style: format markdown files with remark-lint

diff --git a/docs/ds/persistent-in-seg.md b/docs/ds/persistent-in-seg.md
index 24db2a8..d9eaa9e 100644 (file)
--- a/docs/ds/persistent-in-seg.md
+++ b/docs/ds/persistent-in-seg.md
@@ -23,67 +23,69 @@
 给出一种代码实现：
 
 ```cpp
-#include<cstdio>
-#include<cstring>
-#include<algorithm>
-#include<map>
-#include<set>
-#define LC o<<1
-#define RC o<<1|1
+#include <algorithm>
+#include <cstdio>
+#include <cstring>
+#include <map>
+#include <set>
+#define LC o << 1
+#define RC o << 1 | 1
 using namespace std;
-const int maxn=1000010;
-int n,m,a[maxn],u[maxn],x[maxn],l[maxn],r[maxn],k[maxn],cur,cur1,cur2,q1[maxn],q2[maxn],v[maxn];
+const int maxn = 1000010;
+int n, m, a[maxn], u[maxn], x[maxn], l[maxn], r[maxn], k[maxn], cur, cur1, cur2,
+    q1[maxn], q2[maxn], v[maxn];
 char op[maxn];
-set<int>ST;
-map<int,int>mp;
-struct segment_tree//封装的动态开点权值线段树 
+set<int> ST;
+map<int, int> mp;
+struct segment_tree  //封装的动态开点权值线段树
 {
-    int cur,rt[maxn*4],sum[maxn*60],lc[maxn*60],rc[maxn*60];
-    void build(int&o){o=++cur;}
-    void print(int o,int l,int r)
-    {
-        if(!o)return;
-        if(l==r&&sum[o])printf("%d ",l);
-        int mid=(l+r)>>1;
-        print(lc[o],l,mid);
-        print(rc[o],mid+1,r);
-    }
-    void update(int&o,int l,int r,int x,int v)
-    {
-        if(!o)o=++cur;
-        sum[o]+=v;
-        if(l==r)return;
-        int mid=(l+r)>>1;
-        if(x<=mid)update(lc[o],l,mid,x,v);
-        else update(rc[o],mid+1,r,x,v);
-    }
-}st;
-//树状数组实现 
-inline int lowbit(int o){return(o&(-o));}
-void upd(int o,int x,int v){for(;o<=n;o+=lowbit(o))st.update(st.rt[o],1,n,x,v);}
-void gtv(int o,int*A,int&p){p=0;for(;o;o-=lowbit(o))A[++p]=st.rt[o];}
-int qry(int l,int r,int k)
-{
-    if(l==r)return l;
-    int mid=(l+r)>>1,siz=0;
-    for(int i=1;i<=cur1;i++)siz+=st.sum[st.lc[q1[i]]];
-    for(int i=1;i<=cur2;i++)siz-=st.sum[st.lc[q2[i]]];
-    //printf("j %d %d %d %d\n",cur1,cur2,siz,k);
-    if(siz>=k)
-    {
-        for(int i=1;i<=cur1;i++)q1[i]=st.lc[q1[i]];
-        for(int i=1;i<=cur2;i++)q2[i]=st.lc[q2[i]];
-        return qry(l,mid,k);
-    }
+  int cur, rt[maxn * 4], sum[maxn * 60], lc[maxn * 60], rc[maxn * 60];
+  void build(int& o) { o = ++cur; }
+  void print(int o, int l, int r) {
+    if (!o) return;
+    if (l == r && sum[o]) printf("%d ", l);
+    int mid = (l + r) >> 1;
+    print(lc[o], l, mid);
+    print(rc[o], mid + 1, r);
+  }
+  void update(int& o, int l, int r, int x, int v) {
+    if (!o) o = ++cur;
+    sum[o] += v;
+    if (l == r) return;
+    int mid = (l + r) >> 1;
+    if (x <= mid)
+      update(lc[o], l, mid, x, v);
     else
-    {
-        for(int i=1;i<=cur1;i++)q1[i]=st.rc[q1[i]];
-        for(int i=1;i<=cur2;i++)q2[i]=st.rc[q2[i]];
-        return qry(mid+1,r,k-siz);
-    }
+      update(rc[o], mid + 1, r, x, v);
