标签：its include file head def name inpu row return
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) (1≤l≤r≤n1≤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) (1≤l≤r≤n1≤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 first line contains integers nn and mm (1≤n≤2?1051≤n≤2?105, 1≤m≤n1≤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 (1≤pi≤n1≤pi≤n). Each integer between 11 and nn occurs in pp exactly once.
Print the required number.
5 4
2 4 5 3 1
4
5 5
1 2 3 4 5
1
15 8
1 15 2 14 3 13 4 8 12 5 11 6 10 7 9
48
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; }
CF1005E1 Median on Segments (Permutations Edition) 思维
标签：its include file head def name inpu row return
原文地址：https://www.cnblogs.com/l609929321/p/9313856.html
