↵
↵
<spoiler summary="author's code:">↵
~~~~~↵
#include<cstdio>↵
const int MAX_V = 201;↵
bool achieve[MAX_V];↵
void solve() {↵
int n, x;↵
scanf("%d%d", &n, &x);↵
for(int i = 1; i <= n + x; i++) {↵
achieve[i] = false;↵
}↵
for(int i = 1; i <= n; i++) {↵
int ranking;↵
scanf("%d", &ranking);↵
achieve[ranking] = true;↵
}↵
for(int k = n + x; k > 0; k--) {↵
int v = 0;↵
for(int i = 1; i <= k; i++) {↵
if(!achieve[i]) v++;↵
}↵
if(v <= x) {↵
printf("%d\n", k);↵
return;↵
}↵
}↵
}↵
int main() {↵
int T;↵
scanf("%d", &T);↵
while(T--) solve();↵
return 0;↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
↵
[tutorial:1330B]↵
↵
↵
↵
<spoiler summary="author's code">↵
~~~~~↵
#include<cstdio>↵
const int SIZE = 200000;↵
int p[SIZE];↵
int ans[SIZE][2];↵
int ans_cnt;↵
bool judge(int a[], int n){↵
static int used[SIZE+1];↵
for(int i = 1; i <= n; i++) used[i] = 0;↵
for(int i = 0; i < n; i++) used[a[i]] = 1;↵
for(int i = 1; i <= n; i++) {↵
if(!used[i]) return 0;↵
}↵
return 1;↵
}↵
bool judge(int len1, int n){↵
return judge(p, len1) && judge(p + len1, n - len1);↵
}↵
int main() {↵
int t = 0;↵
scanf("%d", &t);↵
while(t--) {↵
ans_cnt = 0;↵
int n;↵
scanf("%d", &n);↵
int ma=0;↵
for(int i = 0; i < n; i++) {↵
scanf("%d", &p[i]);↵
if(ma < p[i]) ma = p[i];↵
}↵
if(judge(n - ma,n)) {↵
ans[ans_cnt][0] = n - ma;↵
ans[ans_cnt++][1] = ma;↵
}↵
if(ma * 2 != n && judge(ma,n)) {↵
ans[ans_cnt][0] = ma;↵
ans[ans_cnt++][1] = n - ma;↵
}↵
printf("%d\n", ans_cnt);↵
for(int i = 0; i < ans_cnt; i++) {↵
printf("%d %d\n", ans[i][0], ans[i][1]);↵
}↵
}↵
return 0;↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
[tutorial:1329A]↵
↵
<spoiler summary="author's code">↵
~~~~~↵
#include<bits/stdc++.h>↵
const int SIZE = 100002;↵
int len[SIZE];↵
long long suffix_sum[SIZE];↵
void err() {puts("-1");}↵
void solve() {↵
int N, M;↵
scanf("%d%d", &N, &M);↵
for(int i = 1; i <= M; i++) {↵
scanf("%d", &len[i]);↵
if(len[i] + i - 1 > N) {↵
err();↵
return;↵
}↵
}↵
for(int i = M; i > 0; i--) {↵
suffix_sum[i] = suffix_sum[i + 1] + len[i];↵
}↵
if(suffix_sum[1] < N) {↵
err();↵
return;↵
}↵
for(int i = 1; i <= M; i++) {↵
printf("%lld", std::max((long long)i, N - suffix_sum[i] + 1));↵
if(i < M) putchar(' ');↵
else puts("");↵
}↵
}↵
int main() {↵
solve();↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:1329B]↵
↵
<spoiler summary="author's code">↵
~~~~~↵
#include<bits/stdc++.h>↵
void solve(){↵
int d, m;↵
scanf("%d%d",&d, &m);↵
long long answer=1;↵
for(int i = 0; i < 30; i++) {↵
if(d < (1 << i)) break;↵
answer = answer * (std::min((1 << (i+1)) - 1, d) - (1 << i) + 2) % m;↵
