线段树-----C - A Simple Problem with Integers

You have N integers, A1, A2, … , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, … , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
“C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.
“Q a b” means querying the sum of Aa, Aa+1, … , Ab.

Output
You need to answer all Q commands in order. One answer in a line.

Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
4
55
9
15
Hint
The sums may exceed the range of 32-bit integers.

线段树模板题，区间更新时候那个延迟标记要注意，求区间和，每次val值要加上区间长度乘以addval值，求区间最大值，每次val值直接加上addval值即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
const int N = 2 * 1e5 + 5;

typedef struct Node{
    LL val,addval;
}Node;
Node tree[N << 2];
LL a[N];

void build(int root,int l,int r)
{
    tree[root].addval = 0;
    if(l == r){
        tree[root].val = a[l];
        return ;
    }
    int mid = (l + r) / 2;
    build(root * 2 + 1,l,mid);
    build(root * 2 + 2,mid + 1,r);
    tree[root].val = tree[root * 2 + 1].val + tree[root * 2 + 2].val;
}

void pushdown(int root,int m)
{
    if(tree[root].addval != 0){
        tree[root * 2 + 1].addval += tree[root].addval;
        tree[root * 2 + 2].addval += tree[root].addval;
        tree[root * 2 + 1].val += (m - (m >> 1)) * tree[root].addval;
        tree[root * 2 + 2].val += (m >> 1) * tree[root].addval;
        tree[root].addval = 0;
    }
}

void update(int root,int l,int r,int pl,int pr,int val)
{
    if(r < pl || l > pr){
        return ;
    }
    if(l >= pl && r <= pr){
        tree[root].addval += val;
        tree[root].val += (r - l + 1) * val;
        return ;
    }
    pushdown(root,r - l + 1);
    int mid = (l + r) / 2;
    if(pl <= mid){
        update(root * 2 + 1,l,mid,pl,pr,val);
    }
    if(pr > mid){
        update(root * 2 + 2,mid + 1,r,pl,pr,val);
    }
    tree[root].val = tree[root * 2 + 1].val + tree[root * 2 + 2].val;
}

LL query(int root,int l,int r,int pl,int pr)
{
    if(r < pl || l > pr){
        return 0;
    }
    if(l >= pl && r <= pr){
        return tree[root].val;
    }
    pushdown(root,r - l + 1);
    int mid = (l + r) / 2;
    LL sum = 0;
    if(pl <= mid){
        sum += query(root * 2 + 1,l,mid,pl,pr);
    }
    if(pr > mid){
        sum += query(root * 2 + 2,mid + 1,r,pl,pr);
    }
    return sum;
}

int main()
{
    int n,q;
    while(~scanf("%d %d",&n,&q))
    {
        for(int i = 1;i <= n;++i){
            scanf("%lld",&a[i]);
        }
        build(1,1,n);
        for(int i = 0;i < q;++i){
            char s[5];
            scanf("%s",s);
            if(s[0] == 'Q'){
                int x,y;
                scanf("%d %d",&x,&y);
                printf("%lld\n",query(1,1,n,x,y));
            }else{
                int x,y,z;
                scanf("%d %d %d",&x,&y,&z);
                update(1,1,n,x,y,z);
            }
        }
    }
    return 0;
}