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2224: Boring Counting
Time Limit: 3 Sec Memory Limit: 128 MB
Description
In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you task is to answer a list of queries, for each query, please tell us among [L, R], how many Pi is not less than A and not greater than B( L<= i <= R). In other words, your task is to count the number of Pi (L <= i <= R, A <= Pi <= B).
Input
In the first line there is an integer T (1 < T <= 50), indicates the number of test cases.
For each case, the first line contains two numbers N and M (1 <= N, M <= 50000), the size of sequence P, the number of queries. The second line contains N numbers Pi(1 <= Pi <= 10^9), the number sequence P. Then there are M lines, each line contains four number L, R, A, B(1 <= L, R <= n, 1 <= A, B <= 10^9)
Output
For each case, at first output a line ‘Case #c:’, c is the case number start from 1. Then for each query output a line contains the answer.
Sample Input
1 13 5 6 9 5 2 3 6 8 7 3 2 5 1 4 1 13 1 10 1 13 3 6 3 6 3 6 2 8 2 8 1 9 1 9
Sample Output
Case #1: 13 7 3 6 9
HINT
Source
题意：求静态的区间[l,r]介于[A,B]的数的个数
这题其实是一道离线树状数组的水题，估计是因为我并不是很具备离线的思维吧，一般的离线都想不到。
好在主席树也能够完成这个操作，并且也只花了20分钟不到就1A了。
二分k的值，然后判断利用第k大来判断出A，B分别是第几大。然后随便搞。
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 #include <cmath>
6 #include <cstdlib>
7 #include <vector>
8 using namespace std;
9 #define w(i) T[i].w
10 #define ls(i) T[i].ls
11 #define rs(i) T[i].rs
12 #define MAXN 100010
13 int p[MAXN],a[MAXN],b[MAXN],root[MAXN];
14 struct node{
15 int ls,rs,w;
16 node(){ls=rs=w=0;}
17 }T[MAXN*20];
18 int tot=0;
19 void Insert(int &i,int l,int r,int x){
20 T[++tot]=T[i];
21 i=tot;
22 w(i)++;
23 if(l==r)return;
24 int mid=(l+r)>>1;
25 if(x<=mid)Insert(ls(i),l,mid,x);
26 else Insert(rs(i),mid+1,r,x);
27 }
28 int query(int lx,int rx,int l,int r,int k){
29 if(l==r)return l;
30 int ret=w(ls(rx))-w(ls(lx));
31 int mid=(l+r)>>1;
32 if(ret>=k)return query(ls(lx),ls(rx),l,mid,k);
33 else return query(rs(lx),rs(rx),mid+1,r,k-ret);
34 }
35 bool cmp(int i,int j){
36 return a[i]<a[j];
37 }
38 int main()
39 {
40 ios::sync_with_stdio(false);
41 tot=0;
42 int n,m;
43 int t;
44 int cas=1;
45 scanf("%d",&t);
46 while(t--){
47 scanf("%d%d",&n,&m);
48 for(int i=1;i<=n;i++){
49 scanf("%d",a+i);
50 p[i]=i;
51 }
52 tot=0;
53 root[0]=0;
54 sort(p+1,p+n+1,cmp);
55 for(int i=1;i<=n;i++)b[p[i]]=i;
56 for(int i=1;i<=n;i++){
57 root[i]=root[i-1];
58 Insert(root[i],1,n,b[i]);
59 }
60 printf("Case #%d:\n",cas++);
61 while(m--){
62 int l,r,x,y;
63 scanf("%d%d%d%d",&l,&r,&x,&y);
64 int lx=1,rx=r-l+1;
65 int fx=-1;
66 int ans=0;
67 int flag =0;
68 int tmpx;
69 while(lx<=rx){
70 int mid = (lx+rx)>>1;
71 tmpx=a[p[query(root[l-1],root[r],1,n,mid)]];
72 if(tmpx<=y){
73 fx=mid;
74 lx=mid+1;
75 }else rx=mid-1;
76 }
77 if(fx==-1)flag=1;
78 else ans+=fx;
79 lx=1,rx=r-l+1;
80 fx=-1;
81 while(lx<=rx){
82 int mid = (lx+rx)>>1;
83 tmpx=a[p[query(root[l-1],root[r],1,n,mid)]];
84 if(tmpx>=x){
85 fx=mid;
86 rx=mid-1;
87 }else lx=mid+1;
88 }
89 if(fx==-1)flag=1;
90 else ans=ans-fx+1;
91 if(flag)ans=0;
92 printf("%d\n",ans);
93 }
94 }
95 return 0;
96 }