}↵
answer--;↵
if(answer < 0) answer += m;↵
printf("%lld\n",answer);↵
}↵
int main() {↵
int T;↵
scanf("%d", &T);↵
while(T--) {↵
solve();↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
[tutorial:1329C]↵
↵
[a super simple solution which is differet to this blog](https://mirror.codeforces.com/blog/entry/75559?#comment-597934) provided by [user:Swistakk,2020-04-05].↵
↵
↵
↵
<spoiler summary="author's code (from bottom to top with min heap)">↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
const int SIZE = 1<<20;↵
int INF = 1000000001;↵
pair<int, int> a[SIZE];↵
int final_v[SIZE];↵
bool used[SIZE];↵
int h, g;↵
void pull(int id) {↵
while(a[id] > min(a[id<<1], a[(id<<1)|1])) {↵
if(a[id<<1] < a[(id << 1) | 1]) {↵
swap(a[id<<1], a[id]);↵
id <<= 1;↵
}↵
else {↵
swap(a[(id<<1)|1], a[id]);↵
id = (id << 1) | 1;↵
}↵
if(id >= (1 << h)) return;↵
}↵
}↵
void solve() {↵
scanf("%d%d", &h, &g);↵
long long an = 0;↵
for(int i = 1; i < (1 << h); i++) {↵
used[i] = 0;↵
scanf("%d", &a[i].first);↵
a[i].second = i;↵
final_v[i] = 0;↵
}↵
h--;↵
for(int lv = h - 1; lv >= 0; lv--) {↵
int ll = 1 << lv;↵
int rr = 1 << (lv + 1);↵
for(int i = ll; i < rr; i++) {↵
pair<int, int> tmp = a[i];↵
int bot = i << (h - lv);↵
a[i] = a[bot];↵
a[bot] = make_pair(INF, 0);↵
pull(i);↵
if(lv < g) {↵
int need_mi = max(final_v[i * 2], final_v[i * 2 + 1]);↵
while(a[i].first < need_mi) {↵
a[i] = make_pair(INF, 0);↵
pull(i);↵
}↵
an += final_v[i] = a[i].first;↵
used[a[i].second] = 1;↵
a[i] = tmp;↵
pull(i);↵
}↵
else {↵
a[bot] = tmp;↵
}↵
}↵
}↵
printf("%lld\n", an);↵
bool first_time = true;↵
for(int i = (1 << (h + 1)) - 1; i > 0; i--) {↵
if(!used[i]) {↵
if(!first_time) printf(" ");↵
else first_time = false;↵
printf("%d", i);↵
}↵
}↵
puts("");↵
}↵
int main(){↵
int T;↵
scanf("%d", &T);↵
while(T--) solve();↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="isaf27's code(from bottom to top with sorted array)">↵
~~~~~↵
//#pragma GCC optimize("O3")↵
#include <bits/stdc++.h>↵
↵
using namespace std;↵
↵
//defines↵
typedef long long ll;↵
typedef long double ld;↵
#define TIME clock() * 1.0 / CLOCKS_PER_SEC↵
#define prev _prev↵
#define y0 _y0↵
#define kill _kill↵
↵
//permanent constants↵
const ld pi = acos(-1.0);↵
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};↵
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};↵
const int dx[4] = {0, 1, 0, -1};↵
const int dy[4] = {1, 0, -1, 0};↵
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};↵
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};↵
const int alf = 26;↵
const int dig = 10;↵
