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, ... , 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.

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*];

 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 = ;

     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)>>;
     Build(d<<,l,mid);
     Build(d<<|,mid+,r);

     //更新当前节点的信息
     a[d].sum =  a[d<<].sum + a[d<<|].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)/;
     a[d].sum += a[d].inc*(a[d].R - a[d].L + );

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

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

 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+);
     }

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

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

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

Freecode : www.cnblogs.com/yym2013