433B. Kuriyama Mirai's Stones
dp, implementation, sorting, 1200
https://codeforces.com/problemset/problem/433/B
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
- She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her
. - Let
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 (
The third line contains an integer m (
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.
Examples
input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6output
24
9
28input
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 2output
10
15
5
15
5
5
2
12
3
5Note
Please note that the answers to the questions may overflow 32-bit integer type.
def precompute_prefix_sums(arr):
prefix_sums = [0] * (len(arr) + 1)
for i in range(1, len(arr) + 1):
prefix_sums[i] = prefix_sums[i - 1] + arr[i - 1]
return prefix_sums
# Read input
n = int(input().strip())
v = list(map(int, input().strip().split()))
m = int(input().strip())
# Precompute prefix sums for the original and sorted lists
original_prefix_sums = precompute_prefix_sums(v)
sorted_v = sorted(v)
sorted_prefix_sums = precompute_prefix_sums(sorted_v)
# Process each query
results = []
for _ in range(m):
query = list(map(int, input().strip().split()))
q_type, l, r = query[0], query[1], query[2]
if q_type == 1:
result = original_prefix_sums[r] - original_prefix_sums[l - 1]
else:
result = sorted_prefix_sums[r] - sorted_prefix_sums[l - 1]
results.append(result)
# Print results
for result in results:
print(result)