poj 3468:A Simple Problem with Integers（线段树，区间修改求和）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 58269		Accepted: 17753
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

　　线段树，区间修改求和。
　　题意：
　　


　　思路：
　　


　　代码：

#include <iostream>
#include <stdio.h>
using namespace std;

#define MAXN 100010

struct Node{
long long  L,R;
long long sum;    //当前区间的所有数的和
long long inc;    //累加量
}a[MAXN*3];

void Build(long long  d,long long  l,long long  r)    //建立线段树
{

//初始化当前节点的信息
a[d].L = l;
a[d].R = r;
a[d].inc = 0;

if(l==r){    //找到叶子节点
scanf("%I64d",&a[d].sum);
return ;
}

//建立线段树
long long  mid = (l+r)>>1;
Build(d<<1,l,mid);
Build(d<<1|1,mid+1,r);

//更新当前节点的信息
a[d].sum =  a[d<<1].sum + a[d<<1|1].sum;
}

void Updata(long long  d,long long  l,long long  r,long long  v)    //更新区间[l,r]的累加量为v
{
if(a[d].L==l && a[d].R==r){    //找到终止节点
a[d].inc += v;
return ;
}

long long  mid = (a[d].L+a[d].R)/2;
a[d].sum += a[d].inc*(a[d].R - a[d].L + 1);

if(mid>=r){    //左孩子找
Updata(d<<1,l,r,v);
}
else if(mid<l){    //右孩子找
Updata(d<<1|1,l,r,v);
}
else{    //左孩子、右孩子都找
Updata(d<<1,l,mid,v);
Updata(d<<1|1,mid+1,r,v);
}

a[d].sum = a[d<<1].sum + a[d<<1|1].sum
+  a[d<<1].inc*(a[d<<1].R - a[d<<1].L + 1)
+  a[d<<1|1].inc*(a[d<<1|1].R - a[d<<1|1].L + 1);
}

long long  Query(long long  d,long long  l,long long  r)    //查询区间[l,r]的所有数的和
{
if(a[d].L==l && a[d].R==r){    //找到终止节点
return  a[d].sum + a[d].inc * (r-l+1);
}

long long  mid = (a[d].L+a[d].R)/2;
//更新每个节点的sum
a[d].sum += a[d].inc * (a[d].R - a[d].L + 1);
a[d<<1].inc += a[d].inc;
a[d<<1|1].inc += a[d].inc;
a[d].inc = 0;

//Updata(d<<1,a[d<<1].L,a[d<<1].R,a[d].inc);
//Updata(d<<1|1,a[d<<1|1].L,a[d<<1|1].R,a[d].inc);

if(mid>=r){    //左孩子找
return Query(d<<1,l,r);
}
else if(mid<l){    //右孩子找
return Query(d<<1|1,l,r);
}
else{    //左孩子、右孩子都找
return Query(d<<1,l,mid) + Query(d<<1|1,mid+1,r);
}
a[d].sum = a[d<<1].sum + a[d<<1|1].sum
+  a[d<<1].inc*(a[d<<1].R - a[d<<1].L + 1)
+  a[d<<1|1].inc*(a[d<<1|1].R - a[d<<1|1].L + 1);
}

int  main()
{
long long  n,q,A,B;
long long  v;
scanf("%I64d%I64d",&n,&q);
Build(1,1,n);
while(q--){    //q次询问
char c[10];
scanf("%s",&c);
switch(c[0]){
case 'Q':
scanf("%I64d%I64d",&A,&B);
printf("%I64d\n",Query(1,A,B));    //输出区间[A,B]所有数的和
break;
case 'C':
scanf("%I64d%I64d%I64d",&A,&B,&v);
Updata(1,A,B,v);
break;
default:break;
}
}
return 0;
}

Freecode : www.cnblogs.com/yym2013