HDU3468:A Simple Problem with Integers(线段树区间更新+lazy)

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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

题意：Q是询问区间和，C是在区间内每个节点加上一个值

思路：这题如果直接更新到叶子节点会超时，所以只要找到一个适合的区间就用一个增量来记录它，当下一次询问时，如果这个范围正好合适询问的范围，就直接是这个节点的sum值加上这个区间长度*lnc，再加到总和上去，若这个节点的范围不适合所要查询的范围，那么就要查询它的子节点，这个时候再把增量传给她的子节点，并且清空父亲节点的增量。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 100000+10;
int n,m,sum;

struct node
{
    int l,r;
    __int64 n,sum;//sum为增量
} a[maxn<<2];

void init(int l,int r,int i)
{
    a[i].l = l;
    a[i].r = r;
    a[i].n = 0;
    a[i].sum = 0;
    if(l!=r)
    {
        int mid = (l+r)>>1;
        init(l,mid,2*i);
        init(mid+1,r,2*i+1);
    }
}

void insert(int i,int l,int r,__int64 m)
{
    a[i].n+=(r-l+1)*m;
    if(a[i].l >= l && a[i].r <= r)
        a[i].sum+=m;////若此节点所在区段被包含在要插入的区段中，就将插入值存在sum

    else
    {
        int mid = (a[i].l+a[i].r)>>1;
        if(r<=mid)
            insert(2*i,l,r,m);
        else if(l>mid)
            insert(2*i+1,l,r,m);
        else
        {
            insert(2*i,l,mid,m);
            insert(2*i+1,mid+1,r,m);
        }
    }
}

__int64 find(int i,int l,int r)
{
    if(a[i].l == l && a[i].r == r)
        return a[i].n;
    else
    {
        int mid = (a[i].l+a[i].r)>>1;
        if(a[i].sum)
        { //若上面if条件不成立，则要询问它的子节点，此时增量要下传，并且要更新其本身的sum;
            a[2*i].sum += a[i].sum;
            a[2*i].n+=a[i].sum*(a[2*i].r-a[2*i].l+1);
            a[2*i+1].sum += a[i].sum;
            a[2*i+1].n+=a[i].sum*(a[2*i+1].r-a[2*i+1].l+1);
            a[i].sum = 0;
        }
        if(r<=mid)
            return find(2*i,l,r);
        else if(l>mid)
            return find(2*i+1,l,r);
        else
        {
            return find(2*i,l,mid)+find(2*i+1,mid+1,r);
        }
    }
}

int main()
{
    int i,j,x,y;
    __int64 k;
    char str[5];
    while(~scanf("%d%d",&n,&m))
    {
        init(1,n,1);
        for(i = 1; i<=n; i++)
        {
            scanf("%I64d",&k);
            insert(1,i,i,k);
        }
        while(m--)
        {
            scanf("%s%d%d",str,&x,&y);
            if(str[0] == 'C')
            {
                scanf("%I64d",&k);
                insert(1,x,y,k);
            }
            else if(str[0] == 'Q')
                printf("%I64d\n", find(1,x,y));
        }
    }

    return 0;
}