+  }
+} st;
+//树状数组实现
+inline int lowbit(int o) { return (o & (-o)); }
+void upd(int o, int x, int v) {
+  for (; o <= n; o += lowbit(o)) st.update(st.rt[o], 1, n, x, v);
+}
+void gtv(int o, int* A, int& p) {
+  p = 0;
+  for (; o; o -= lowbit(o)) A[++p] = st.rt[o];
+}
+int qry(int l, int r, int k) {
+  if (l == r) return l;
+  int mid = (l + r) >> 1, siz = 0;
+  for (int i = 1; i <= cur1; i++) siz += st.sum[st.lc[q1[i]]];
+  for (int i = 1; i <= cur2; i++) siz -= st.sum[st.lc[q2[i]]];
+  // printf("j %d %d %d %d\n",cur1,cur2,siz,k);
+  if (siz >= k) {
+    for (int i = 1; i <= cur1; i++) q1[i] = st.lc[q1[i]];
+    for (int i = 1; i <= cur2; i++) q2[i] = st.lc[q2[i]];
+    return qry(l, mid, k);
+  } else {
+    for (int i = 1; i <= cur1; i++) q1[i] = st.rc[q1[i]];
+    for (int i = 1; i <= cur2; i++) q2[i] = st.rc[q2[i]];
+    return qry(mid + 1, r, k - siz);
+  }
 }
 /*
-线段树实现 
+线段树实现
 void build(int o,int l,int r)
 {
     st.build(st.rt[o]);
@@ -135,38 +137,36 @@ int query(int l,int r,int k)
     }
 }
 */
-int main()
-{
-    scanf("%d%d",&n,&m);
-    for(int i=1;i<=n;i++)scanf("%d",a+i),ST.insert(a[i]);
-    for(int i=1;i<=m;i++)
-    {
-       scanf(" %c",op+i);
-       if(op[i]=='C')scanf("%d%d",u+i,x+i),ST.insert(x[i]);
-       else scanf("%d%d%d",l+i,r+i,k+i);
-    }
-    for(set<int>::iterator it=ST.begin();it!=ST.end();it++)mp[*it]=++cur,v[cur]=*it;
-    for(int i=1;i<=n;i++)a[i]=mp[a[i]];
-    for(int i=1;i<=m;i++)if(op[i]=='C')x[i]=mp[x[i]];
-    n+=m;
-    //build(1,1,n);
-    for(int i=1;i<=n;i++)upd(i,a[i],1);
-    //print(1,1,n);
-    for(int i=1;i<=m;i++)
-    {
-       if(op[i]=='C')
-       {
-               upd(u[i],a[u[i]],-1);
-               upd(u[i],x[i],1);
-               a[u[i]]=x[i];
-        }
-        else
-        {
-            gtv(r[i],q1,cur1);
-            gtv(l[i]-1,q2,cur2);
-            printf("%d\n",v[qry(1,n,k[i])]);
-        }
+int main() {
+  scanf("%d%d", &n, &m);
+  for (int i = 1; i <= n; i++) scanf("%d", a + i), ST.insert(a[i]);
+  for (int i = 1; i <= m; i++) {
+    scanf(" %c", op + i);
+    if (op[i] == 'C')
+      scanf("%d%d", u + i, x + i), ST.insert(x[i]);
+    else
+      scanf("%d%d%d", l + i, r + i, k + i);
+  }
+  for (set<int>::iterator it = ST.begin(); it != ST.end(); it++)
+    mp[*it] = ++cur, v[cur] = *it;
+  for (int i = 1; i <= n; i++) a[i] = mp[a[i]];
+  for (int i = 1; i <= m; i++)
+    if (op[i] == 'C') x[i] = mp[x[i]];
+  n += m;
+  // build(1,1,n);
+  for (int i = 1; i <= n; i++) upd(i, a[i], 1);
+  // print(1,1,n);
+  for (int i = 1; i <= m; i++) {
+    if (op[i] == 'C') {
+      upd(u[i], a[u[i]], -1);
+      upd(u[i], x[i], 1);
+      a[u[i]] = x[i];
+    } else {
+      gtv(r[i], q1, cur1);
+      gtv(l[i] - 1, q2, cur2);
+      printf("%d\n", v[qry(1, n, k[i])]);
     }
-    return 0;
+  }
+  return 0;
 }
-``` 
+```