Group

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1959    Accepted Submission(s): 1006

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

15 23 1 2 5 41 52 4

Sample Output

12

/*hdu 4638 树状数组 区间内连续区间的个数(尽可能长)给你n个数,让你查询区间[l,r]内最长连续区间的个数。求的是区间长度平方的和,所以区间长度越长越好3 1 2 5 4  在[1,5]上:1,2,3,4,5   1个           在[2,4]上:1,2 和 4    2个首先我们把每个数看成独立的,即每个数赋值为1对于当前出现的a[i],如果前面出现了a[i-1],a[i+1]那么就能组成一队,但是我们是从左往右递推出来的,所以保存的值应尽可能在右边以方便查询,所以如果前面出现了a[i-1],a[i+1].则在它们的位置上-1,删除它们的独立值.hhh-2016-04-04 17:05:44*/#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;#define lson  (i<<1)#define rson  ((i<<1)|1)typedef long long ll;const int mod = 10007;const int maxn = 100050;int s[maxn];int a[maxn];int p[maxn];int ans[maxn];struct node{    int l,r;    int id;} op[maxn];int n,m;bool cmp(node a,node b){    return a.r < b.r;}int lowbit(int x){    return x&(-x);}void add(int x,int val){    while(x <= n)    {        s[x] += val;        x += lowbit(x);    }}int sum(int x){    int cnt = 0;    while(x)    {        cnt += s[x];        x -= lowbit(x);    }    return cnt;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        memset(s,0,sizeof(s));        for(int i = 1; i <= n; i++)        {            scanf("%d",&a[i]);            p[a[i]] = i;        }        for(int i =0; i< m; i++)        {            scanf("%d%d",&op[i].l,&op[i].r);            op[i].id = i;        }        sort(op,op+m,cmp);        for(int i = 1,cur = 0; i <= n; i++)        {            add(i,1);            if(a[i] > 1 && p[a[i]-1] < i)                add(p[a[i]-1],-1);            if(a[i] < n && p[a[i]+1] < i)                add(p[a[i]+1],-1);            while(i == op[cur].r && cur < m)            {                ans[op[cur].id] = sum(op[cur].r)-sum(op[cur].l-1);                cur++;            }        }        for(int i = 0; i < m; i++)            printf("%d\n",ans[i]);    }    return 0;}