A Simple Problem with Integers

A Simple Problem with Integers

You have N integers, A1, A2, … , AN. You need to deal with two kindsof operations. One type of operation is to add some given number toeach 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.

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;

#define INF 0x3f3f3f3f
#define MAX_LEN 100005

ll add[MAX_LEN<<2];

struct node{
    int ln,rn; //记录左右节点
    ll sum; //节点的值
}tree[MAX_LEN*4];

void push_Up(int root)  //数据上浮更新
{
    tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;
}

void push_Down(int root,int length)  //数据下沉，传入下沉结点和下沉区间差值
{
    if(add[root]){		//如果root对应区间需要更新
        add[root<<1]+=add[root];		//左右子叶状态更新
        add[root<<1|1]+=add[root];
        tree[root<<1].sum+=add[root]*(length-(length>>1));		//左右子叶元素和更新
        tree[root<<1|1].sum+=add[root]*(length>>1);
        add[root]=0;		//标记root区间状态已更新
    }
}

void build_Tree(int root,int L,int R) //建树
{
    add[root]=0; 		//初始化所有结点对应区间均不需要更新
    tree[root].ln=L;
    tree[root].rn=R;
    if(tree[root].ln==tree[root].rn){		//最底层接收数据
        scanf("%I64d",&tree[root].sum);
        return;
    }
    int mid=(L+R)>>1;
    build_Tree(root<<1,L,mid);		//建立左子叶
    build_Tree(root<<1|1,mid+1,R);		//建立右子叶
    push_Up(root);		//建立后的子叶开始数据上浮
}

void update(int root,int left,int right,int num) //更新
{
    if(left<=tree[root].ln&&right>=tree[root].rn){  //当前区间完全处于待更新区间内
        add[root]+=num;
        tree[root].sum+=(long long int)num*(tree[root].rn-tree[root].ln+1);
        return;
    }
    push_Down(root,tree[root].rn-tree[root].ln+1);
    int mid=(tree[root].ln+tree[root].rn)>>1;
    if(right<=mid)   //如果待查找区间完全在当前区间左子叶部分
        update(root<<1,left,right,num);
    else if(left>mid)   //如果待查找区间完全在当前区间右子叶部分
        update(root<<1|1,left,right,num);
    else{    //如果待查找区间横跨当前区间
        update(root<<1,left,right,num);
        update(root<<1|1,left,right,num);
    }
    push_Up(root);
}

ll query(int root,int left,int right)  //查询
{
    if(left<=tree[root].ln&&right>=tree[root].rn)  //当前区间完全在查询区间范围内
        return tree[root].sum;   	//返回当前区间总和
    push_Down(root,tree[root].rn-tree[root].ln+1);		//由于是延迟更新，所以数据需要下沉一层
    int mid=(tree[root].ln+tree[root].rn)>>1;
    ll sum=0;
    if(right<=mid)		//待查找区间完全在当前区间左子叶部分
        sum+=query(root<<1,left,right);
    else if(mid<left)		//待查找区间完全在当前区间右子叶部分
        sum+=query(root<<1|1,left,right);
    else{		//待查找区间分别在区间左右子叶
        sum+=query(root<<1,left,right);
        sum+=query(root<<1|1,left,right);
    }
    return sum;
}

int main()
{
    char s[10];
    int n,q;
    int x,y,z;
    cin>>n>>q;
    build_Tree(1,1,n);
    while(q--){
        cin>>s;
        if(s[0]=='Q'){
            cin>>x>>y;
            cout<<query(1,x,y)<<endl;
        }
        else if(s[0]=='C'){
            cin>>x>>y>>z;
            update(1,x,y,z);
        }
    }
    return 0;
}