const int two = 2;↵
const int th = 3;↵
const ll prost = 239;↵
const ll btc = 30;↵
const ld eps = 1e-8;↵
const ll INF = (ll)(1e18 + 239);↵
const int BIG = (int)(1e9 + 239);↵
const int MOD = 1e9 + 7; //↵
↵
//random↵
mt19937 rnd(239); //(chrono::high_resolution_clock::now().time_since_epoch().count());↵
↵
//constants↵
const int M = (int)(2e5 + 239);↵
const int N = (int)(2e3 + 239);↵
const int L = 20;↵
const int T = (1 << 20) + 239;↵
const int B = 500;↵
const int X = 35;↵
const int T2 = 2 * T;↵
↵
int h, g, a[T2], n;↵
bool mark[T];↵
int id[T], buf[T], timer;↵
↵
bool cmp(int i, int j)↵
{↵
return (a[i] < a[j]);↵
}↵
↵
int pos;↵
↵
void dfs_write(int p)↵
{↵
id[pos++] = p;↵
if (2 * p + 1 <= n)↵
{↵
dfs_write(2 * p);↵
dfs_write(2 * p + 1);↵
}↵
}↵
↵
int dfs(int p, int sz)↵
{↵
if ((1 << (g - 1)) <= p && p < (1 << g))↵
{↵
int l = timer;↵
timer += sz;↵
pos = l;↵
dfs_write(p);↵
sort(id + l, id + l + sz, cmp);↵
mark[id[l]] = true;↵
return id[l];↵
}↵
int l = timer;↵
int new_sz = ((sz - 1) / 2);↵
int s1 = dfs(2 * p, new_sz);↵
int s2 = dfs(2 * p + 1, new_sz);↵
merge(id + l, id + l + new_sz, id + l + new_sz, id + l + new_sz + new_sz, buf, cmp);↵
buf[sz - 1] = p;↵
for (int i = 0; i < sz; i++)↵
id[i + l] = buf[i];↵
timer++;↵
if (cmp(s1, s2))↵
s1 = s2;↵
for (int i = 0; i < sz; i++)↵
if (cmp(s1, id[i + l]))↵
{↵
mark[id[i + l]] = true;↵
return id[i + l];↵
}↵
return 0;↵
}↵
↵
void rm(int i)↵
{↵
if (a[2 * i] == 0 && a[2 * i + 1] == 0)↵
{↵
a[i] = 0;↵
mark[i] = false;↵
return;↵
}↵
if (a[2 * i] > a[2 * i + 1])↵
{↵
a[i] = a[2 * i];↵
mark[i] = mark[2 * i];↵
rm(2 * i);↵
}↵
else↵
{↵
a[i] = a[2 * i + 1];↵
mark[i] = mark[2 * i + 1];↵
rm(2 * i + 1);↵
}↵
}↵
↵
void dfs_ans(int p)↵
{↵
while (a[p] != 0 && !mark[p])↵
{↵
cout << p << " ";↵
rm(p);↵
}↵
if (a[p] == 0)↵
return;↵
dfs_ans(2 * p);↵
dfs_ans(2 * p + 1);↵
}↵
↵
void solve()↵
{↵
cin >> h >> g;↵
n = (1 << h) - 1;↵
for (int i = 1; i <= n; i++)↵
cin >> a[i];↵
for (int i = 1; i <= n; i++)↵
mark[i] = false;↵
timer = 0;↵
dfs(1, n);↵
ll ans = 0;↵
for (int i = 1; i <= n; i++)↵
if (mark[i])↵
ans += a[i];↵
cout << ans << "\n";↵
dfs_ans(1);↵
cout << "\n";↵
for (int i = 0; i <= n; i++)↵
a[i] = 0;↵
}↵
↵
int32_t main()↵
{↵
#ifdef ONPC↵
freopen("input", "r", stdin);↵
#endif↵
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);↵
int t;↵
cin >> t;↵
while (t--)↵
solve();↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
<spoiler summary="author's code (from top to bottom)">↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
const int SIZE = 1<<20;↵
int INF = 1000000001;↵
