POJ 3468 A Simple Problem with Integers 线段树区间更新-优快云博客

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 67718		Accepted: 20897
Case Time Limit: 2000MS

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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

给你n个数，Q（a,b)代表查询区间[a,b]数字之和

C（a,b,c)代表在区间[a,b]增加值c

在点更新的基础上加上一个mark标记。

//11412 KB	2704 ms
#include<stdio.h>
#include<algorithm>
#include<string.h>
#define ll __int64
#define M 100007
using namespace std;
struct node
{
    ll l,r,mid,val,mark;
}tree[M<<2];
ll s[M];
void build(ll left,ll right,ll i)//建树
{
    tree[i].l=left;tree[i].r=right;
    tree[i].mid=(left+right)>>1;tree[i].mark=0;
    if(left==right){tree[i].val=s[left]; return;}
    build(left,tree[i].mid,i*2);
    build(tree[i].mid+1,right,i*2+1);
    tree[i].val=tree[i*2].val+tree[i*2+1].val;
}
void update(int left,int right,ll val,int i)//区间更新
{
    if(tree[i].l==left&&tree[i].r==right){tree[i].mark+=val;return;}
    tree[i].val+=val*(right-left+1);
    if(tree[i].mid<left)update(left,right,val,2*i+1);
    else if(tree[i].mid>=right)update(left,right,val,2*i);
    else
    {
        update(left,tree[i].mid,val,2*i);
        update(tree[i].mid+1,right,val,2*i+1);
    }
}
ll query(int left,int right,int i)//区间查询
{
    if(tree[i].l==left&&tree[i].r==right)  return tree[i].val+tree[i].mark*(right-left+1);
    if(tree[i].mark!=0)
    {
        tree[i*2].mark+=tree[i].mark;tree[i*2+1].mark+=tree[i].mark;
        tree[i].val+=(tree[i].r-tree[i].l+1)*tree[i].mark;tree[i].mark=0;
    }
    if(tree[i].mid>=right){return query(left,right,i*2);}
    else if(tree[i].mid<left){return query(left,right,i*2+1);}
    else{return query(left,tree[i].mid,i*2)+query(tree[i].mid+1,right,i*2+1);}
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%I64d",&s[i]);
        build(0,M,1);
        char z[2];
        int a,b;
        ll c;
        while(m--)
        {
            scanf("%s",z);
            if(z[0]=='Q')
            {
                scanf("%d%d",&a,&b);
                printf("%I64d\n",query(a,b,1));
            }
            else
            {
                scanf("%d%d%I64d",&a,&b,&c);
                update(a,b,c,1);
            }
        }
    }
    return 0;
}