poj 3468 线段树成段更新

A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 54012   Accepted: 16223
Case Time Limit: 2000MS

Description

You have N integers, A1A2, ... , 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 A1A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of AaAa+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.
 
#include<iostream>
#include<cstdio>
using namespace std;

#define lson left,mid,i<<1
#define rson mid+1,right,i<<1|1

typedef __int64 LL;
int a[100005];

struct IntervalTree
{
    int left,right;
    LL sum,add;
}f[300005];

void pushdown(int i)
{
    if(f[i].add != 0)
    {
        f[i<<1].add+=f[i].add;
        f[i<<1|1].add+=f[i].add;
        f[i<<1].sum+=(f[i<<1].right-f[i<<1].left+1)*f[i].add;
        f[i<<1|1].sum+=(f[i<<1|1].right-f[i<<1|1].left+1)*f[i].add;
        f[i].add=0;
    }
}

void pushup(int i)
{
    f[i].sum=f[i<<1].sum+f[i<<1|1].sum;
}

void bulid(int left,int right,int i)
{
    f[i].left=left,f[i].right=right,f[i].add=0;
    if(left==right)
    {
        f[i].sum=a[left];return ;
    }
    int mid=(left+right)>>1;
    bulid(lson);
    bulid(rson);
    pushup(i);
    return ;
}

void update(int left,int right,int add,int i)
{
    if(f[i].left==left && f[i].right==right)
    {
        f[i].add+=add;
        f[i].sum+=(right-left+1)*add;
        return ;
    }
    pushdown(i);
    if(f[i<<1].right>=right) update(left,right,add,i<<1);
    else if(f[i<<1|1].left<=left) update(left,right,add,i<<1|1);
    else { update(left,f[i<<1].right,add,i<<1);update(f[i<<1|1].left,right,add,i<<1|1);}
    pushup(i);
    return ;
}

LL query(int left,int right,int i)
{
    if(f[i].left==left && f[i].right==right) return f[i].sum;
    pushdown(i);
    if(f[i<<1].right>=right) return query(left,right,i<<1);
    else if (f[i<<1|1].left<=left) return query(left,right,i<<1|1);
    else return query(left,f[i<<1].right,i<<1)+query(f[i<<1|1].left,right,i<<1|1);
}

int main()
{
    int i,n,m,ai,bi,d;
    char ch[2];
    while(~scanf("%d %d",&n,&m))
    {
        for(i=1;i<=n;i++) scanf("%d",a+i);
        bulid(1,n,1);
        while(m--)
        {
            scanf("%s",ch);
            if(strcmp(ch,"C")==0)
            {
                scanf("%d %d %d",&ai,&bi,&d);
                update(ai,bi,d,1);
            }
            else
            {
                scanf("%d %d",&ai,&bi);
                printf("%I64d
",query(ai,bi,1));
            }
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/xiong-/p/3590251.html