int a[SIZE], ops[SIZE];↵
int h, g;↵
int qq[24], qn;↵
int pull(int id) {↵
int tmp = a[id];↵
a[id] = 0;↵
qn = 0;↵
qq[qn++] = id;↵
while(id * 2 < (1 << h) && a[id] < max(a[id<<1], a[(id<<1)|1])) {↵
if(a[id<<1] > a[(id << 1) | 1]) {↵
swap(a[id<<1], a[id]);↵
id <<= 1;↵
}↵
else {↵
swap(a[(id<<1)|1], a[id]);↵
id = (id << 1) | 1;↵
}↵
qq[qn++] = id;↵
}↵
if(id < (1 << g)) {↵
for(int i = qn - 1; i > 0; i--) {↵
a[qq[i]] = a[qq[i - 1]];↵
}↵
a[qq[0]] = tmp;↵
return 0;↵
}↵
return tmp;↵
}↵
void solve() {↵
scanf("%d%d", &h, &g);↵
long long an = 0;↵
for(int i = 1; i < (1 << h); i++) {↵
scanf("%d", &a[i]);↵
an += a[i];↵
}↵
int need = 0;↵
for(int i = 1; i < (1 << g); i++) {↵
while(1) {↵
int v = pull(i);↵
if(v) {↵
an -= v;↵
ops[need++] = i;↵
}↵
else break;↵
}↵
}↵
printf("%lld\n", an);↵
for(int i = 0; i < need; i++) printf("%d%c", ops[i], " \n"[i == need - 1]);↵
}↵
int main(){↵
int T;↵
scanf("%d", &T);↵
while(T--) solve();↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:1329D]↵
↵
↵
<spoiler summary="author's code">↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
const int SIZE = 2e5+10;↵
char s[SIZE];↵
int cnt[SIZE];↵
int cc[26];↵
int all_cnt;↵
int ma;↵
void update_ma() {↵
while(ma > 0 && !cnt[ma]) ma--;↵
}↵
void dec1(int id) {↵
cnt[cc[id]]--;↵
cc[id]--;↵
cnt[cc[id]]++;↵
}↵
pair<int, int> stk[SIZE];↵
int sn;↵
int m;↵
int now;↵
int last_len;↵
void add(int i, bool flag) {↵
if(flag) {↵
dec1(stk[sn - 1].second);↵
dec1(stk[i].second);↵
all_cnt -= 2;↵
printf("%d %d\n",now + 1, now + stk[i].first);↵
update_ma();↵
sn--;↵
now -= stk[sn].first;↵
if(i + 1 < m) {↵
stk[i + 1].first += stk[sn].first;↵
}↵
else{↵
last_len += stk[sn].first;↵
}↵
}↵
else {↵
stk[sn++] = stk[i];↵
now += stk[i].first;↵
}↵
}↵
void solve(){↵
scanf("%s", s);↵
int n = strlen(s);↵
all_cnt = 0;↵
int lt = 0;↵
m = 0;↵
for(int i = 1; i < n; i++) {↵
if(s[i] == s[i - 1]) {↵
cc[s[i] - 'a']++;↵
all_cnt++;↵
stk[m++] = make_pair(i - lt, (int)(s[i] - 'a'));↵
lt = i;↵
}↵
}↵
last_len = n - lt;↵
ma = 0;↵
for(int i = 0; i < 26; i++) {↵
cnt[cc[i]]++;↵
ma = max(ma, cc[i]);↵
}↵
printf("%d\n", 1 + max(ma, (all_cnt + 1) / 2));↵
if(ma * 2 < all_cnt) {↵
sn = now = 0;↵
for(int i = 0; i < m; i++) {↵
add(i, sn && stk[sn - 1].second != stk[i].second && ma * 2 < all_cnt);↵
}↵
m = sn;↵
}↵
int main_id = -1;↵
for(int i = 0; i < 26; i++) {↵
if(cc[i] == ma) main_id = i;↵
}↵
sn = now = 0;↵
for(int i = 0; i < m; i++) {↵
add(i, sn && ((stk[sn - 1].second == main_id) ^ (stk[i].second == main_id)));↵
}↵
for(int i = 0; i < sn; i++) {↵
printf("%d %d\n",1 ,stk[i].first);↵
}↵
