#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

namespace dts
{
    struct st_node
    {
        int l,r,mid,suml,sumr;
    };
    
    int n,m;
    st_node st[(1<<18)+2];
    #define lc(now) ((now)<<1)
    #define rc(now) (((now)<<1)|1)
    
    void st_build(int now,int l,int r)
    {
        st[now].l=l,st[now].r=r;
        st[now].suml=st[now].sumr=0;
        if (l<r)
        {
            st[now].mid=(l+r)>>1;
            st_build(lc(now),l,st[now].mid);
            st_build(rc(now),st[now].mid+1,r);
        }
    }
    
    void st_update(int now,int pos,int task)//l:task=1 r:task=-1
    {
        if (st[now].l==pos&&pos==st[now].r)
        {
            if (task==1)
                st[now].suml++;
            else if (task==-1)
                st[now].sumr--;
        }
        else
        {
            if (pos<=st[now].mid)
                st_update(lc(now),pos,task);
            else if (st[now].mid+1<=pos)
                st_update(rc(now),pos,task);
            if (task==1)
                st[now].suml=st[lc(now)].suml+st[rc(now)].suml;
            else if (task==-1)
                st[now].sumr=st[lc(now)].sumr+st[rc(now)].sumr;
        }
    }
    
    int st_ask(int now,int l,int r,int task)//l:task=1 r:task=-1
    {
        if (st[now].l==l&&r==st[now].r)
        {
            if (task==1)
                return st[now].suml;
            else if (task==-1)
                return st[now].sumr;
        }
        else
        {
            if (r<=st[now].mid)
                return st_ask(lc(now),l,r,task);
            else if (st[now].mid+1<=l)
                return st_ask(rc(now),l,r,task);
            else
                return st_ask(lc(now),l,st[now].mid,task)+st_ask(rc(now),st[now].mid+1,r,task);
        }
    }
    
    void main()
    {
        scanf("%d%d",&n,&m);
        st_build(1,1,n);
        for (int i=1;i<=m;i++)
        {
            int K,l,r;
            scanf("%d%d%d",&K,&l,&r);
            if (K==1)
            {
                st_update(1,l,1);
                if (r+1<=n)
                    st_update(1,r+1,-1);
            }
            else if (K==2)
            {
                int suml=st_ask(1,1,r,1),sumr=st_ask(1,1,l,-1);
                printf("%d\n",suml+sumr);
            }
        }
    }
}

int main()
{
    dts::main();
}

简单的模板题
开两个树状数组维护即可

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=500040;
int a[maxn],b[maxn];
int n,m;
int lowbit(int x)
{
    return x&(-x);
}
void add1(int x,int y)
{
    while(x<=n)
    {
        a[x]+=y;
        x+=lowbit(x);
    }
}
void add2(int x,int y)
{
    while(x<=n)
    {
        b[x]+=y;
        x+=lowbit(x);
    }
}

int sum1(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=a[x];
        x-=lowbit(x);
    }
    return ans;
}
int sum2(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=b[x];
        x-=lowbit(x);
    }
    return ans;
}
int main(){
    cin>>n>>m;
    while(m--)
    {
        int wzx,x,y;
        scanf("%d",&wzx);
        if(wzx==1)
        {
            scanf("%d %d",&x,&y);
            add1(x,1);
            add2(y,1);
        }
        if(wzx==2)
        {
            scanf("%d%d",&x,&y);
            printf("%d\n",sum1(y)-sum2(x-1));
        }
    }
    return 0;
}

#include<stdio.h>

int main()

{

int L, M, i, j, n;

int a[10001], b[10001];

scanf("%d %d",&L, &M); //输入L和M

n = M*2;//循环输入b数组0~n的数据

for(i=0; i<n; i+=2)

{

scanf("%d %d", &b[i], &b[i+1]);

}

for(i=0; i<=L; i++) //循环给a数组L个元素赋值

{

a[i] = i;

}

int r, s;

for(i=0; i<n; i+=2) //遍历访问数组b的各个区间

{

r = b[i]; //区间起始点

s = b[i+1]; //区间终点

for(j=r; j<=s; j++) //把数组b各个区间内元素在数组a中映射为0

{

a[j] = -1;

