Day8 - B - Non-Secret Cypher CodeForces - 190D
Berland starts to seize the initiative on the war with Flatland. To drive the enemy from their native land, the berlanders need to know exactly how many more flatland soldiers are left in the enemy's reserve. Fortunately, the scouts captured an enemy in the morning, who had a secret encrypted message with the information the berlanders needed so much.
The captured enemy had an array of positive integers. Berland intelligence have long been aware of the flatland code: to convey the message, which contained a number m, the enemies use an array of integers a. The number of its subarrays, in which there are at least k equal numbers, equals m. The number k has long been known in the Berland army so General Touristov has once again asked Corporal Vasya to perform a simple task: to decipher the flatlanders' message.
Help Vasya, given an array of integers a and number k, find the number of subarrays of the array of numbers a, which has at least k equal numbers.
Subarray a[i... j] (1 ≤ i ≤ j ≤ n) of array a = (a1, a2, ..., an) is an array, made from its consecutive elements, starting from the i-th one and ending with the j-th one: a[i... j] = (ai, ai + 1, ..., aj).
Input
The first line contains two space-separated integers n, k (1 ≤ k ≤ n ≤ 4·105), showing how many numbers an array has and how many equal numbers the subarrays are required to have, correspondingly.
The second line contains n space-separated integers ai (1 ≤ ai ≤ 109) — elements of the array.
Output
Print the single number — the number of such subarrays of array a, that they have at least k equal integers.
Please do not use the %lld specifier to read or write 64-bit integers in С++. In is preferred to use the cin, cout streams or the %I64d specifier.
Examples
4 2
1 2 1 2
3
5 3
1 2 1 1 3
2
3 1
1 1 1
6
Note
In the first sample are three subarrays, containing at least two equal numbers: (1,2,1), (2,1,2) and (1,2,1,2).
In the second sample are two subarrays, containing three equal numbers: (1,2,1,1,3) and (1,2,1,1).
In the third sample any subarray contains at least one 1 number. Overall they are 6: (1), (1), (1), (1,1), (1,1) and (1,1,1).
思路:双指针问题,和A题相似,一个指针每次递增一,另一个指针随条件改变不止一,一般是维护第一个指针的极值,然后计算数量
const int maxm = 4e5+; int buf[maxm], n, k;
map<int, int> vis; int main() {
ios::sync_with_stdio(false), cin.tie();
cin >> n >> k;
for(int i = ; i < n; ++i)
cin >> buf[i];
LL ans = ;
int l = , r = ;
vis[buf[]]++;
while(l + k - < n) {
while(r < n) {
if(vis[buf[r]] >= k) break;
r++;
vis[buf[r]]++;
}
if(r == n) break;
ans += (LL)(n - r);
vis[buf[l]]--;
l++;
}
cout << ans << "\n";
return ;
}
Day8 - B - Non-Secret Cypher CodeForces - 190D的更多相关文章
- CodeForces 190D Non-Secret Cypher
双指针. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> ...
- Day8 - A - Points on Line CodeForces - 251A
Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. N ...
- [Codeforces] #603 (Div. 2) A-E题解
[Codeforces]1263A Sweet Problem [Codeforces]1263B PIN Code [Codeforces]1263C Everyone is a Winner! [ ...
- CodeForces 490C Hacking Cypher
Hacking Cypher Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Sub ...
- CodeForces 496B Secret Combination
Secret Combination Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u ...
- Codeforces Round #279 (Div. 2) C. Hacking Cypher 前缀+后缀
C. Hacking Cypher time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Codecraft-17 and Codeforces Round #391 (Div. 1 + Div. 2, combined)D. Felicity's Big Secret Revealed
题目连接:http://codeforces.com/contest/757/problem/D D. Felicity's Big Secret Revealed time limit per te ...
- Codeforces Round #279 (Div. 2) C. Hacking Cypher (大数取余)
题目链接 C. Hacking Cyphertime limit per test1 secondmemory limit per test256 megabytesinputstandard inp ...
- 构造+暴力 Codeforces Round #283 (Div. 2) B. Secret Combination
题目传送门 /* 构造+暴力:按照题目意思,只要10次加1就变回原来的数字,暴力枚举所有数字,string大法好! */ /************************************** ...
随机推荐
- NSDateFormatter使用注意事项
NSDateFormatter是用来连接NSDate和NSString之间的桥梁 它的使用方式,不(自)做(行)说(百)明(度) 要说的注意事项就是,NSString转NSDate时,NSDateFo ...
- jupyter配置 nbextension
jupyter contrib nbextension install --user --skip-running-check No module named 'pysqlite2' 解决方法:打开此 ...
- Legal High
不让任何人承担责任,不想看的东西就回避, 但是,如果想夺回值得夸耀的生存方式,就必须看那些不愿意看的现实,必须带着身负重伤的觉悟前进,这才叫做战斗. 有怨言的话去坟墓里说,钱不是全部,钱就是你们向对手 ...
- 吴裕雄 Bootstrap 前端框架开发——Bootstrap 排版:设定文本居中对齐
<!DOCTYPE html> <html> <head> <title>菜鸟教程(runoob.com)</title> <meta ...
- Spring boot 中发送邮件
参考:https://blog.csdn.net/qq_39241443/article/details/81293939 添加依赖: <dependency> <groupId&g ...
- 在webView中的返回键
在写webView中我们按一下返回键,退到上一个我们浏览的网页,到第一个页面时,按两下退出程序,且按一下时提示你在按一下退出程序 只要加上这个方法即可 public void onBackPresse ...
- 奈奎斯特采样定理(Nyquist)
采样定理在1928年由美国电信工程师H.奈奎斯特首先提出来的,因此称为奈奎斯特采样定理. 1933年由苏联工程师科捷利尼科夫首次用公式严格地表述这一定理,因此在苏联文献中称为科捷利尼科夫采样定理. 1 ...
- A letter for NW RDMA configuration
Dear : If you have to use EMC NW NDMA to backup oracle database and want to see what happen when bac ...
- html常用整理
视频链接:https://www.bilibili.com/video/av5862916?from=search&seid=12139895566389560177 我的第一个html &l ...
- luogu P4014 分配问题
简单的费用流问题,每个人对每个任务连边,每个任务对汇点连,源点对每个人连,最大费用取反即可 #include<bits/stdc++.h> using namespace std; #de ...