简单线段树模板

为什么需要线段树

#include <stdio.h>
#include <iostream>
#include <string.h>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 5000;
int arr[6] = {2,6,8,9,5,3};
int size = 6;
int tree[15] = {0};

void build(int node,int start,int end){
    if(start == end){
        tree[node] = arr[start];
    }
    else{
        int mid = (start + end) / 2;
        int left_node  = 2 * node + 1;
        int right_node = 2 * node + 2;
        build(left_node,start,mid);
        build(right_node,mid + 1,end);
        tree[node] = tree[left_node] + tree[right_node];
    }


}

void update(int node,int start,int end,int idx,int val){
    if(start == end){
        tree[node] = val;
        arr[idx] = val;
        return;
    }
    else{
        int mid  = (start + end) / 2;
        int left_node  = 2 * node + 1;
        int right_node = 2 * node + 2;
        if(idx <= mid )update(left_node,start,mid,idx,val);
        else update(right_node,mid+1,end,idx,val);
        tree[node] = tree[left_node] + tree[right_node];
    }
}

int query(int node,int start,int end,int L,int R){
    printf("start = %d\n",start);
    printf("end = %d\n",end);
    printf("\n");
    if(R < start || L > end){
        return 0;
    }
    if(start == end){
        return tree[node];
    }
    else if(L <= start&& end <= R){
        return tree[node];
    }
    else{
        int mid = (start + end) / 2;
        int left_node  = 2 * node + 1;
        int right_node = 2 * node + 2;
        int sum_left  = query(left_node,start,mid,L,R);
        int sum_right = query(right_node,mid + 1,end,L,R);
        return sum_left + sum_right;
    }
}
int main()
{
    build(0,0,size-1);
    for(int i = 0; i < 15;++i){
        printf("tree[%d]:%d\n",i,tree[i]);
    }
    printf("\n");
    for(int i = 0; i < size;++i){
        printf("index:%d  val:%d\n",i,arr[i]);
    }
    update(0,0,size-1,3,4);
    for(int i = 0; i < 15;++i){
        printf("tree[%d]:%d\n",i,tree[i]);
    }

    printf("\n");
    for(int i = 0; i < size;++i){
        printf("index:%d  val:%d\n",i,arr[i]);
    }
    int r = query(0,0,size-1,2,5);
    printf("%d\n",r);

        return 0;
}

带懒惰标记的求和
（和懒加载的思想差不多，只有当你真正想要的时候，我再做，再返回给你）

#include <stdio.h>
#include <iostream>
#include <string.h>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
struct T{
    int L,R;
    ll x,lazy;
}tree[4 * maxn];

void pushup(int node){
    tree[node].x = tree[node<<1].x + tree[node<<1|1].x;
}
void pushdown(int node){
    if(tree[node].lazy){
        tree[node<<1].x   += (tree[node<<1].R - tree[node<<1].L +1) * tree[node].lazy;
        tree[node<<1|1].x += (tree[node<<1|1].R - tree[node<<1|1].L + 1) * tree[node].lazy;
        tree[node<<1].lazy += tree[node].lazy;
        tree[node<<1|1].lazy += tree[node].lazy;
        tree[node].lazy = 0;
    }
}
void build(int node,int L,int R){
    tree[node].L = L;
    tree[node].R = R;
    tree[node].lazy = 0;
    if(L == R){
        scanf("%lld",&tree[node].x);
        return;
    }
    int mid = (L+R)/2;
    build(node<<1,L,mid);
    build(node<<1|1,mid+1,R);
    pushup(node);
}

//区间修改
void update(int node,int L,int R,long long x){
    if(R < tree[node].L || L > tree[node].R)return;
    if(L <= tree[node].L && tree[node].R <= R){
        tree[node].x += (R - L)*x;
        tree[node].lazy += x;
        return;
    }
    pushdown(node);
    update(node<<1,L,R,x);
    update(node<<1|1,L,R,x);
    pushup(node);
}

long long query(int node,int L,int R){
    if(L <= tree[node].L && R >= tree[node].R){
        return tree[node].x;
    }
    if(R < tree[node].L || L > tree[node].R){
        return 0;
    }
    pushdown(node);
    return query(node<<1,L,R) + query(node<<2|1,L,R);

}

int main()
{
    build(1,1,6);

    for(int i = 0; i < 16;++i){
        printf("tree[%d]:%d\n",i,tree[i].x);
    }
    printf("\n");
    update(1,1,3,2);
    for(int i = 0; i < 16;++i){
        printf("tree[%d]:%d\n",i,tree[i].x);
    }
    long long r = query(1,1,2);
    printf("\n");
    for(int i = 0; i < 16;++i){
        printf("tree[%d]:%d\n",i,tree[i].x);
    }
    printf("%d\n",r);

        return 0;
}