}

}

int k=0; 

for(i=0; i<=L; i++)

{

if(a[i] != -1)

{

k++; 

}

}

printf("%d", k);

return 0;

}

这题看起来和树状数组没什么关系，不过我们通过一定的转化，可以利用树状数组很好地解决这个问题。

我们不妨把所有线段的端点看成括号序列，即把询问的区间[lq,rq]看成在横坐标lq处的一个‘[’和rq处的‘]’, 即把插入的线段[li,ri]看成在横坐标li处的一个‘(’和ri处的‘)’。

稍作分析，我们不难发现，最后的答案等于‘]’左边‘(’的个数减去‘[’左边‘)’的个数。

那么我们现在做的就是对某个点做修改，对某个前缀求和。我们就可以很容易想到树状数组的做法：

1.建立两个树状数T1和T2，分别维护‘(’和‘)’。

2.若K=1，读入li,ri，plusT1(li,1)，plusT2(ri,1)。

3.若K=2，读入lq,rq，sumT1(rq)-sumT2(lq-1)

#include<bits/stdc++.h>
using namespace std;
int a[500005],b[500005],n,m;
int lowbit(int x){return x&(-x);}
void add(int x,int y,int *k){while(x<=n){k[x]+=y;x+=lowbit(x);}}
int sum(int x,int *k){int ans=0;while(x>0){ans+=k[x];x-=lowbit(x);}return ans;}
int main(){
cin>>n>>m;
while(m--){
int opt,x,y;
cin>>opt;
if(opt==1){
cin>>x>>y;
add(x,1,a);
add(y,1,b);
}else{
cin>>x>>y;
cout<<sum(y,a)-sum(x-1,b)<<endl;
}
}
return 0;
}

#include<bits/stdc++.h>
using namespace std;int n,m,k,l,r;int a1[50005],a2[50005];int c(int x){return x&-x;}void update1(int x,int v){while(x<=n){a1[x]+=v;x+=c(x);}}void update2(int x,int v){while(x<=n){a2[x]+=v;x+=c(x);}}int sum1(int x){int res=0;while(x>0){res+=a1[x];x-=c(x);}return res;}int sum2(int x){int res=0;while(x>0){res+=a2[x];x-=c(x);}return res;}int main(){cin>>n>>m;for(int i=1;i<=m;i++){cin>>k>>l>>r;if(k==1)update1(l,1),update2(r,1);else cout<<sum1(r)-sum2(l-1)<<endl;}}
自己看吧

参考https://www.cnblogs.com/shadowland/p/5870395.html
这篇写的很好
用树状数组和线段树都可以，只要可以维护区间加法查询就可以了 - 原理是用两个数据结构分别维护每个树种区间的左端点和右端点，
设查询的区间为tl, tr
一个查询区间里存在的树种的个数（多少个不同的树种区间在我们的区间里）= [1, tr]上左端点的个数 - [1, tl - 1]上右端点的个数。这样就是所有[1, tr]上的区间里有所覆盖的树区间，去掉 只出现在[1, tl - 1]上的树区间，剩下的区间就是所有涉及到[tl, tr]的树区间辣！压力马斯内！（赞赏）

经验教训
1. 又没有好好读题，只有单个更新和区间查询，和懒更新毛线的关系都没有，不要多写。。。
2. 刚开始Runtime Error,后来一想线段树至少要开数据规模x4吧，居然没写。。。
3. 快读写了居然用了cin??? 哈 哈。对比一下快了20多ms，快读还是狠滴

代码

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

int n, m;

int read() {
    int f = 1, x = 0;
    char c; c = getchar();
    if (c == '-') { f = -1; c = getchar(); }
    while ('0' <= c && c <= '9') {
        x *= 10;
        x += c - '0';
        c = getchar();
    }
    return x * f;
}

struct Node {
    int l, r, val;
    Node& operator++() { 
        val++; 
        return *this;
    }
};

const int MAXN = (5e4 + 500) * 10;

struct SegmentTree {
    int cnt, root;
    Node nodes[MAXN];

