Get sum of some arrays' elements C++ - c++

Question

How could I possibly get sum of elements between two given points in array if given:
n - Array length
m - Number of questions about array
a[n] - array numbers
m questions in format x y
Example
Input
3 3
-1 2 0
3 3
1 3
1 2
Output
0
1
1
Here is my code
#include <iostream>
#include <numeric>
using namespace std;
int main()
{
int n,m,x,y;
cin>>n>>m;
int a [n];
int s [m];
for (int i = 0; i < n; i++){
cin>>a[i];
}
for (int i = 0; i < m ; i++){
cin>>x>>y;
if(x == y){
s[i] = a[x-1];
continue;
}
s[i] = accumulate(a + x - 1, a + y,0);
}
for (int i = 0; i < m; i++){
cout<<s[i]<<endl;
}
return 0;
}
This is not a school homework, just a task from internet - it will be run on test server using Standard streams for input and result check.
I need this program to be run in less than 0.2 seconds with:
0 < n,m <= 50000
-1000 <= a[i] <= 1000
1<=x<=y<=n
How can I make it run more time efficiently?

Answer 1

A lot of time can be spent in the accummulate function when n and m are large.
Let's use following data in our example:
const int n = 10;
int a[n] = {6, 1, 3, 2, 9, 8, 7, 2, 0, 1};
Let's create an auxiliary array and fill it with sums from a[0] to a[i] for each i. This can be made in a single loop.
int s[n+1];
void precalculate() {
s[0] = 0;
for (int i = 0; i < n; ++i) {
s[i+1] = s[i] + a[i];
}
}
Then the whole accummulate function collapses to a single subtraction:
int accummulate(int i, int j) {
return s[j] - s[i-1];
}
It works well as it can be easily demonstrated:
int main() {
precalculate();
assert(accummulate(1, 1) == 6);
assert(accummulate(10, 10) == 1);
assert(accummulate(1, 10) == 39);
return 0;
}