思路：优先队列求出每个人的最短用时，归并排序求最短用时的逆序对

归并排序将两段区间 [l, mid] 和 [mid + 1, r] 合并时，如左区间中 a[i] > a[j]，a[j]与 a[i] ~ a[mid] 都构成逆序对，ans += mid - i + 1;

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
ll b[100010], t[100010];
 
class Solution {
public:
    ll ans = 0;
    void _merge(int l, int mid, int r)  ///合并
    {
        int p = l, q = mid + 1;
        for(int i = l; i <= r; ++i)
        {
            if(p <= mid && (q > r || t[p] <= t[q]))
            {
                b[i] = t[p];
                p++;
            }
            else
            {
                b[i] = t[q];
                q++;
                ans += mid - p + 1;
            }
        }
        for(int i = l; i <= r; ++i)
            t[i] = b[i];
        return ;
    }
 
    void mergesort(int l, int r)
    {
        if(l >= r) return;
        int mid = (l + r) / 2;
        mergesort(l, mid);
        mergesort(mid + 1, r);
        _merge(l, mid, r);
    }
 
    long long getNumValidPairs(int n, int m, vector<int>& a) {
        ans = 0;
        priority_queue<long long, vector<long long>, greater<long long> >q;
        for(int i = 0; i < min(n, m); ++i)
        {
            t[i] = 1ll * a[i];
            q.push(t[i]);
        }
        for(int i = m; i < n; ++i)
        {
            ll tp = q.top();
            q.pop();
            t[i] = tp + 1ll * a[i];
            q.push(t[i]);
        }
        mergesort(0, n - 1);
        return ans;
    }
};