普通莫队

P3901 数列找不同

Thinking

一定是用可以用莫队来写题，这点是不用质疑的，所以那就简单了，只需要判断每次询问的区间是否满足$r-l+1==num$就行了。

$Codin{g}_1$

莫队写法

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10;int n, m, a[N], num[N], ans[N], block, sum, k;struct Node {int l, r, id;// bool operator < (const Node & t) const {//   return r < t.r;// }
}query[N];bool cmp1(Node x, Node y) {return ((x.l / block) == (y.l / block)) ? x.r < y.r : x.l < y.l;
}//按照块排序bool cmp2(Node x, Node y) {return ((x.l / block) != (y.l / block)) ? x.l < y.l : ((x.l / block) & 1) ? x.r < y.r : x.r > y.r;
}//按照块的奇偶排序void add(int x) {num[a[x]]++;if(num[a[x]] == 1)  sum++;// sum += 2 * num[a[x]] - 1;
}void del(int x) {num[a[x]]--;if(num[a[x]] == 0)  sum--;// sum -= 2 * num[a[x]] + 1;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read();for(int i = 1; i <= n; i++) a[i] = read();block = sqrt(n);for(int i = 1; i <= m; i++) {query[i].l = read(), query[i].r = read();query[i].id = i;}sort(query + 1, query + 1 + m, cmp2);int l = 0, r = 0;for(int i = 1; i <= m; i++) {while(r > query[i].r) del(r--);while(r < query[i].r) add(++r);while(l < query[i].l) del(l++);while(l > query[i].l) add(--l);ans[query[i].id] = (sum == query[i].r - query[i].l + 1);}for(int i = 1; i <= m; i++)puts(ans[i] ? "Yes" : "No");return 0;
}

$Codin{g}_2$

树状数组写法。

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10;int n, m, a[N], pos[N], tree[N], ans[N];struct Node {int l, r, id;bool operator < (const Node & t) const {return r < t.r;}
}query[N];void add(int x, int value) {while(x < N) {tree[x] += value;x += x & (-x);}
}int ask(int x) {int ans = 0;while(x) {ans += tree[x];x -= x & (-x);}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read();for(int i = 1; i <= n; i++) a[i] = read();for(int i = 1; i <= m; i++) {query[i].l = read(), query[i].r = read();query[i].id = i;}sort(query + 1, query + 1 + m);int l = 1;for(int i = 1; i <= m; i++) {for(int j = l; j <= query[i].r; j++) {if(pos[a[j]])add(pos[a[j]], -1);pos[a[j]] = j;add(j, 1);}l = query[i].r + 1;ans[query[i].id] = (ask(query[i].r) - ask(query[i].l - 1) == query[i].r - query[i].l + 1);}for(int i = 1; i <= m; i++)puts(ans[i] ? "Yes" : "No");return 0;
}

D. Tree and Queries

Thinking

利用$dfs$序的特性，同一颗子树上的节点会连成一片，所以很好的将一个树上问题转换成了区间问题，这个时候就可以用上莫队来写了。

Coding

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e5 + 10;int head[N], to[N << 1], nex[N << 1], cnt = 1;
int a[N], n, m, block, sum[N], ans[N], Ans[N];
int pre[N], suc[N], rk[N], num;struct Node {int l, r, id, minn;Node(int _l = 0, int _r = 0, int _id = 0, int _minn = 0) :l(_l), r(_r), id(_id), minn(_minn) {}
}ask[N];bool cmp(Node a, Node b) {return ((a.l / block) != (b.l / block)) ? a.l < b.l : ((a.l / block) & 1) ? a.r < b.r : a.r > b.r;
}void add_edge(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void dfs(int rt, int f) {pre[rt] = ++num;rk[num] = rt;for(int i = head[rt]; i; i = nex[i]) {if(to[i] == f)  continue;dfs(to[i], rt);}suc[rt] = num;
}void add(int x) {Ans[++sum[a[x]]]++;
}void del(int x) {Ans[sum[a[x]]--]--;
}int main() {// freopen("in.txt", "r", stdin); // freopen("out.txt", "r", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read();block = sqrt(n);for(int i = 1; i <= n; i++)  a[i] = read();for(int i = 1; i < n; i++) {int x = read(), y = read();add_edge(x, y);add_edge(y, x);}dfs(1, 0);for(int i = 1; i <= m; i++) {int x = read(), y = read();ask[i] = Node(pre[x], suc[x], i, y);}sort(ask + 1, ask + 1 + m, cmp);int l = 0, r = 0;for(int i = 1; i <= m; i++) {while(r < ask[i].r) add(rk[++r]);while(r > ask[i].r) del(rk[r--]);while(l > ask[i].l) add(rk[--l]);while(l < ask[i].l) del(rk[l++]);ans[ask[i].id] = Ans[ask[i].minn];}for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);return 0;
}