printf("%d %d\n", 1, last_len);↵
memset(cc, 0, sizeof(cc));↵
memset(cnt, 0, sizeof(int) * (ma + 1));↵
}↵
int main(){↵
int T;↵
scanf("%d", &T);↵
for(int i = 1; i <= T; i++) solve();↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
↵
[tutorial:1329E]↵
↵
<spoiler summary="isaf27's solution">↵
↵
~~~~~↵
//#pragma GCC optimize("O3")↵
#include <bits/stdc++.h>↵
↵
using namespace std;↵
↵
//defines↵
typedef long long ll;↵
typedef long double ld;↵
#define TIME clock() * 1.0 / CLOCKS_PER_SEC↵
#define prev _prev↵
#define y0 _y0↵
#define kill _kill↵
↵
//permanent constants↵
const ld pi = acos(-1.0);↵
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};↵
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};↵
const int dx[4] = {0, 1, 0, -1};↵
const int dy[4] = {1, 0, -1, 0};↵
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};↵
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};↵
const int alf = 26;↵
const int dig = 10;↵
const int two = 2;↵
const int th = 3;↵
const ll prost = 239;↵
const ll btc = 30;↵
const ld eps = 1e-8;↵
const ll INF = (ll)(1e18 + 239);↵
const int BIG = (int)(1e9 + 239);↵
const int MOD = 1e9 + 7; //↵
↵
//random↵
mt19937 rnd(239); //(chrono::high_resolution_clock::now().time_since_epoch().count());↵
↵
//constants↵
const int M = (int)(4e5 + 239);↵
const int N = (int)(2e3 + 239);↵
const int L = 20;↵
const int T = (1 << 18) + 239;↵
const int B = 500;↵
const int X = 210;↵
↵
ll up(ll a, ll b)↵
{↵
return (a + b - 1) / b;↵
}↵
↵
ll down(ll a, ll b)↵
{↵
return a / b;↵
}↵
↵
int m;↵
ll l[M], n, k;↵
↵
void solve()↵
{↵
cin >> n >> m >> k;↵
ll pr = 0;↵
for (int i = 0; i < m; i++)↵
{↵
ll p;↵
cin >> p;↵
l[i] = p - pr;↵
pr = p;↵
}↵
l[m++] = n - pr;↵
//for (int i = 0; i < m; i++)↵
// cerr << l[i] << " ";↵
//cerr << "\n";↵
ll sum = k + m;↵
ll bl = 1;↵
ll br = n + 1;↵
while (br - bl > 1)↵
{↵
ll h = (bl + br) >> 1LL;↵
ll cur = 0;↵
for (int i = 0; i < m; i++)↵
cur += down(l[i], h);↵
if (sum <= cur)↵
bl = h;↵
else↵
br = h;↵
}↵
ll MIN = bl;↵
bl = 0;↵
br = n + 1;↵
while (br - bl > 1)↵
{↵
ll h = (bl + br) >> 1LL;↵
ll cur = 0;↵
for (int i = 0; i < m; i++)↵
cur += up(l[i], h);↵
if (sum >= cur)↵
br = h;↵
else↵
bl = h;↵
}↵
ll MAX = br;↵
vector<pair<ll, ll>> ban;↵
for (int i = 0; i < m; i++)↵
{↵
if (up(l[i], MAX) <= down(l[i], MIN))↵
continue;↵
ll bl = 0;↵
ll br = MIN;↵
while (br - bl > 1)↵
{↵
ll h = (bl + br) >> 1LL;↵
if (down(l[i], h) == down(l[i], MIN))↵
br = h;↵
else↵
bl = h;↵
}↵
ll first = MIN - br;↵
bl = MAX;↵
br = INF;↵
while (br - bl > 1)↵
{↵
ll h = (bl + br) >> 1LL;↵
if (up(l[i], MAX) == up(l[i], h))↵