    SegmentTree() {
        getNode(root);
    }

    int getNode(int& x) {
        if (!x) { return (x = ++cnt); }
        return x;
    }

    bool exist(int x) {
        if (!x) return false;
        return true;
    }

    void add(int p, int t, int l, int r) {
        Node& curr = nodes[p];
        if (r < t || t < l || r < l) { return; }
        if (l == r && l == t) { ++curr; return; }
        else {
            int mid = (l + r) / 2;
            if (t <= mid) { add(getNode(curr.l), t, l, mid); }
            else if (t > mid) { add(getNode(curr.r), t, mid + 1, r); }
        }
        curr.val = nodes[curr.l].val + nodes[curr.r].val;
    }

    int query(int p, int tl, int tr, int l, int r) {
        Node& curr = nodes[p];
        if (tl <= l && r <= tr) return curr.val;
        if (r < tl || tr < l || !exist(p) || r < l) return 0;
        else {
            int mid = (l + r) / 2, sum = 0;
            if (tl <= mid) { sum += query(curr.l, tl, tr, l, mid); }
            if (tr > mid) { sum += query(curr.r, tl, tr, mid + 1, r); }
            return sum;
        }
    }

} lb, rb; //left bracket, right bracket

int main() {
    n = read(); m = read();
    for (int i = 1; i <= m; i++) {
        int c, param1, param2;
        c = read(); param1 = read(); param2 = read();
        if (c == 1) {
                lb.add(lb.root, param1, 1, n);
                rb.add(rb.root, param2, 1, n);
        } else {
            cout << lb.query(lb.root, 1, param2, 1, n) - rb.query(rb.root, 1, param1 - 1, 1, n) << endl;
        }
    }
    return 0;
}

树状数组解

#include <iostream>
#include <cstdio>
#define ll long long
#define rll register long long
using namespace std;
const int MAXN=5e4+5;
ll n,m,c1[MAXN],c2[MAXN];
inline ll lowbit(rll i){return i&(-i);}
inline void update1(rll x,rll value)
{
    for(rll i=x;i<=n;i+=lowbit(i))
        c1[i]+=value;
}
inline ll find1(rll x)
{
    rll res=0;
    for(rll i=x;i;i-=lowbit(i))
        res+=c1[i];
    return res;
}
inline void update2(rll x,rll value)
{
    for(rll i=x;i<=n;i+=lowbit(i))
        c2[i]+=value;
}
inline ll find2(rll x)
{
    rll res=0;
    for(rll i=x;i;i-=lowbit(i))
        res+=c2[i];
    return res;
}
int main()
{
    scanf("%lld%lld",&n,&m);
    for(rll i=1,k,l,r;i<=m;i++)
    {
        scanf("%lld%lld%lld",&k,&l,&r);
        if(k==1)
        {
            update1(l,1);
            update2(r,1);
        }
        else
            printf("%lld\n",find1(r)-find2(l-1));
    }
    return 0；
}

左右括号法
线段树||树状数组......
不存在的
#include<bits/stdc++.h>
#define F(i,j) for(int i=1;i<=j;i++)
using namespace std;
int n,m,l[50001]={0},r[50001]={0},k,x,y;;
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++){
scanf("%d%d%d",&k,&x,&y);
if(k==1)l[min(x,y)]++,r[max(x,y)]++;
else{
int ans=0,lh=0,rh=0;
F(j,x-1)lh+=r[j];
F(j,y)rh+=l[j];
ans=rh-lh;
printf("%d\n",ans);
}
}
}

JRX2015U43：树状数组。
http://blog.csdn.net/qq_31640513/article/details/61435432

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;

const int N = 50000 + 5;

int n, m, L[N], R[N];

int main() {
scanf("%d%d", &n, &m);
while (m--) {
int k, l, r;
scanf("%d%d%d", &k, &l, &r);
if (k == 1) {
for (int x = r; x <= n; x += x & -x) R[x]++;
for (int x = l; x <= n; x += x & -x) L[x]++;
} else {
int ans = 0;
for (int x = r; x; x -= x & -x) ans += L[x];
for (int x = l - 1; x; x -= x & -x) ans -= R[x];
printf("%d\n", ans);
}
}
return 0;
}

