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;template <typename TYPE>
struct SumTreap {struct Node {TYPE data, sum;int rnd, siz, dup, son[2];Node() { data = sum = rnd = siz = dup = son[0] = son[1] = 0; }Node(const TYPE &d, int cnt=1) { init(d, cnt); }void init(const TYPE & d, int cnt) {data = d; dup = siz = cnt; sum = d * cnt; rnd = rand(); son[0] = son[1] = 0;}};SumTreap(int multi=1):multiple(multi) { reset(); }void setmulti(int multi) { multiple = multi; }int newnode(const TYPE & d, int cnt) {if (!reuse.empty()) {int r = reuse.front();reuse.pop_front();node[r].init(d, cnt);return r;}node.push_back({d, cnt});return node.size() - 1;}void reset() { root = 0; node.resize(1); reuse.clear(); }void maintain(int x) {node[x].siz = node[x].dup;node[x].sum = node[x].data * node[x].dup;if (node[x].son[0]) {node[x].siz += node[node[x].son[0]].siz;node[x].sum += node[node[x].son[0]].sum;}if (node[x].son[1]) {node[x].siz += node[node[x].son[1]].siz;node[x].sum += node[node[x].son[1]].sum;}}void rotate(int d, int &r) {int k = node[r].son[d^1];node[r].son[d^1] = node[k].son[d];node[k].son[d] = r;maintain(r);maintain(k);r = k;}void insert(const TYPE &data, int cnt, int &r) {if (r) {if (!(data < node[r].data) && !(node[r].data < data)) {if (multiple) {node[r].dup += cnt;maintain(r);}} else {int d = data < node[r].data ? 0 : 1;int u = node[r].son[d];insert(data, cnt, u);node[r].son[d] = u;if (node[node[r].son[d]].rnd > node[r].rnd) {rotate(d^1, r);} else {maintain(r);}}} else {r = newnode(data, cnt);}}TYPE kth(int k) {assert(1 <= k && k <= size());for (int r = root; r; ) {int x = node[r].son[0] ? node[node[r].son[0]].siz : 0;int y = node[r].dup;if (k <= x) {r = node[r].son[0];} else if (k <= x + y) {return node[r].data;} else {k -= x + y;r = node[r].son[1];}}assert(k == 0);return 0;}TYPE Kth(int k) {assert(1 <= k && k <= size());for (int r = root; r; ) {int x = node[r].son[1] ? node[node[r].son[1]].siz : 0;int y = node[r].dup;if (k <= x) {r = node[r].son[1];} else if (k <= x + y) {return node[r].data;} else {k -= x + y;r = node[r].son[0];}}assert(k == 0);return 0;}TYPE ksum(int k) {assert(0 <= k && k <= node[root].siz);TYPE ans = 0;for (int r = root; r && k; ) {int x = node[r].son[0] ? node[node[r].son[0]].siz : 0;int y = node[r].dup;if (k <= x) {r = node[r].son[0];} else if (k <= x + y) {ans += node[node[r].son[0]].sum + node[r].data * (k - x);k = 0;} else {ans += node[node[r].son[0]].sum + node[r].data * y;k -= x + y;r = node[r].son[1];}}return ans;}TYPE kSum(int k) {assert(0 <= k && k <= node[root].siz);TYPE ans = 0;for (int r = root; r && k; ) {int x = node[r].son[1] ? node[node[r].son[1]].siz : 0;int y = node[r].dup;if (k <= x) {r = node[r].son[1];} else if (k <= x + y) {ans += node[node[r].son[1]].sum + node[r].data * (k - x);k = 0;} else {ans += node[node[r].son[1]].sum + node[r].data * y;k -= x + y;r = node[r].son[0];}}return ans;}void erase(const TYPE& data, int cnt, int & r) {if (r == 0) return;int d = -1;if (data < node[r].data) {d = 0;} else if (node[r].data < data) {d = 1;}if (d == -1) {node[r].dup -= cnt;if (node[r].dup > 0) {maintain(r);} else {if (node[r].son[0] == 0) {reuse.push_back(r);r = node[r].son[1];} else if (node[r].son[1] == 0) {reuse.push_back(r);r = node[r].son[0];} else {int dd = node[node[r].son[0]].rnd > node[node[r].son[1]].rnd ? 1 : 0;rotate(dd, r);erase(data, cnt, node[r].son[dd]);}}} else {erase(data, cnt, node[r].son[d]);}if (r) maintain(r);}int ltcnt(const TYPE &data) {int ans = 0;for (int r = root; r; ) {int x = node[r].son[0] ? node[node[r].son[0]].siz : 0;if (node[r].data < data) {ans += node[r].dup + x;r = node[r].son[1];} else if (data < node[r].data) {r = node[r].son[0];} else {ans += x;break;}}return ans;}int gtcnt(const TYPE &data) {int ans = 0;for (int r = root; r; ) {int x = node[r].son[1] ? node[node[r].son[1]].siz : 0;if (node[r].data < data) {r = node[r].son[1];} else if (data < node[r].data) {ans += node[r].dup + x;r = node[r].son[0];} else {ans += x;break;}}return ans;}int count(const TYPE &data) {for (int r = root; r; ) {if (data < node[r].data) {r = node[r].son[0];} else if (node[r].data < data) {r = node[r].son[1];} else {return node[r].dup;}}return 0;}std::pair<bool,TYPE> prev(const TYPE &data) {std::pair<bool,TYPE> ans = {false, 0};for (int r = root; r; ) {if (node[r].data < data) {if (ans.first) {ans.second = std::max(ans.second, node[r].data);} else {ans = {true, node[r].data};}r = node[r].son[1];} else {r = node[r].son[0];}}return ans;}       std::pair<bool,TYPE> next(const TYPE &data) {std::pair<bool,TYPE> ans = {false, 0};for (int r = root; r; ) {if (data < node[r].data) {if (ans.first) {ans.second = std::min(ans.second, node[r].data);} else {ans = {true, node[r].data};}r = node[r].son[0];} else {r = node[r].son[1];}}return ans;}std::vector<Node> node;std::deque<int> reuse;int root, multiple;void insert(const TYPE& data, int cnt=1) { insert(data, cnt, root); }void erase(const TYPE& data, int cnt=1) { erase(data, cnt, root); }int size() const { return root ? node[root].siz : 0; }int lecnt(const TYPE& data) { return size() - gtcnt(data, root); }int gecnt(const TYPE& data) { return size() - ltcnt(data, root); }
};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() {int t = 1;while (t--) solve();return 0;
}