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
【分析】
很不错的数据结构，真的很不错....
跟树状数组有点像..
相比与普通线段树,zkw线段树具有标记永久化，简单好写的特点。
但zkw线段树占用的空间会相对大些，毕竟是2的阶乘...，然后应该不能可持久化吧(别跟我说用可持久化线段树维护可持久化数组....)
这样就不能动态分配内存了。。不过总体来说还是有一定的实用价值...
1 /*
2 宋代陆游
3 《临安春雨初霁》
4
5 世味年来薄似纱，谁令骑马客京华。
6 小楼一夜听春雨，深巷明朝卖杏花。
7 矮纸斜行闲作草，晴窗细乳戏分茶。
8 素衣莫起风尘叹，犹及清明可到家。
9 */
10 #include <iostream>
11 #include <cstdio>
12 #include <algorithm>
13 #include <cstring>
14 #include <vector>
15 #include <utility>
16 #include <iomanip>
17 #include <string>
18 #include <cmath>
19 #include <queue>
20 #include <assert.h>
21 #include <map>
22 #include <ctime>
23 #include <cstdlib>
24 #include <stack>
25 #define LOCAL
26 const int INF = 0x7fffffff;
27 const int MAXN = 100000 + 10;
28 const int maxnode = 300000;
29 using namespace std;
30 struct Node{
31 int l, r;
32 long long sum, d;
33 }tree[maxnode];
34 int data[MAXN], M;//M为总结点个数
35
36 //初始化线段树
37 void build(int l, int r){
38 for (int i = 2 * M - 1; i > 0; i--){
39 if (i >= M){
40 tree[i].sum = data[i - M];
41 tree[i].l = tree[i].r = (i - M);//叶子节点
42 }
43 else {
44 tree[i].sum = tree[i<<1].sum + tree[(i<<1)+1].sum;
45 tree[i].l = tree[i<<1].l;
46 tree[i].r = tree[(i<<1)+1].r;
47 }
48 }
49 }
50 long long query(int l, int r){
51 long long sum = 0;//sum记录和
52 int numl = 0, numr = 0;
53 l += M - 1;
54 r += M + 1;
55 while ((l ^ r) != 1){
56 if (~l&1){//l取反，表示l是左节点
57 sum += tree[l ^ 1].sum;
58 numl += (tree[l ^ 1].r - tree[l ^ 1].l + 1);
59 }
60 if (r & 1){
61 sum += tree[r ^ 1].sum;
62 numr += (tree[r ^ 1].r - tree[r ^ 1].l + 1);
63 }
64 l>>=1;
65 r>>=1;
66 //numl,numr使用来记录标记的
67 sum += numl * tree[l].d;
68 sum += numr * tree[r].d;
69 }
70 for (l >>= 1; l > 0; l >>= 1) sum += (numl + numr) * tree[l].d;
71 return sum;
72 }
73
74 void add(int l, int r, int val){
75 int numl = 0, numr = 0;
76 l += M - 1;
77 r += M + 1;
78 while ((l ^ r) != 1){
79 if (~l&1){//l取反
80 tree[l ^ 1].d += val;
81 tree[l ^ 1].sum += (tree[l ^ 1].r - tree[l ^ 1].l + 1) * val;
82 numl += (tree[l ^ 1].r - tree[l ^ 1].l + 1);
83 }
84 if (r & 1){
85 //更新另一边
86 tree[r ^ 1].d += val;
87 tree[r ^ 1].sum += (tree[r ^ 1].r - tree[r ^ 1].l + 1) * val;
88 numr += (tree[r ^ 1].r - tree[r ^ 1].l + 1);
89 }
90 l>>=1;
91 r>>=1;
92 tree[l].sum += numl * val;
93 tree[r].sum += numr * val;
94 }
95 //不要忘了往上更新
96 for (l >>= 1; l > 0; l >>= 1) tree[l].sum += (numl+numr) * val;
97 }
98 int n, m;
99
100 void init(){
101 scanf("%d%d", &n, &m);
102 for (int i = 1; i <= n; i++) scanf("%d", &data[i]);
103 M = 1;while (M < (n + 2)) M <<= 1;//找到最大的段
104 build(0, M - 1);
105 }
106 void work(){
107 while (m--){
108 char str[2];
109 scanf("%s", str);
110 if (str[0] == ‘Q‘){
111 int l, r;
112 scanf("%d%d", &l, &r);
113 printf("%lld\n", query(l, r));
114 }
115 else{
116 int val, l, r;
117 scanf("%d%d%d", &l, &r, &val);
118 if (val != 0) add(l, r, val);
119 }
120 }
121 }
122
123 int main(){
124
125 init();
126 work();
127 return 0;
128 }