D. Powerful array

Thinking

同样是一个区间问题，我们考虑用莫队如何维护。

假设当前数字$x$出现的次数是$n$次，有$su{m}_1=n^{2}x$。

我们假设其增加一则变成了$su{m}_2=(n+1{)}^{2}x=n^{2}x+(2n+1)x$，其增量是$(n<<1|1)x$

我们再假设其减一则变成了$su{m}_3=(n-1{)}^{2}x=n^{2}x-(2(n-1)+1)x$，其减少量是$((n-1)<<1|1)x$

由此我们就推出了答案变换的公式了。

Coding

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10;int n, m, block, a[N];
ll num[N], ans[N], now;struct Node {int l, r, id;bool operator < (const Node & t) const {return (l / block) != (t.l / block) ? l < t.l : ((l / block) & 1) ? r < t.r : r > t.r;}
}ask[N];void add(int x) {now += (num[a[x]]++ << 1 | 1) * a[x];
}
void del(int x) {now -= (--num[a[x]] << 1 | 1) * a[x];
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "r", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read();for(int i = 1; i <= n; i++) a[i] = read();block = sqrt(n);for(int i = 1; i <= m; i++) {ask[i].l = read(), ask[i].r = read();ask[i].id = i;}sort(ask + 1, ask + 1 + m);int l = 1, r = 0;for(int i = 1; i <= m; i++) {while(r < ask[i].r) add(++r);while(r > ask[i].r) del(r--);while(l < ask[i].l) del(l++);while(l > ask[i].l) add(--l);ans[ask[i].id] = now;}for(int i = 1; i <= m; i++) printf("%lld\n", ans[i]);return 0;
}

P1533 可怜的狗狗

Thinking

权值线段树加莫队。

题目给的数据表意不明，所以我们最好还是给$a_i$离散化后，再进行权值的插入删除操作。

这里的权值标记的是第$i$只够是否在区间里，如果在就标记为$1$，否则就删去这个点变成$0$，对每个区间都完成了相应的操作，接下来我们就是要去查询第$K$大的是什么，这里通过树状数组的前缀和性质对其进行二分，得到第$k$大的$id$，然后通过这个得到答案。

Coding

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 3e5 + 10;int a[N], b[N], id[N], n, m, block;
int tree[N], ans[N];struct Node {int l, r, k, id;Node(int _l = 0, int _r = 0, int _k = 0, int _id = 0) : l(_l), r(_r), k(_k), id(_id) {}void input() {l = read(), r = read(), k = read();}bool operator < (const Node & t) const {return (l / block) != (t.l / block) ? l < t.l : ((l / block) & 1) ? r < t.r : r > t.r;}
}ask[N];void update(int x, int value) {while(x <= n) {tree[x] += value;x += x & (-x);}
}int query(int x) {int ans = 0;while(x) {ans += tree[x];x -= x & (-x);}return ans;
}int find(int value) {int l = 1, r = n;while(l < r) {int mid = l + r >> 1;if(query(mid) < value) l = mid + 1;else r = mid;}return l;
}void del(int x) {if(id[x]) update(id[x], -1);//if(id[x])是为了放置树状数组死循环而加上了的。
}void add(int x) {if(id[x]) update(id[x], 1);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read(), block = sqrt(n);for(int i = 1; i <= n; i++) a[i] = b[i] = read();sort(b + 1, b + 1 + n);for(int i = 1; i <= n; i++) id[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;for(int i = 1; i <= m; i++) {ask[i].input();ask[i].id = i;}sort(ask + 1, ask + 1 + m);int l = 0, r = 0;for(int i = 1; i <= m; i++) {while(r > ask[i].r) del(r--);while(r < ask[i].r) add(++r);while(l < ask[i].l) del(l++);while(l > ask[i].l) add(--l);ans[ask[i].id] = find(ask[i].k);}for(int i = 1; i <= m; i++) printf("%d\n", b[ans[i]]);return 0;
}

