编辑收藏举报

刷新页面返回顶部

kaike

没有人知道未来会是怎么样的 你的未来是你。

Test on 11/17/2018

D：Cutting Out

Description

Input

Output

Examples

Input

Output

Input

Output

Input

Output

Hint

正确解法：

E：Thematic Contests

Description

Input

Output

Examples

Input

Output

Input

Output

Input

Output

Hint

正确解法：

F1:Pictures with Kittens

Description

Input

Output

Examples

Input

Output

Input

Output

Input

Output

正确解法：

公告

没有人知道未来会是怎么样的你的未来是你。