ACM POJ 3468 A Simple Problem with Integers（线段树） by kuangbin

题目链接：http://poj.org/problem?id=3468

本文作者：kuangbin

博客地址：http://www.cnblogs.com/kuangbin/

题目：

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 22796		Accepted: 6106
Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... , 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 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.
Source

POJ Monthly--2007.11.25, Yang Yi

[b][b]简单的线段树练习题。[/b][/b]
[b][b][b]主要是树的节点存储哪些信息。[/b][/b][/b]
[b][b][b][b]每个一个数都更新到叶子节点不是明智的做法，很消耗时间。[/b][/b][/b][/b]
[b][b][b][b][b]故每个节点加一个Inc来记录增量的累加。[/b][/b][/b][/b][/b]
[b][b][b][b][b][b]具体看代码：[/b][/b][/b][/b][/b][/b]

/*
POJ 3468 A Simple Problem with Integers
题目意思：
给定Q个数：A1,A2,```,AQ,以及可能多次进行下列两个操作：
1）对某个区间Ai```Aj的数都加n（n可变）
2）对某个区间Ai```Aj求和
*/
#include<stdio.h>
#include<algorithm>
#include<iostream>
usingnamespace std;
constint MAXN=100000;
int num[MAXN];
struct Node
{
int l,r;//区间的左右端点
longlong nSum;//区间上的和
longlong Inc;//区间增量的累加
}segTree[MAXN*3];
void Build(int i,int l,int r)
{
segTree[i].l=l;
segTree[i].r=r;
segTree[i].Inc=0;
if(l==r)
{
segTree[i].nSum=num[l];
return;
}
int mid=(l+r)>>1;
Build(i<<1,l,mid);
Build(i<<1|1,mid+1,r);
segTree[i].nSum=segTree[i<<1].nSum+segTree[i<<1|1].nSum;
}
void Add(int i,int a,int b,longlong c)//在结点i的区间（a,b）上增加c
{
if(segTree[i].l==a&&segTree[i].r==b)
{
segTree[i].Inc+=c;
return;
}
segTree[i].nSum+=c*(b-a+1);
int mid=(segTree[i].l+segTree[i].r)>>1;
if(b<=mid)  Add(i<<1,a,b,c);
elseif(a>mid)  Add(i<<1|1,a,b,c);
else
{
Add(i<<1,a,mid,c);
Add(i<<1|1,mid+1,b,c);
}
}
longlong Query(int i,int a,int b)//查询a-b的总和
{
if(segTree[i].l==a&&segTree[i].r==b)
{
return segTree[i].nSum+(b-a+1)*segTree[i].Inc;
}
segTree[i].nSum+=(segTree[i].r-segTree[i].l+1)*segTree[i].Inc;
int mid=(segTree[i].l+segTree[i].r)>>1;
Add(i<<1,segTree[i].l,mid,segTree[i].Inc);
Add(i<<1|1,mid+1,segTree[i].r,segTree[i].Inc);
segTree[i].Inc=0;
if(b<=mid)  return Query(i<<1,a,b);
elseif(a>mid)  return Query(i<<1|1,a,b);
elsereturn Query(i<<1,a,mid)+Query(i<<1|1,mid+1,b);
}
int main()
{
int n,q;
int i;
int a,b,c;
char ch;
while(scanf("%d%d",&n,&q)!=EOF)
{
for(i=1;i<=n;i++)  scanf("%d",&num[i]);
Build(1,1,n);
for(i=1;i<=q;i++)
{
cin>>ch;
if(ch=='C')
{
scanf("%d%d%d",&a,&b,&c);
Add(1,a,b,c);
}
else
{
scanf("%d%d",&a,&b);
printf("%I64d\n",Query(1,a,b));
}
}
}
return0;
}