Chika and Friendly Pairs

Thinking

这题与上一题的思想有异曲同工之处，离散化后莫队，通过插入，删除，区间查询操作，来得到我们最后的答案。

Coding

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 3e4 + 10;int a[N], b[N], tree[N], n, m, k, block, all;
ll sum, ans[N];struct Node {int l, r, id;bool operator < (const Node & t) const {return (l / block) != (t.l / block) ? l < t.l : ((l / block) & 1) ? r < t.r : r > t.r; }
}ask[N];void update(int x, int value) {while(x <= n) {tree[x] += value;x += x & (-x);}
}int query(int x) {int ans = 0;while(x) {ans += tree[x];x -= x & (-x);}return ans;
}void del(int x) {//删除点，以及对答案进行的影响操作。//我们不难发现出了这个点外在区间[b[x] - k, b[x] + k]里的点都会对答案进行贡献影响。//所以我们删除这个点后查询query(r - 1) - query(l - 1)就是对答案的减小影响。update(x, -1);int l = lower_bound(b + 1, b + 1 + all, b[x] - k) - b;//这个点应该不用多说。int r = upper_bound(b + 1, b + 1 + all, b[x] + k) - b;//r是第一个大于b[x] + k的，所以r - 1是最后一个 <= b[x] + k的点。sum -= query(r - 1) - query(l - 1);
}void add(int x) {update(x, 1);int l = lower_bound(b + 1, b + 1 + all, b[x] - k) - b;int r = upper_bound(b + 1, b + 1 + all, b[x] + k) - b;sum += query(r - 1) - query(l - 1) - 1;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read(), k = read(), block = sqrt(n);for(int i = 1; i <= n; i++) a[i] = b[i] = read();sort(b + 1, b + 1 + n);all = unique(b + 1, b + 1 + n) - (b + 1);for(int i = 1; i <=  n; i++)  a[i] = lower_bound(b + 1, b + 1 + all, a[i]) - b;for(int i = 1; i <= m; i++) {ask[i].l = read(), ask[i].r = read();ask[i].id = i;}sort(ask + 1, ask + 1 + m);int l = 1, r = 0;for(int i = 1; i <= m; i++) {while(r > ask[i].r) del(a[r--]);while(r < ask[i].r) add(a[++r]);while(l > ask[i].l) add(a[--l]);while(l < ask[i].l) del(a[l++]);ans[ask[i].id] = sum;}for(int i = 1; i <= m; i++) print(ans[i]), putchar('\n');return 0;
}

可修改莫队

P1903 [国家集训队]数颜色 / 维护队列

Thinking

无非就是加了一个时间标记变成了有三个条件束缚，分别对$l,r,t$三个变量进行排序，接下来就是莫队的基本操作了。

Coding

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(int x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e6 + 10;int a[N], num[N], ans[N], n, q_time, c_time, block, m, sum;struct Change {int pos, pre, cur;
}change[N];struct Query {int l, r, id, t;Query(int _l = 0, int _r = 0, int _id = 0, int _t = 0) : l(_l), r(_r), id(_id), t(_t) {}bool operator < (const Query & temp) const {return  (l / block) != (temp.l / block) ? l < temp.l : (r / block) != (temp.r / block) ? r < temp.r : t < temp.t;}
}ask[N];void add(int x) {sum += !num[a[x]]++;
}void del(int x) {sum -= !--num[a[x]];
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read();for(int i = 1; i <= n; i++)   a[i] = read();for(int i = 1; i <= m; i++) {char op = getchar();while(op != 'Q' && op != 'R') op = getchar();if(op == 'Q') {q_time++;ask[q_time].l = read(), ask[q_time].r = read();ask[q_time].id = q_time, ask[q_time].t = c_time;}else {c_time++;change[c_time].pos = read(), change[c_time].cur = read();change[c_time].pre = a[change[c_time].pos];a[change[c_time].pos] = change[c_time].cur;//这点一定要注意把a[pos]变成我们要改变的，因为后面可能再次改变这个点，所以他的pre应该是前一次已经改变后的状态。}}for(int i = c_time; i >= 1; i--)  a[change[i].pos] = change[i].pre;//注意从后向前恢复原状。block = ceil(exp((log(n) + log(q_time)) / 3));int l = 0, r = 0, t = 0;sort(ask + 1, ask + 1 + q_time);for(int i = 1; i <= q_time; i++) {while(r > ask[i].r) del(r--);while(r < ask[i].r) add(++r);while(l > ask[i].l) add(--l);while(l < ask[i].l) del(l++);while(t < ask[i].t) {t++;int pos = change[t].pos;if(l <= pos && pos <= r)//同时满足在l , r之间才能进行区间的筛选操作。del(pos);a[pos] = change[t].cur;//改变其值。if(l <= pos && pos <= r)add(pos);}while(t > ask[i].t) {int pos = change[t].pos;if(l <= pos && pos <= r)del(pos);a[pos] = change[t].pre;if(l <= pos && pos <= r)add(pos);t--;}ans[ask[i].id] = sum;}for(int i = 1; i <= q_time; i++) print(ans[i]), putchar('\n');return 0;
}

