CF1005E1 Median on Segments (Permutations Edition) 思维

Posted l609929321

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了CF1005E1 Median on Segments (Permutations Edition) 思维相关的知识,希望对你有一定的参考价值。

Median on Segments (Permutations Edition)
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given a permutation p1,p2,,pnp1,p2,…,pn. A permutation of length nn is a sequence such that each integer between 11 and nn occurs exactly once in the sequence.

Find the number of pairs of indices (l,r)(l,r) (1lrn1≤l≤r≤n) such that the value of the median of pl,pl+1,,prpl,pl+1,…,pr is exactly the given number mm.

The median of a sequence is the value of the element which is in the middle of the sequence after sorting it in non-decreasing order. If the length of the sequence is even, the left of two middle elements is used.

For example, if a=[4,2,7,5]a=[4,2,7,5] then its median is 44 since after sorting the sequence, it will look like [2,4,5,7][2,4,5,7] and the left of two middle elements is equal to 44. The median of [7,1,2,9,6][7,1,2,9,6] equals 66 since after sorting, the value 66 will be in the middle of the sequence.

Write a program to find the number of pairs of indices (l,r)(l,r) (1lrn1≤l≤r≤n) such that the value of the median of pl,pl+1,,prpl,pl+1,…,pr is exactly the given number mm.

Input

The first line contains integers nn and mm (1n2?1051≤n≤2?105, 1mn1≤m≤n) — the length of the given sequence and the required value of the median.

The second line contains a permutation p1,p2,,pnp1,p2,…,pn (1pin1≤pi≤n). Each integer between 11 and nn occurs in pp exactly once.

Output

Print the required number.

Examples
input
Copy
5 4
2 4 5 3 1
output
Copy
4
input
Copy
5 5
1 2 3 4 5
output
Copy
1
input
Copy
15 8
1 15 2 14 3 13 4 8 12 5 11 6 10 7 9
output
Copy
48
Note

In the first example, the suitable pairs of indices are: (1,3)(1,3), (2,2)(2,2), (2,3)(2,3) and (2,4)(2,4).

 

题意:给你n个数和m,问在这n个数中以m为中位数的区间有多少个?

因为要使m为中位数,肯定是m的值位于这些数的中间的部分,即要有比m大的数和比m小的数,且大的数和小的数要相等或者大的数比小的数多一

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll a[maxn], vis[maxn];
int main() {
    ll n, m;
    while( cin >> n >> m ) {
        map<ll,ll> mm;
        ll  pos;
        for( ll i = 1; i <= n; i ++ ) {
            cin >> a[i];
            if( a[i] == m ) {
                pos = i;
            }
        }
        ll cnt = 0;
        for( ll i = pos; i <= n; i ++ ) {
            if( a[i] > m ) {
                cnt ++;
            } else if( a[i] < m ) {
                cnt --;
            }
            mm[cnt] ++;
        }
        ll ans = 0;
        cnt = 0;
        for( ll i = pos; i >= 1; i -- ) {
            if( a[i] > m ) {
                cnt ++;
            } else if( a[i] < m ) {
                cnt --;
            }
            ans += mm[-cnt];
            ans += mm[1-cnt];  //个数为偶数,中位数在中间两位的左边一位
        }
        cout << ans << endl;
    }
    return 0;
}

 




以上是关于CF1005E1 Median on Segments (Permutations Edition) 思维的主要内容,如果未能解决你的问题,请参考以下文章

Codeforces #496 E1. Median on Segments (Permutations Edition)

E1. Median on Segments (Permutations Edition)

Codeforces 1005E1&2 Median on Segments (General Case & Permutations Edition)

CF1360H Binary Median

[CF1526F]Median Queries

CF1486D Max Median