[kuangbin带你飞]专题七 线段树——C（延迟标记）

C - A Simple Problem with Integers

  POJ - 3468 

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

Hint

The sums may exceed the range of 32-bit integers.

线段树延迟标记模板题

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e5+5;
long long sum[maxn*4];
long long lazy[maxn*4];
long long c;
int n,m,x,y;
string M;

void pushup(int rt)
{
    sum[rt]=sum[rt*2]+sum[rt*2+1];
}

void pushdown(int rt,int len)
{
    if(lazy[rt])
    {
        lazy[rt*2]+=lazy[rt];
        lazy[rt*2+1]+=lazy[rt];
        sum[rt*2]+=lazy[rt]*(len-len/2);
        sum[rt*2+1]+=lazy[rt]*(len/2);
        lazy[rt]=0;
    }
}

void build(int l,int r,int rt)
{
    lazy[rt]=0;
    if(l==r)
    {
        scanf("%lld",&sum[rt]);
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,2*rt);
    build(mid+1,r,2*rt+1);
    pushup(rt);
}

void update(int l,int r,int rt)
{
    if(x<=l&&r<=y)
    {
        lazy[rt]+=c;
        sum[rt]+=c*(r-l+1);
        return;
    }
    pushdown(rt,r-l+1);
    int mid=(l+r)/2;
    if(x<=mid) update(l,mid,2*rt);
    if(y>mid) update(mid+1,r,2*rt+1);
    pushup(rt);
}

long long query(int l,int r,int rt)
{
    if(x<=l&&r<=y)
    {
        return sum[rt];
    }
    pushdown(rt,r-l+1);
    int mid=(l+r)/2;
    long long left=0;
    long long right=0;
    if(x<=mid) left+=query(l,mid,2*rt);
    if(y>mid) right+=query(mid+1,r,2*rt+1);
    return left+right;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        build(1,n,1);
        while(m--)
        {
            cin>>M;
            if(M=="Q")
            {
                scanf("%d%d",&x,&y);
                printf("%lld\n",query(1,n,1));
            }
            else
            {
                scanf("%d%d%lld",&x,&y,&c);
                update(1,n,1);
            }
        }
    }
    return 0;
}