433B.Kuriyama Mirai's Stones
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.
刚开始没有注意下面的Note,用int类型的数组WA了。后来改用long long型。我之前也没去算过OJ可以接受多大的数组,就这道AC的代码来看4W的long long是没有问题的。
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include<unordered_set>
#define ll long long
using namespace std;
int dir[][] = { {,},{-,},{,},{,-} }; int main()
{
int n;
cin >> n;
vector<ll> v(n),a(n);
for (int i = ; i < n; i++)
{
int t;
cin >> t;
v[i] = t;
a[i] = t;
}
sort(a.begin(), a.end());
for (int i = ; i < n; i++)
{
v[i] += v[i - ];
a[i] += a[i - ];
} int m;
cin >> m;
while (m--)
{
int q, l, r;
cin >> q >> l >> r;
if (q == )
{
if (l == )
cout << v[r - ] << endl;
else
cout << v[r - ] - v[l - ] << endl;
}
else
{
if (l == )
cout << a[r - ] << endl;
else
cout << a[r - ] - a[l - ] << endl;
}
}
//system("pause");
return ;
}
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