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  • CF1005E1 Median on Segments (Permutations Edition) 思维

    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 (1n21051≤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;
    }
    彼时当年少,莫负好时光。
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  • 原文地址:https://www.cnblogs.com/l609929321/p/9313856.html
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