[Codeforces Round #248 (Div. 2)] B. Kuriyama Mirai's Stones

B. Kuriyama Mirai's Stones

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:

1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her .
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her .

For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 10^9) — costs of the stones.

The third line contains an integer m (1 ≤ m ≤ 10^5) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.

Output

Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.

Sample test(s)

input

6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6

output

24
9
28

input

4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2

output

10
15
5
15
5
5
2
12
3
5

Note

Please note that the answers to the questions may overflow 32-bit integer type.

 

题解：维护两个前缀和，原序sum1[]和有序sum2[],因为n (1 ≤ n ≤ 10^5)，(1 ≤ vi ≤ 10^9) ，所以n*v<=10^14，所以sum[]为long long 类型即可。剩下的m次询问，每次运行时间复杂度就为O(1),然后本题就得到解决。

代码：

 

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdbool.h>

 4 #include<stdlib.h>

 5 #include<math.h>

 6 #include<ctype.h>

 7 #include<time.h>

 8 

 9 #define rep(i,a,b)  for(i=(a);i<=(b);i++)

10 #define red(i,a,b)  for(i=(a);i>=(b);i--)

11 #define sqr(x)      ((x)*(x))

12 #define clr(x,y)    memset(x,y,sizeof(x))

13 #define LL          long long

14 

15 

16 int i,j,n,m,

17     a[110000];

18 

19 LL sum1[110000],sum2[110000];

20 

21 void pre()

22 {

23     clr(a,0);

24     clr(sum1,0);

25     clr(sum2,0);

26 }

27 

28 void sort(int head,int tail)

29 {

30     int i,j,x;

31     i=head;j=tail;

32     x=a[head];

33     

34     while(i<j)

35     {

36         while((i<j)&&(a[j]>=x)) j--;

37         a[i]=a[j];

38         while((i<j)&&(a[i]<=x)) i++;

39         a[j]=a[i];

40     }

41     

42     a[i]=x;

43     

44     if(head<(i-1)) sort(head,i-1);

45     if((i+1)<tail) sort(i+1,tail);

46 }

47 

48 int init()

49 {

50     int i;

51     scanf("%d",&n);

52     rep(i,1,n){

53         scanf("%d",&a[i]);

54         sum1[i]=sum1[i-1]+a[i];

55     }

56     sort(1,n);

57     

58     rep(i,1,n)

59         sum2[i]=sum2[i-1]+a[i];

60     return 0;

61 }

62 

63 int main()

64 {

65     int i,ty,x,y;

66     pre();

67     init();

68     

69     scanf("%d",&m);

70    

71     rep(i,1,m){

72         scanf("%d%d%d",&ty,&x,&y);

73         if(ty==1) printf("%lld\n",sum1[y]-sum1[x-1]);

74         else printf("%lld\n",sum2[y]-sum2[x-1]);

75     }

76    

77     return 0;

78 }