树上莫队

SP10707 COT2 - Count on a tree II

Thinking

设定两个数组$pre，suc$数组，$pr{e}_i$数组记录的是节点$i$第一次进入$dfs$的时间戳，$su{c}_i$记录的是节点$i$出$dfs$的时间戳。

如这个例子：

有节点权值及树的形状，满足如下

105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8

有欧拉序的排列如下。

1 4 8 8 4 3 7 7 6 6 5 5 3 2 2 1

求2， 8的间的，不同的元素。

有$lca(2,8)==1!=2,!=8$，有$pre[2]=15>pre[8]=3$，所以我们取区间$suc[8],pre[2]$也就是

{8 4 3 7 7 6 6 5 5 3 2}​，成对出现的舍弃，最后{8 4 2} + lca(2, 8)刚好是$2->8$的最短路。所以我们维护的东西就简单了。

再看一种情况，求1, 8之间，显然有$pre[1]<pre[8],lca(1,8)=1=mi{n}_{id}(pre[x],pre[y])$，所以选定区间$pre[1],pre[8]$即{1, 4, 8}得到了$1->8$的最短路。所以我们只需要特判$lca$就行了。

Coding

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_backusing namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e5 + 10;int a[N], b[N], id[N], ans[N], num[N], n, m, block, sum;
int fa[N], son[N], sz[N], dep[N], top[N], pre[N], suc[N], pos[N], tot;
int head[N], to[N], nex[N], cnt = 1;
bool flag[N];struct Node {int l, r, id, lca;bool operator < (const Node & t) const {return (l / block) != (t.l / block) ? l < t.l : (l / block) & 1 ? r < t.r : r > t.r;}
}ask[N];void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void dfs1(int rt, int f) {pre[rt] = ++tot;pos[tot] = rt;sz[rt] = 1, fa[rt] = f;dep[rt] = dep[f] + 1;for(int i = head[rt]; i; i = nex[i]) {if(to[i] == f)  continue;dfs1(to[i], rt);if(!son[rt] || sz[to[i]] > sz[son[rt]])son[rt] = to[i];sz[rt] += sz[to[i]];}suc[rt] = ++tot;pos[tot] = rt;
}void dfs2(int rt, int tp) {top[rt] = tp;if(!son[rt])  return ;dfs2(son[rt], tp);for(int i = head[rt]; i; i = nex[i]) {if(to[i] == fa[rt] || to[i] == son[rt]) continue;dfs2(to[i], to[i]);}
}int lca(int x, int y) {while(top[x] != top[y]) {if(dep[top[x]] < dep[top[y]]) swap(x, y);x = fa[top[x]];}return dep[x] < dep[y] ? x : y;
}void add(int x) {sum += (++num[a[x]] == 1);
}void del(int x) {sum -= (--num[a[x]] == 0);
}void calc(int x) {if(!flag[x])  add(x);else del(x);flag[x] ^= 1;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m =read(), block = sqrt(2 * n);for(int i = 1; i <= n; i++) a[i] = b[i] = read();sort(b + 1, b + 1 + n);int all = unique(b + 1, b + 1 + n) - (b + 1);for(int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + 1 + all, a[i]) - b;for(int i = 1; i < n; i++) {int x = read(), y = read();add(x, y);add(y, x);}dfs1(1, 0);dfs2(1, 1);for(int i = 1; i <= m; i++) {int x = read(), y = read();if(pre[x] > pre[y]) swap(x, y);ask[i].id = i, ask[i].lca = lca(x, y);if(ask[i].lca == x) {ask[i].l = pre[x];ask[i].r = pre[y];ask[i].lca = 0;}else {ask[i].l = suc[x];ask[i].r = pre[y];}}sort(ask + 1, ask + 1 + m);int l = 1, r = 0;for(int i = 1; i <= m; i++) {while(r > ask[i].r) calc(pos[r--]);while(r < ask[i].r) calc(pos[++r]);while(l > ask[i].l) calc(pos[--l]);while(l < ask[i].l) calc(pos[l++]);//lca不在区间里，所以add后需要马上del。if(ask[i].lca)      calc(pos[pre[ask[i].lca]]);ans[ask[i].id] = sum;if(ask[i].lca)      calc(pos[pre[ask[i].lca]]);}for(int i = 1; i <= m; i++) print(ans[i]), putchar('\n');return 0;
}