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

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　　线段树，区间修改求和。

　　题意：

　　

　　思路：

　　

 

　　代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;  4 
 5 #define MAXN 100010
 6 
 7 struct Node{  8     long long L,R;  9     long long sum;    //当前区间的所有数的和
 10     long long inc;    //累加量
 11 }a[MAXN*3];  12 
 13 void Build(long long  d,long long  l,long long  r)    //建立线段树
 14 {  15 
 16     //初始化当前节点的信息
 17     a[d].L = l;  18     a[d].R = r;  19     a[d].inc = 0;  20 
 21     if(l==r){    //找到叶子节点
 22         scanf("%I64d",&a[d].sum);  23         return ;  24  }  25 
 26     //建立线段树
 27     long long  mid = (l+r)>>1;  28     Build(d<<1,l,mid);  29     Build(d<<1|1,mid+1,r);  30 
 31     //更新当前节点的信息
 32     a[d].sum =  a[d<<1].sum + a[d<<1|1].sum;  33 }  34 
 35 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
 36 {  37     if(a[d].L==l && a[d].R==r){    //找到终止节点
 38         a[d].inc += v;  39         return ;  40  }  41 
 42     long long  mid = (a[d].L+a[d].R)/2;  43     a[d].sum += a[d].inc*(a[d].R - a[d].L + 1);  44 
 45     if(mid>=r){    //左孩子找
 46         Updata(d<<1,l,r,v);  47  }  48     else if(mid<l){    //右孩子找
 49         Updata(d<<1|1,l,r,v);  50  }  51     else{    //左孩子、右孩子都找
 52         Updata(d<<1,l,mid,v);  53         Updata(d<<1|1,mid+1,r,v);  54  }  55 
 56     a[d].sum = a[d<<1].sum + a[d<<1|1].sum  57             +  a[d<<1].inc*(a[d<<1].R - a[d<<1].L + 1)  58             +  a[d<<1|1].inc*(a[d<<1|1].R - a[d<<1|1].L + 1);  59 }  60 
 61 long long  Query(long long  d,long long  l,long long  r)    //查询区间[l,r]的所有数的和
 62 {  63     if(a[d].L==l && a[d].R==r){    //找到终止节点
 64         return  a[d].sum + a[d].inc * (r-l+1);  65  }  66 
 67     long long  mid = (a[d].L+a[d].R)/2;  68     //更新每个节点的sum
 69     a[d].sum += a[d].inc * (a[d].R - a[d].L + 1);  70     a[d<<1].inc += a[d].inc;  71     a[d<<1|1].inc += a[d].inc;  72     a[d].inc = 0;  73 
 74     //Updata(d<<1,a[d<<1].L,a[d<<1].R,a[d].inc);  75     //Updata(d<<1|1,a[d<<1|1].L,a[d<<1|1].R,a[d].inc);
 76 
 77     if(mid>=r){    //左孩子找
 78         return Query(d<<1,l,r);  79  }  80     else if(mid<l){    //右孩子找
 81         return Query(d<<1|1,l,r);  82  }  83     else{    //左孩子、右孩子都找
 84         return Query(d<<1,l,mid) + Query(d<<1|1,mid+1,r);  85  }  86     a[d].sum = a[d<<1].sum + a[d<<1|1].sum  87             +  a[d<<1].inc*(a[d<<1].R - a[d<<1].L + 1)  88             +  a[d<<1|1].inc*(a[d<<1|1].R - a[d<<1|1].L + 1);  89 }  90 
 91 int main()  92 {  93     long long n,q,A,B;  94     long long v;  95     scanf("%I64d%I64d",&n,&q);  96     Build(1,1,n);  97     while(q--){    //q次询问
 98         char c[10];  99         scanf("%s",&c); 100         switch(c[0]){ 101         case 'Q': 102             scanf("%I64d%I64d",&A,&B); 103             printf("%I64d\n",Query(1,A,B));    //输出区间[A,B]所有数的和
104             break; 105         case 'C': 106             scanf("%I64d%I64d%I64d",&A,&B,&v); 107             Updata(1,A,B,v); 108             break; 109         default:break; 110  } 111  } 112     return 0; 113 }

 

Freecode : www.cnblogs.com/yym2013