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
(来自http://poj.org/problem?id=3468)
j讲一下题目大意，就是有n个数和m个操作，操作有2种，一种是C还有一种是Q，C是将[a,b]这个区间内每一个数增加c，而Q是查询[a,b]这个区间的和
这道题没什么可以讲的，线段树直接上
1 /**
2 * poj.org
3 * Problem#3468
4 */
5 #include<iostream>
6 #include<cstdio>
7 using namespace std;
8 typedef long long ll;
9 ll *a;
10 /**
11 * 树节点
12 */
13 typedef class TreeNode{
14 private:
15 void init(){
16 left = NULL;
17 right= NULL;
18 sum = 0;
19 state= 0;
20 }
21 public:
22 int from; //区间的开始
23 int end; //区间的结束
24 TreeNode *left; //左子树
25 TreeNode *right; //右子树指针
26 ll sum; //这一段的和
27 ll state; //延时标记，当非0表示有更新
28 TreeNode(){
29 init();
30 }
31 TreeNode(int from, int end){
32 init();
33 this->from = from;
34 this->end = end;
35 }
36 }TreeNode;
37 /**
38 * 树结构
39 */
40 typedef class Tree{
41 public:
42 TreeNode* root;
43 Tree():root(NULL){}
44 Tree(int len){
45 root = build(root, 1, len);
46 }
47 void pushUp(TreeNode* node){
48 if(node->left != NULL && node->right != NULL)
49 node->sum = node->left->sum + node->right->sum;
50 }
51 void pushDown(TreeNode* node){
52
53 node->left->state += node->state;
54 node->left->sum += (node->left->end - node->left->from + 1) * node->state;
55
56 node->right->state += node->state;
57 node->right->sum += (node->right->end - node->right->from + 1) * node->state;
58
59 node->state = 0;
60
61 }
62 TreeNode* build(TreeNode* root,int from, int end){
63 if(from == end){
64 root = new TreeNode(from, end);
65 root->sum = a[from];
66 return root;
67 }
68 root = new TreeNode(from, end);
69 int mid = (from + end)/2;
70 root->left = build(root->left, from, mid);
71 root->right = build(root->right, mid + 1, end);
72 pushUp(root);
73 return root;
74 }
75 void updata(TreeNode* now, int from, int end, ll data){
76
77 if(from <= now->from && now->end <= end ){
78 now->state += data;
79 now->sum += ( now->end - now->from + 1) * data;
80 return ;
81 }
82 now->sum += (end - from + 1) * data;
83 if(now->state != 0) pushDown(now);
84 int mid = (now->from + now->end)/2;
85 if(end <= mid) updata(now->left, from, end, data);
86 else if(from > mid) updata(now->right, from, end,data);
87 else{
88 updata(now->left, from, mid, data);
89 updata(now->right, mid + 1, end, data);
90 }
91
92 }
93 ll query(TreeNode* now, int from, int end){
94
95 if(from <= now->from && now->end <= end ) return now->sum;
96 if(now->state != 0) pushDown(now);
97 int mid = (now->from + now->end)/2;
98 if(end <= mid) return query(now->left, from, end);
99 else if(from > mid) return query(now->right, from, end);
100 else{
101 return (query(now->left, from, mid) +
102 query(now->right, mid + 1, end));
103 }
104
105 }
106 }Tree;
107 int n,m;
108 Tree MyTree;
109 char c;
110 int buffer[3];
111 int main(){
112 scanf("%d%d",&n,&m);
113 a = new ll[(const int)(n + 1)];
114 for(int i = 1;i <= n;i++){
115 scanf("%lld",&a[i]);
116 }
117 MyTree = Tree(n);
118 for(int i = 1;i <= m;i++){
119 cin>>c;
120 if(c == ‘C‘){
121 scanf("%d%d%d",&buffer[0],&buffer[1],&buffer[2]);
122 MyTree.updata(MyTree.root, buffer[0], buffer[1], buffer[2]);
123 }else{
124 scanf("%d%d",&buffer[0],&buffer[1]);
125 printf("%lld\n",MyTree.query(MyTree.root, buffer[0], buffer[1]));
126 }
127 }
128 return 0;
129 }
另外，原数组的类型最好要用long long
后记:
一个神奇的问题:原数组int过不了，改成long long提交一次，过了,接着改成int，就Accepted