bl = h;↵
else↵
br = h;↵
}↵
ll second = bl - MAX;↵
ban.push_back(make_pair(first, second));↵
}↵
ll add = MAX - MIN;↵
if (ban.empty())↵
{↵
cout << add << "\n";↵
return;↵
}↵
sort(ban.begin(), ban.end());↵
vector<pair<ll, ll>> a;↵
for (int i = 0; i < (int)ban.size(); i++)↵
{↵
while (!a.empty() && a.back().second <= ban[i].second)↵
a.pop_back();↵
a.push_back(ban[i]);↵
}↵
ll ans = INF;↵
ans = min(a[0].second + 1, ans);↵
ans = min(a.back().first + 1, ans);↵
for (int i = 0; i + 1 < (int)a.size(); i++)↵
ans = min(ans, a[i].first + a[i + 1].second + 2);↵
cout << ans + add << "\n";↵
}↵
↵
int32_t main()↵
{↵
#ifdef ONPC↵
freopen("input", "r", stdin);↵
#endif↵
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);↵
int t;↵
cin >> t;↵
while (t--)↵
solve();↵
return 0;↵
}↵
↵
~~~~~↵
</spoiler>↵
↵
<spoiler summary="author's solution">↵
↵
~~~~~↵
/*{{{*/↵
#include<cstdio>↵
#include<cstdlib>↵
#include<cstring>↵
#include<cmath>↵
#include<algorithm>↵
#include<string>↵
#include<iostream>↵
#include<sstream>↵
#include<set>↵
#include<map>↵
#include<queue>↵
#include<bitset>↵
#include<vector>↵
#include<limits.h>↵
#include<assert.h>↵
#define SZ(X) ((int)(X).size())↵
#define ALL(X) (X).begin(), (X).end()↵
#define REP(I, N) for (int I = 0; I < (N); ++I)↵
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)↵
#define FOR(I, A, B) for (int I = (A); I <= (B); ++I)↵
#define FORS(I, S) for (int I = 0; S[I]; ++I)↵
#define RS(X) scanf("%s", (X))↵
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))↵
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())↵
#define CASET int ___T; scanf("%d", &___T); for(int cs=1;cs<=___T;cs++)↵
#define MP make_pair↵
#define PB push_back↵
#define MS0(X) memset((X), 0, sizeof((X)))↵
#define MS1(X) memset((X), -1, sizeof((X)))↵
#define LEN(X) strlen(X)↵
#define F first↵
#define S second↵
using namespace std;↵
typedef long long LL;↵
typedef unsigned long long ULL;↵
typedef long double LD;↵
typedef pair<int,int> PII;↵
typedef vector<int> VI;↵
typedef vector<LL> VL;↵
typedef vector<PII> VPII;↵
typedef pair<LL,LL> PLL;↵
typedef vector<PLL> VPLL;↵
template<class T> void _R(T &x) { cin >> x; }↵
void _R(int &x) { scanf("%d", &x); }↵
void _R(LL &x) { scanf("%lld", &x); }↵
void _R(double &x) { scanf("%lf", &x); }↵
void _R(char &x) { scanf(" %c", &x); }↵
void _R(char *x) { scanf("%s", x); }↵
void R() {}↵
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }↵
template<class T> void _W(const T &x) { cout << x; }↵
void _W(const int &x) { printf("%d", x); }↵
void _W(const LL &x) { printf("%lld", x); }↵
