2018HDU多校二 -F 题 Naive Operations(线段树)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6315

In a galaxy far, far away, there are two integer sequence a and b of length n.
b is a static permutation of 1 to n. Initially a is filled with zeroes.
There are two kind of operations:
1. add l r: add one for

a_{l}, a_{l + 1} . . . a_{r}

InputThere are multiple test cases, please read till the end of input file.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form 'add l r' or 'query l r', representing an operation.

1 \leq n, q \leq 100000

OutputOutput the answer for each 'query', each one line.
Sample Input

5 12
1 5 2 4 3
add 1 4
query 1 4
add 2 5
query 2 5
add 3 5
query 1 5
add 2 4
query 1 4
add 2 5
query 2 5
add 2 2
query 1 5

Sample Output

题意：题目给你N个数，Q个操作，另外有个数组a,a 的初始值都是0，然后Q个操作，若是add 则在区间x~y之间的a[]都加一，query 就查找l~r之间   ∑ri=l⌊ai/bi⌋题解： 由于是取下界，我们可以求每个区间内距离该位置上b[i]值最近的数，然后没加一，就把b[i]减一，如果最小值为零，就出现了a[i]/b[i]==1的情况，就将区间的sum加一，对于
每个查询操作，我们只要求区间的sun和即可；

参考代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+10;
 4 int n,q,x,y,b[maxn];
 5 char str[10];
 6 
 7 struct Node{
 8     int l,r,sum,tag,minnumm;
 9 } tree[maxn<<2];
10 
11 void build(int l,int r,int pos)
12 {
13     tree[pos].l=l,tree[pos].r=r;
14     if(l==r)
15     {
16         tree[pos].minnumm=b[l];
17         tree[pos].sum=0;
18         tree[pos].tag=0;
19         return ;
20     }
21     int mid=(l+r)>>1;
22     build(l,mid,pos<<1);
23     build(mid+1,r,pos<<1|1);
24     tree[pos].minnumm=min(tree[pos<<1].minnumm,tree[pos<<1|1].minnumm);
25     tree[pos].sum=tree[pos<<1].sum+tree[pos<<1|1].sum;
26     tree[pos].tag=tree[pos<<1].tag+tree[pos<<1|1].tag;
27 }
28 
29 void pushdown(int pos)
30 {
31     tree[pos<<1].minnumm+=tree[pos].tag;
32     tree[pos<<1|1].minnumm+=tree[pos].tag;
33     tree[pos<<1].tag+=tree[pos].tag;
34     tree[pos<<1|1].tag+=tree[pos].tag;
35     tree[pos].tag=0;
36 }
37 
38 void update(int pos,int l,int r,bool temp)
39 {
40     if(tree[pos].l==l&&tree[pos].r==r)
41     {
42         if(temp)
43         {
44             tree[pos].tag--;
45             tree[pos].minnumm--;
46         }
47         if(tree[pos].minnumm>0) return ;
48         if(tree[pos].l==tree[pos].r)
49         {
50             if(tree[pos].minnumm==0) tree[pos].minnumm=b[tree[pos].l],tree[pos].sum++;
51             return ;    
52         } 
53         temp=false;
54     }
55     
56     if(tree[pos].tag) pushdown(pos);
57     
58     int mid=(tree[pos].l+tree[pos].r)>>1;
59     if(r<=mid) update(pos<<1,l,r,temp);
60     else if(l>=mid+1) update(pos<<1|1,l,r,temp);
61     else update(pos<<1,l,mid,temp),update(pos<<1|1,mid+1,r,temp); 
62     
63     tree[pos].minnumm=min(tree[pos<<1].minnumm,tree[pos<<1|1].minnumm);
64     tree[pos].sum=tree[pos<<1].sum+tree[pos<<1|1].sum; 
65 }
66 
67 int query(int pos,int l,int r)
68 {
69     if(tree[pos].tag) pushdown(pos);
70     if(tree[pos].l==l&&tree[pos].r==r) return tree[pos].sum;
71     int mid=(tree[pos].l+tree[pos].r)>>1,ans=0;
72     if(r<=mid) ans+=query(pos<<1,l,r);
73     else if(l>=mid+1) ans+=query(pos<<1|1,l,r);
74     else ans+=query(pos<<1,l,mid)+query(pos<<1|1,mid+1,r);
75     return ans;
76 }
77 
78 int main()
79 {
80     while(~scanf("%d%d",&n,&q))
81     {
82         for(int i=1;i<=n;i++) scanf("%d",b+i);
83         build(1,n,1);
84         while(q--)
85         {
86             scanf("%s%d%d",str,&x,&y);
87             if(str[0]=='a') update(1,x,y,true);
88             else if(str[0]=='q') printf("%d
",query(1,x,y));
89         }
90     }
91     return 0;
92 }

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