Least Prefix Sum

By xiaruize

分析

令 M=∑i=1maiM=\sum\limits^{m}_{i=1}a_i , Sk=∑i=1kaiS_k=\sum\limits^{k}_{i=1}a_i

当 Sk<MS_k<M 时，分两类考虑

  1. k<mk<m

    此时，对 a1⋯aka_1 \cdots a_k 执行题目所述的操作，都会同时使 SkS_k 和 MM 增加 , 不会使得当前局面合法

    那么只会对 ak+1⋯ama_{k+1} \cdots a_m 这些数执行操作

    考虑怎样操作最优，容易发现，可以贪心的选择尽可能大的数进行操作

    每次操作会使 M=M−ax×2M=M-a_x\times 2

  2. k>mk>m

    同上分析，可以发现，这次的操作对象为 am+1⋯aka_{m+1} \cdots a_k ， 优先选择小的数进行操作

实现

从 mm 开始，倒序循环至 11 ，用 priority_queue 维护最大值

同理，再从 m→nm \to n 正序循环 ，维护最小值

Code

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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int t;
int n, m;
int a[N];
// bool en;

void solve()
{
    int res = 0;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    int sum = 0;
    for (int i = 1; i <= m; i++)
        sum += a[i];
    int tot = 0;
    tot = sum - a[m];
    priority_queue<int> q;
    q.push(a[m]);
    for (int i = m - 1; i >= 1; i--)
    {
        if (tot < sum)
        {
            while (tot < sum)
            {
                sum -= q.top() * 2ll;
                q.pop();
                res++;
            }
        }
        q.push(a[i]);
        tot -= a[i];
    }
    priority_queue<int, vector<int>, greater<int>> p;
    tot = sum;
    for (int i = m + 1; i <= n; i++)
    {
        p.push(a[i]);
        tot += a[i];
        if (tot < sum)
        {
            while (tot < sum)
            {
                tot -= p.top() * 2ll;
                p.pop();
                res++;
            }
        }
    }
    cout << res << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}