#include <bits/stdc++.h>
#define maxN 200010 

using namespace std ;
typedef long long QAQ ;

struct Tree{int l , r ; QAQ liml , limr ;};

Tree tr[maxN<<2];

void Build_Tree ( int x , int y , int i ){
    tr[i].l = x ;
    tr[i].r = y ;
    if( x==y )return ;
    else {
        QAQ mid = (tr[i].l + tr[i].r) >> 1 ;
        Build_Tree ( x , mid , i<<1);
        Build_Tree ( mid+1 , y , i<<1|1);
        return ;
    }
}

void Update_left ( int w , int i ){
    if( w==tr[i].l && w==tr[i].r )tr[i].liml++;
    else {
        QAQ mid = (tr[i].l + tr[i].r) >> 1 ;
        if( w>mid )Update_left( w , i<<1|1);else
        if( w<=mid)Update_left( w , i<<1 );
        tr[i].liml = tr[i<<1].liml + tr[i<<1|1].liml ;
    }
}

void Update_right ( int w , int i ){
    if( w==tr[i].l && w==tr[i].r )tr[i].limr++;
    else {
        QAQ mid = (tr[i].l + tr[i].r) >> 1 ;
        if( w>mid )Update_right( w , i<<1|1);else
        if( w<=mid)Update_right( w , i<<1 );
        tr[i].limr = tr[i<<1].limr + tr[i<<1|1].limr ;
    }
}

QAQ Query_left ( int q ,int w ,int i ){
    if( q<=tr[i].l && w>=tr[i].r )return tr[i].liml ;
    else {
        QAQ mid = (tr[i].l + tr[i].r) >> 1 ;
        if ( q>mid )return Query_left ( q , w , i<<1|1);
        else if ( w<=mid ) return Query_left ( q , w , i<<1);
        else return Query_left ( q , w , i<<1|1)+Query_left ( q , w , i<<1);
    }
}

QAQ Query_right ( int q ,int w ,int i ){
    if( q<=tr[i].l && w>=tr[i].r )return tr[i].limr ;
    else {
        QAQ mid = (tr[i].l + tr[i].r) >> 1 ;
        if ( q>mid )return Query_right ( q , w , i<<1|1);
        else if ( w<=mid ) return Query_right ( q , w , i<<1);
        else return Query_right ( q , w , i<<1|1)+Query_right ( q , w , i<<1);
    }
}

int main(){
    int N,M,op,ll,rr ;
    scanf("%d %d",&N,&M);
    Build_Tree ( 1 , N , 1 ) ;
    while(M--){
        scanf("%d%d%d",&op,&ll,&rr);
        if( op==1 ){
            Update_left ( ll , 1);
            Update_right ( rr , 1 );
        }
        else {
            QAQ ans = Query_left( 1 , rr , 1);
            if (ll!=1)ans-=Query_right(1 , ll-1 , 1);
            printf("%I64d\n",ans);
        }
    }
    return 0 ;
}

Horizontal Rule

线段树括号法

代码

var
l,r:array[1..50001]of longint;
n,m,i,j,x,y,k:longint;

begin
readln(n,m);
for i:=1to m do
begin
read(k);
if k=1 then begin
readln(x,y);
for j:=1to x do inc(l[j]);
for j:=1to y do inc(r[j]);
end
else begin
readln(x,y);
writeln(r[x]-l[y+1]);
end;
end;
end.

