A Simple Problem with Integers
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
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 报告题目大意： 一串数，“C a b c" 表示在区间[a,b]中每个数 + c "Q a b"表示查询区间[a,b] 的和 线段树： 学习了线段树后的第一道题。这道题包含了线段树的Pushdown,Pushup,Query,Build,Update几个函数，很好地让初学者了解了线段树的用法和精髓--延迟更新，即每次更新时只是做一个标记，当要查找时才继续向下更新。还要注意的是，有些题不需要向下回溯（pushup），也不需要延迟更新（pushdown），甚至连更新都不用。但是需要根据题意改变要查找的内容和更新的内容，有的需要找最大值，有的需要找和，等等，需要随机应变。线段树能解决的问题，树状数组不一定能解决，但是树状数组能解决的，线段树一定能解决。 思路： 便不多说。上代码。
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #define L(u) (u<<1)
6 #define R(u) (u<<1|1)//没有分号
7 using namespace std;
8 int n,q;
9 long long c;
10 long long a[100005];
11 struct node{
12 int l,r;
13 long long add,sum;
14 };
15 char s[2];
16 node pp[400005];/**数据范围**/
17 void pushup(int u)
18 {
19 pp[u].sum=pp[L(u)].sum+pp[R(u)].sum;
20 return ;
21 }
22 void pushdown(int u)
23 {
24 pp[L(u)].add+=pp[u].add;
25 pp[L(u)].sum+=(pp[L(u)].r-pp[L(u)].l+1)*pp[u].add;
26 pp[R(u)].add+=pp[u].add;
27 pp[R(u)].sum+=(pp[R(u)].r-pp[R(u)].l+1)*pp[u].add;
28 pp[u].add=0;/*！！！*/
29 }
30 void update(int u,int left,int right,long long t)
31 {
32 if (left<=pp[u].l&&pp[u].r<=right)
33 {
34 pp[u].add+=t;
35 pp[u].sum+=(pp[u].r-pp[u].l+1)*t;
36 return ;
37 }
38 pp[u].sum+=(right-left+1)*t;
39 if (pp[u].add) pushdown(u);
40 int mid=(pp[u].l+pp[u].r)>>1;
41 if (right<=mid) update(L(u),left,right,t);
42 else /***/if (left>mid) update(R(u),left,right,t);
43 else
44 {
45 update(L(u),left,mid,t);
46 update(R(u),mid+1,right,t);
47 }
48 //pushup(u);
49 }
50
51 void build(int u,int left,int right)
52 {
53 pp[u].l=left;
54 pp[u].r=right;
55 pp[u].add=0;
56 if (pp[u].l==pp[u].r)
57 {
58 pp[u].sum=a[left];//*****
59 return ;
60 }
61 int mid=(pp[u].l+pp[u].r)>>1;
62 build(L(u),left,mid);
63 build(R(u),mid+1,right);
64 pushup(u);
65 }
66 long long query(int u,int left,int right)
67 {
68 if (left==pp[u].l&&pp[u].r==right)
69 return pp[u].sum;
70 if (pp[u].add) pushdown(u);
71 int mid=(pp[u].l+pp[u].r)>>1;
72 if (right<=mid) return query(L(u),left,right);
73 if (left>mid) return query(R(u),left,right);
74 else
75 return (query(L(u),left,mid)+query(R(u),mid+1,right));
76 }
77 int main()
78 {
79 //freopen("tree.in","r",stdin);
80 cin>>n>>q;
81 for (int i=1;i<=n;i++)
82 scanf("%lld",&a[i]);//long long 读数
83 build(1,1,n);
84 for (int i=1;i<=q;i++)
85 {
86 scanf("%s",s);
87 if (s[0]==‘Q‘)
88 {
89 int k,b;
90 scanf("%d%d",&k,&b);
91 cout<<query(1,k,b)<<endl;
92 }
93 else
94 {
95 int k,b;
96 cin>>k>>b>>c;
97 update(1,k,b,c);
98 }
99 }
100
101 return 0;
102 }
注意事项：1、数据范围：线段树一般定义4*n 2、#difine 的使用 function() ( ** ) 注意括号没有分号； 3、runtime error ：运行错误 long long a[i], %lld(linux) %I64d （windows) 4、模板答题 顺便抱怨一下，poj的评测系统太慢了，waiting了好久.......