Algorithm
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:
- She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her
. - 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.
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — 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.
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.
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
24
9
28
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
10
15
5
15
5
5
2
12
3
5
Please note that the answers to the questions may overflow 32-bit integer type.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int const N = 1e5 + 1;
long long scum[N], uscum[N];
vector<int> arr;
int main() {
int n;
cin >> n;
arr.resize(n);
for(int i = 0; i < n; i++) cin >> arr[i];
uscum[0] = 0;
for(int i = 1; i <= n; i++)
uscum[i] = uscum[i-1] + arr[i-1];
sort(arr.begin(), arr.end());
scum[0] = 0;
for(int i = 1; i <= n; i++)
scum[i] = scum[i-1] + arr[i-1];
int q;
cin >> q;
int t, a, b;
while(q--) {
cin >> t >> a >> b;
if(t == 1)
cout << uscum[b] - uscum[a-1] << endl;
else
cout << scum[b] - scum[a-1] << endl;
}
}Input
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
9
28
Demonstration
Codeforces Solution-Kuriyama Mirai's Stones-Solution in C, C++, Java, Python