虽然说可以用线段树做，但还是树状数组比较短
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<iostream>
#include<string>
int t1[50001],t2[50001],n,m,k,l,r,ans;
int main()
{
scanf("%d%d",&n,&m);
while(m--)
{
scanf("%d%d%d",&k,&l,&r);
if(k==1)
{
for(int i=l;i<=n;i+=(i&((~i)+1)))
++t1[i];
for(int i=r;i<=n;i+=(i&((~i)+1)))
++t2[i];
}
else
{
ans=0;
for(int i=r;i;i-=(i&((~i)+1)))
ans+=t1[i];
for(int i=l-1;i;i-=(i&((~i)+1)))
ans-=t2[i];
printf("%d\n",ans);
}
}
return 0;
}

树状数组还是很牛逼的

记录信息
评测状态 Accepted
题目 P1448 校门外的树
递交时间 2015-10-11 16:36:36
代码语言 C++
评测机 VijosEx
消耗时间 1012 ms
消耗内存 916 KiB
评测时间 2015-10-11 16:36:37
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 916 KiB, score = 1
测试数据 #1: Accepted, time = 0 ms, mem = 916 KiB, score = 1
测试数据 #2: Accepted, time = 62 ms, mem = 916 KiB, score = 1
测试数据 #3: Accepted, time = 78 ms, mem = 916 KiB, score = 1
测试数据 #4: Accepted, time = 156 ms, mem = 916 KiB, score = 1
测试数据 #5: Accepted, time = 140 ms, mem = 916 KiB, score = 1
测试数据 #6: Accepted, time = 140 ms, mem = 916 KiB, score = 1
测试数据 #7: Accepted, time = 140 ms, mem = 916 KiB, score = 1
测试数据 #8: Accepted, time = 140 ms, mem = 912 KiB, score = 1
测试数据 #9: Accepted, time = 156 ms, mem = 916 KiB, score = 1
Accepted, time = 1012 ms, mem = 916 KiB, score = 10
代码
#include <iostream>
#include <stdio.h>
using namespace std;
int h[50010],t[50010];
int n,k;
void add(int a[],int k)
{
while(k<=n){
a[k]++;
k+=k&(-k);
}
}
int search(int a[],int k)
{
int tot=0;
while(k){
tot+=a[k];
k-=k&(-k);
}
return tot;
}
int main()
{

scanf("%d%d",&n,&k);
for(int i=1;i<=k;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
if(a==1){
add(h,b);
add(t,c);
}
else printf("%d\n",search(h,c)-search(t,b-1));
}
return 0;
}

#include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;

int n , m;
int sta;
int t , p;
int i;
int a[ 50000 + 10 ];
int b[ 50000 + 10 ];

int lowbit( int x )
{
return x & ( -x );
}

void modify( int q , int delta )
{
while( q <= n )
{
a[ q ] += delta;
q += lowbit( q );
}
return;
}

void add( int q , int delta )
{
while( q <= n )
{
b[ q ] += delta;
q += lowbit( q );
}
return;
}

int sum( int q )
{
int ans = 0;
while( q > 0 )
{
ans += a[ q ];
q -= lowbit( q );
}
return ans;
}

int value( int q )
{
int ans = 0;
while( q > 0 )
{
ans += b[ q ];
q -= lowbit( q );
}
return ans;
}

int main()
{
scanf( "%d %d" , &n , &m );
for( i = 0 ; i < m ; i++ )
{
scanf( "%d %d %d" , &sta , &t , &p );
if( sta == 1 )
{
modify( t , 1 );
modify( p , -1 );
add( t , 1 );
}
else
printf( "%d\n" , sum( t - 1 ) + value( p ) - value( t - 1 ) );
}
return 0;
}

var
a:array[1..50000]of longint;
n,m,i,j,s,b,c,max:longint;
begin
read(n,m);
fillchar(a,sizeof(a),0);
for i:=1 to m do
begin
read(s,b,c);
if s=1 then for j:=b to c do inc(a[j])
else begin
max:=0;
for j:=b to c do
if a[j]>max then max:=a[j];
writeln(max);
end;
end;

求解为什么我暴力只过一个点其余的都WA了，是题意我搞错了吗？？？？
end.

