# Group

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2483    Accepted Submission(s): 1272

Problem Description
There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i-1 are friends, Whose ID is i and i+1 are friends. These n men stand in line. Now we select an interval of men to make some group. K men in a group can create K*K value. The value of an interval is sum of these value of groups. The people of same group's id must be continuous. Now we chose an interval of men and want to know there should be how many groups so the value of interval is max.

Input
First line is T indicate the case number.
For each case first line is n, m(1<=n ,m<=100000) indicate there are n men and m query.
Then a line have n number indicate the ID of men from left to right.
Next m line each line has two number L,R(1<=L<=R<=n),mean we want to know the answer of [L,R].

Output
For every query output a number indicate there should be how many group so that the sum of value is max.

Sample Input
1
5 2
3 1 2 5 4
1 5
2 4

Sample Output
1
2

Source
题意：t组数据 给你一个长度为n的序列 m个询问[l,r] 问 l到r 的值可以组成多少个连续的段
题解：例如序列3 1 2 5 4

 1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 #include<map>
7 #include<queue>
8 #include<stack>
9 #include<vector>
10 #include<set>
11 #define ll __int64
12 using namespace std;
13 int n,m;
14 struct node
15 {
16     int l,r,id;
17 } N[100005];
18 int p[100005];
19 int block;
20 int a[100005];
21 int x[100005];
22 int mp[100005];
23 ll ans=0;
24 ll re[100005];
25 int cmp(struct node aa,struct node bb)
26 {
27     if(p[aa.l]==p[bb.l])
28         return  aa.r<bb.r;
29     else
30         return p[aa.l]<p[bb.l];
31 }
32 void update(int w,int h)
33 {
34     if(h==1)
35     {
36         mp[a[w]]=1;
37         if(mp[a[w]-1]==0&&mp[a[w]+1]==0)
38             ans++;
39         if(mp[a[w]-1]==1&&mp[a[w]+1]==1)
40             ans--;
41     }
42     else
43     {
44         mp[a[w]]=0;
45         if(mp[a[w]-1]==1&&mp[a[w]+1]==1)
46             ans++;
47         if(mp[a[w]-1]==0&&mp[a[w]+1]==0)
48             ans--;
49     }
50 }
51 int t;
52 int main()
53 {
54     scanf("%d",&t);
55     for(int o=1;o<=t;o++)
56     {
57         for(int i=0;i<100004;i++)
58             mp[i]=0;
59         scanf("%d %d",&n,&m);
60         for(int i=1; i<=n; i++)
61             scanf("%d",&a[i]);
62         for(int i=1; i<=m; i++)
63         {
64             scanf("%d %d",&N[i].l,&N[i].r);
65             N[i].id=i;
66         }
67         block=(int)sqrt((double)n);
68         for(int i=1; i<=n; i++)
69             p[i]=(i-1)/block+1;
70         sort(N+1,N+1+m,cmp);
71         ans=0;
72         for(int i=1,l=1,r=0; i<=m; i++)
73         {
74             for(; r<N[i].r; r++) update(r+1,1);
75             for(; l>N[i].l; l--) update(l-1,1);
76             for(; r>N[i].r; r--) update(r,-1);
77             for(; l<N[i].l; l++) update(l,-1);
78             re[N[i].id]=ans;
79         }
80         for(int i=1; i<=m; i++)
81             printf("%I64d\n",re[i]);
82     }
83     return 0;
84 }