线段树区间求和树节点不能只存和，只存和，会导致每次加数的时候都要更新到叶子节点，速度太慢(O(nlogn))。所以我们要存两个量，一个是原来的和nSum，一个是累加的增量Inc。
在增加时，如果要加的区间正好覆盖一个节点，则增加其节点的Inc值，不再往下走，否则要更新nSum(加上本次增量),再将增量往下传，这样更新的复杂度就是O(log(n))。
在查询时，如果待查区间不是正好覆盖一个节点，就将节点的Inc往下带，然后将Inc清0，接下来再往下查询。 Inc往下带的过程也是区间分解的过程，复杂度也是O(log(n))。
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 79334 | Accepted: 24455 | |
| 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
1 #include <iostream>
2 #include<cstdio>
3 using namespace std;
4 struct CNode
5 {
6 int L ,R;
7 CNode * pLeft, * pRight;
8 long long nSum; //原来的和
9 long long Inc; //增量c的累加
10 };
11 CNode Tree[200010]; // 2倍叶子节点数目就够
12 int nCount = 0;
13 int Mid( CNode * pRoot)
14 {
15 return (pRoot->L + pRoot->R)/2;
16 }
17 void BuildTree(CNode * pRoot,int L, int R)
18 {
19 pRoot->L = L;
20 pRoot->R = R;
21 pRoot->nSum = 0;
22 pRoot->Inc = 0;
23 if( L == R)
24 return;
25 nCount ++;
26 pRoot->pLeft = Tree + nCount;
27 nCount ++;
28 pRoot->pRight = Tree + nCount;
29 BuildTree(pRoot->pLeft,L,(L+R)/2);
30 BuildTree(pRoot->pRight,(L+R)/2+1,R);
31 }
32 void Insert( CNode * pRoot,int i, int v)
33 {
34 if( pRoot->L == i && pRoot->R == i)
35 {
36 pRoot->nSum = v;
37 return ;
38 }
39 pRoot->nSum += v;
40 if( i <= Mid(pRoot))
41 Insert(pRoot->pLeft,i,v);
42 else
43 Insert(pRoot->pRight,i,v);
44 }
45 void Add( CNode * pRoot, int a, int b, long long c)
46 {
47 if( pRoot->L == a && pRoot->R == b)
48 {
49 pRoot->Inc += c;
50 return ;
51 }
52 pRoot->nSum += c * ( b - a + 1) ;
53 if( b <= (pRoot->L + pRoot->R)/2)
54 Add(pRoot->pLeft,a,b,c);
55 else if( a >= (pRoot->L + pRoot->R)/2 +1)
56 Add(pRoot->pRight,a,b,c);
57 else
58 {
59 Add(pRoot->pLeft,a,
60 (pRoot->L + pRoot->R)/2 ,c);
61 Add(pRoot->pRight,
62 (pRoot->L + pRoot->R)/2 + 1,b,c);
63 }
64 }
65 long long QuerynSum( CNode * pRoot, int a, int b)
66 {
67 if( pRoot->L == a && pRoot->R == b)
68 return pRoot->nSum +
69 (pRoot->R - pRoot->L + 1) * pRoot->Inc ;
70 pRoot->nSum += (pRoot->R - pRoot->L + 1) * pRoot->Inc ;
71 Add( pRoot->pLeft,pRoot->L,Mid(pRoot),pRoot->Inc);
72 Add( pRoot->pRight,Mid(pRoot) + 1,pRoot->R,pRoot->Inc);
73 pRoot->Inc = 0;
74 if( b <= Mid(pRoot))
75 return QuerynSum(pRoot->pLeft,a,b);
76 else if( a >= Mid(pRoot) + 1)
77 return QuerynSum(pRoot->pRight,a,b);
78 else
79 {
80 return QuerynSum(pRoot->pLeft,a,Mid(pRoot)) +
81 QuerynSum(pRoot->pRight,Mid(pRoot) + 1,b);
82 }
83 }
84 int main()
85 {
86 int n,q,a,b,c;
87 char cmd[10];
88 scanf("%d%d",&n,&q);
89 int i,j,k;
90 nCount = 0;
91 BuildTree(Tree,1,n);
92 for( i = 1; i <= n; i ++ )
93 {
94 scanf("%d",&a);
95 Insert(Tree,i,a);
96 }
97 for( i = 0; i < q; i ++ )
98 {
99 scanf("%s",cmd);
100 if ( cmd[0] == ‘C‘ )
101 {
102 scanf("%d%d%d",&a,&b,&c);
103 Add( Tree,a,b,c);
104 }
105 else
106 {
107 scanf("%d%d",&a,&b);
108 printf("%I64d\n",QuerynSum(Tree,a,b));
109 }
110 }
111 return 0;
112 }