abc373E How to Win the Election

有N个候选人和总共K张选票，目前第i个候选人的票数为A[i]。在全部选票统计完成后，如果得票数多于自己的人数小于M，则当选，可以多个人同时当选。对于每个人，输出当选需要再获得的最少票数。
1<=M<=N<=2E5, 1<=K<=1E12, 0<=A[i]<=1E12, sum(A[i])<=K

分析：对每个候选人，二分答案，假设需要的票数为x，那么最终得票为A[i]+x，大于该得票的人数记为P，则需要统计另外M-P个人的票数进行补齐，看剩下的票数是否足够。

#include <bits/stdc++.h>
using i64 = long long;

// SumTreap模板...

void solve() {
    i64 N, M, K;
    std::cin >> N >> M >> K;
    SumTreap<i64> tr;
    std::vector<i64> A(N);
    for (int i = 0; i < N; i++) {
        std::cin >> A[i];
        tr.insert(A[i]);
        K -= A[i];
    }

    if (N == M) {
        for (int i = 0; i < N; i++) {
            std::cout << " 0";
        }
        return;
    }

    auto check = [&](i64 cur, i64 add) {
        i64 tar = cur + add;
        int cnt1 = tr.gtcnt(tar);
        if (cnt1 >= M) {
            return false;
        }
        int cnt2 = M - cnt1;
        i64 sum2 = tr.kSum(M) - tr.kSum(cnt1);
        i64 diff = cnt2 * (tar + 1) - sum2;
        return diff + add > K;
    };

    std::vector<i64> ans(N);
    for (int i = 0; i < N; i++) {
        tr.erase(A[i]);
        i64 lo = 0, hi = 1E12, mid;
        while (lo < hi) {
            mid = lo + (hi - lo) / 2;
            if (check(A[i], mid)) {
                hi = mid;
            } else {
                lo = mid + 1;
            }
        }
        if (lo <= K) {
            ans[i] = lo;
        } else {
            ans[i] = -1;
        }
        tr.insert(A[i]);
    }

    for (int i = 0; i < N; i++) {
        std::cout << ans[i] << " ";
    }
}

int main() {
    std::cin.tie(0)->sync_with_stdio(0);
    int t = 1;
    while (t--) solve();
    return 0;
}