void _W(const double &x) { printf("%.16f", x); }↵
void _W(const char &x) { putchar(x); }↵
void _W(const char *x) { printf("%s", x); }↵
template<class T,class U> void _W(const pair<T,U> &x) {_W(x.F); putchar(' '); _W(x.S);}↵
template<class T> void _W(const vector<T> &x) { for (auto i = x.begin(); i != x.end(); _W(*i++)) if (i != x.cbegin()) putchar(' '); }↵
void W() {}↵
template<class T, class... U> void W(const T &head, const U &... tail) { _W(head); putchar(sizeof...(tail) ? ' ' : '\n'); W(tail...); }↵
#ifdef HOME↵
#define DEBUG(...) {printf("# ");printf(__VA_ARGS__);puts("");}↵
#else↵
#define DEBUG(...)↵
#endif↵
int MOD = 1e9+7;↵
void ADD(LL& x,LL v){x=(x+v)%MOD;if(x<0)x+=MOD;}↵
/*}}}*/↵
const int SIZE = 1e6+10;↵
LL a[SIZE],num[SIZE];↵
bool done[SIZE],be_added[SIZE];↵
int n;↵
LL K;↵
LL get_upper_bound(int i){↵
LL v=a[i]/num[i];↵
if(v*num[i]!=a[i])v++;↵
return v;↵
}↵
LL get_next_upper_bound(int i){↵
LL v=a[i]/(num[i]-1);↵
if(v*(num[i]-1)!=a[i])v++;↵
return v;↵
}↵
LL go(){↵
if(K==n){↵
return a[n-1]-a[0];↵
}↵
LL ll=1,rr=a[n-1];↵
while(ll<rr){↵
LL mm=(ll+rr)/2;↵
LL need=0;↵
REP(i,n){↵
if(a[i]>mm)need+=a[i]/mm;↵
else need++;↵
}↵
if(need>=K)ll=mm+1;↵
else rr=mm;↵
}↵
LL low=ll;↵
VPLL pp;↵
LL need=0;↵
REP(i,n){↵
if(a[i]>=low){↵
num[i]=a[i]/low;↵
pp.PB({a[i]/(num[i]+1),i});↵
}↵
else {↵
num[i]=1;↵
}↵
need+=num[i];↵
}↵
{↵
bool fail=0;↵
REP(i,n){↵
if(get_upper_bound(i)>=low){↵
fail=1;↵
break;↵
}↵
}↵
if(!fail) return low-1-min(low-1,a[0]);↵
}↵
REP(i,n){↵
if(a[i]>=low&&a[i]/(num[i]+1)==low-1){↵
LL v=a[i]/(low-1);↵
LL mi=min(v-num[i],K-need);↵
num[i]+=mi;↵
need+=mi;↵
if(need==K)break;↵
}↵
}↵
priority_queue<PLL,VPLL,greater<PLL>>added;↵
priority_queue<PLL>top;↵
LL mi=a[0];↵
REP(i,n){↵
if(num[i]>1){↵
added.push(MP(get_next_upper_bound(i),i));↵
}↵
top.push(MP(get_upper_bound(i),i));↵
mi=min(mi,a[i]/num[i]);↵
}↵
LL an=top.top().F-mi;↵
REP(i,n)be_added[i]=done[i]=0;↵
LL ma=low-1;↵
while(!top.empty()&&!added.empty()){↵
int id=top.top().S;↵
if(be_added[id])break;↵
done[id]=1;↵
top.pop();↵
num[id]++;↵
mi=min(mi,a[id]/num[id]);↵
auto tmp=added.top();↵
added.pop();↵
if(done[tmp.S])break;↵
be_added[tmp.S]=1;↵
num[tmp.S]--;↵
if(num[tmp.S]>1)added.push({get_next_upper_bound(tmp.S),tmp.S});↵
ma=max(ma,tmp.F);↵
an=min(an,(top.empty()?ma:max(ma,top.top().F))-mi);↵
}↵
return an;↵
}↵
void solve(){↵
LL L;↵
//R(n,K,L);↵
R(L,n,K);↵
K+=n+1;↵
LL lt=0;↵
REP(i,n){↵
LL x;↵
R(x);↵
a[i]=x-lt;↵
lt=x;↵
}↵
a[n++]=L-lt;↵
sort(a,a+n);↵
W(go());↵
}↵
int main(){↵
CASET{↵
solve();↵
}↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
