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：
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
