An Average Game (block + Mo Team)

　Alice and Bob has just learned how to find average of some numbers. They got really excited and
decided to come up with a game about finding average.
　The game works like this, at the start of game a sequence of numbers is written.
　Then there will be several rounds in the game. In first round Alice will say a number x and Bob
has to select two index i and j. Let’s say the average of the unique numbers between i-th number and
j-th number is y. Then Alice gets abs(x− y) points in that round. In next round Alice and Bob switch
the role. The game continues this way.
　While Alice and Bob enjoy playing this game, they hate calculating average of unique numbers. So,
they are asking you, their only programmer friend, to write a program that calculates the average for
them.

Input
　First line of input is a number T (T ≤ 100). T test cases follow.
　First line of each test case is an integer n (0 < n < 1e4) representing the length of number sequence.
Next line consists n space separated integer representing the sequence. Each of these integers has
absolute value less than 1e9. Next line has an integer q (q < 1e5). q lines follow, each representing a
query. Each query has two space separated integer i, j (1 ≤ i ≤ j ≤ n).

Output
　For each case output q lines representing average of unique elements of corresponding range rounded 6 digits after decimal points. No case will have output whose rounding changes if the 10e−9 is added to
or subtracted from answer.
Notes:
　Explanation for second query in first sample
　The range is (1,10) which includes ten numbers (1, 2, 3, 4, 4, 3, 2, 1, -1, 0). Unique numbers in this
ranges are (1, 2, 3, 4, -1, 0) whose average is 1.5
　Explanation for third query in first sample
　The range is (3,5) which includes three numbers (3, 4, 4). Unique numbers in this ranges are 3 and
4 whose average is 3.5

Sample Input
2
10
1 2 3 4 4 3 2 1 -1 0
4
1 4
1 10
3 5
8 10
3
1 1 0
1
1 3

Sample Output
Case 1:
2.500000
1.500000
3.500000
0.000000
Case 2:
0.500000

The meaning of problems:
on the array is known, to find the average value of the number of different given interval

Ideas:
Block team with Mo, like Group, is seeking more than a several different numbers

. 1 #include <bits / STDC ++ H.>
 2  
. 3  the using  namespace STD;
 . 4  
. 5 Map < int , int > coun-;        // teammate said, map and consequently the number of stem ,,, here too 1e9, 1e4 number will
 6                   // as nails c language, writing many things, we really have to recognize, c ++ is convenient point 
. 7  struct Node
 . 8  {
 . 9      int left, right, NUM;
 10 } STR [ 100005 ];
 . 11  int a [ 10005 ], belong [ 10005 ], Number;
 12 is  
13 is  BOOL CMP ( struct Node X,struct node y)
14 {
15     if(belong[x.left]==belong[y.left]) return x.right < y.right;
16     else return x.left < y.left;
17 }
18 
19 double add(int n, double sum)
20 {
21     if(coun[n]==0)
22     {
23         sum += n;
24         number++;
25     }
26     coun[n]++;
27     return sum;
28 }
29 
30 double subtract(int n, double sum)
31 {
32     coun[n]--;
33     if(coun[n]==0)
34     {
35         number--;
36         sum -= n;
37     }
38     return sum;
39 }
40 
41 int main()
42 {
43     int n, m, t, i, temp, l, r, j;
44     doubleRe [ 100005 ], sum;    // start sum is defined as int, WA, and into a double on the right, do not know what 
45      Scanf ( " % D " , & T);
 46 is      for (J = . 1 ; J <= T; J ++ )
 47      {
 48          coun.clear ();         // coun- cleared 
49          Scanf ( " % D " , & n-);
 50          TEMP = sqrt (n-);
 51 is          for (I = . 1 ; I <= n-; I ++ )
 52 is          {
 53 is              Scanf ( " % D", &a[i]);
54             belong[i] = (i-1) / temp + 1;
55         }
56         scanf("%d", &m);
57         for(i=1;i<=m;i++)
58         {
59             str[i].num = i;
60             scanf("%d %d", &str[i].left, &str[i].right);
61         }
62         sort(str+1, str+m+1, cmp);
63         l = 1;
64         r = 0;
65         sum = 0;
66         number = 0;
67         for(i=1;i<=m;i++)
68         {
69             while(r<str[i].right)
70             {
71                 r++;
72                 sum = add(a[r], sum);
73             }
74             while(r>str[i].right)
75             {
76                 sum = subtract(a[r], sum);
77                 r--;
78             }
79             while(l<str[i].left)
80             {
81                 sum = subtract(a[l], sum);
82                 l++;
83             }
84             while(l>str[i].left)
85             {
86                 l--;
87                 sum = add(a[l], sum);
88             }
89             re[str[i].num] = 1.0 * sum / number;
90         }
91         printf("Case %d:\n", j);
92         for(i=1;i<=m;i++)
93         {
94             printf("%.6f\n", re[i]);
95         }
96     }
97     return 0;
98 }