测试数据 #0: Accepted, time = 0 ms, mem = 4192 KiB, score = 1

测试数据 #1: Accepted, time = 0 ms, mem = 4188 KiB, score = 1

测试数据 #2: Accepted, time = 140 ms, mem = 4192 KiB, score = 1

测试数据 #3: Accepted, time = 187 ms, mem = 4192 KiB, score = 1

测试数据 #4: Accepted, time = 546 ms, mem = 4188 KiB, score = 1

测试数据 #5: Accepted, time = 530 ms, mem = 4192 KiB, score = 1

测试数据 #6: Accepted, time = 530 ms, mem = 4188 KiB, score = 1

测试数据 #7: Accepted, time = 546 ms, mem = 4192 KiB, score = 1

测试数据 #8: Accepted, time = 546 ms, mem = 4192 KiB, score = 1

测试数据 #9: Accepted, time = 577 ms, mem = 4188 KiB, score = 1

Accepted, time = 3602 ms, mem = 4192 KiB, score = 10

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<math.h>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
int n,m,sz=0,l[500500][2];
int a,b,k;
void work()
{
int tem=0;
for(int i=0;i<sz;i++)
if(l[i][0]<=b&&l[i][1]>=a)tem++;
printf("%d\n",tem);
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&k,&a,&b);
if(k==2)work();
else l[sz++][0]=min(a,b),l[sz-1][1]=max(a,b);
}
//system("pause");
return 0;
}

数据太弱，纯暴力枚举每一次植树是否对所询问区间有影响

好慢！！~~~：
编译成功

测试数据 #0: Accepted, time = 15 ms, mem = 828 KiB, score = 1
测试数据 #1: Accepted, time = 0 ms, mem = 824 KiB, score = 1
测试数据 #2: Accepted, time = 0 ms, mem = 828 KiB, score = 1
测试数据 #3: Accepted, time = 15 ms, mem = 828 KiB, score = 1
测试数据 #4: Accepted, time = 15 ms, mem = 828 KiB, score = 1
测试数据 #5: Accepted, time = 0 ms, mem = 828 KiB, score = 1
测试数据 #6: Accepted, time = 3 ms, mem = 832 KiB, score = 1
测试数据 #7: Accepted, time = 0 ms, mem = 824 KiB, score = 1
测试数据 #8: Accepted, time = 3 ms, mem = 828 KiB, score = 1
测试数据 #9: Accepted, time = 15 ms, mem = 824 KiB, score = 1
Accepted, time = 66 ms, mem = 832 KiB, score = 10

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std ;

#define MAXN 50010
#define lowbit( x ) ( ( - x ) & x )

struct Bit {
int a[ MAXN ] , N ;
void Init( int _N ) {
N = _N ; memset( a , 0 , sizeof( a ) ) ;
}
void Add( int x , int y ) {
for ( int i = x ; i <= N ; i += lowbit( i ) ) a[ i ] += y ;
}
int Sum( int x ) {
int rec = 0 ;
for ( ; x ; x -= lowbit( x ) ) rec += a[ x ] ;
return rec ;
}
} b1 , b2 ;

int n , m ;

void getint( int &t ) {
int ch ;
for ( ch = getchar( ) ; ! ( ch >= '0' && ch <= '9' ) ; ch = getchar( ) ) ;
t = ch - '0' ;
for ( ch = getchar( ) ; ch >= '0' && ch <= '9' ; ch = getchar( ) ) t *= 10 , t += ch - '0' ;
}

void putint( int t ) {
if ( ! t ) putchar( '0' )
; else {
int ans[ 20 ] ;
ans[ 0 ] = 0 ;
for ( ; t ; t /= 10 ) ans[ ++ ans[ 0 ] ] = t % 10 ;
for ( int i = ans[ 0 ] ; i ; i -- ) putchar( '0' + ans[ i ] ) ;
}
putchar( '\n' );
}

int main( ) {
getint( n ) , getint( m ) ;
b1.Init( n ) , b2.Init( n ) ;
while ( m -- ) {
int x , y , z ;
getint( x ) , getint( y ) , getint( z ) ;
if ( x == 1 ) b1.Add( y , 1 ) , b2.Add( z , 1 ) ;
else putint( b1.Sum( z ) - b2.Sum( y - 1 ) ) ;
}
return 0 ;
}

线段树解决，记录suml与sumr