If you have "Time limit exceeded", just use sys.stdin.read()... and it will put load on memory, as I suppose reading huge amount of data via input() takes forever.

Can you suggest what is causing this error.

N = int(input())
s = [0]

for i in range(N):
   k = int(input())
   s.append(int(s[i] + k))

Q = int(input())
while Q:
   l, r = map(int, input().split())
   print(s[r] - s[l-1])
   Q -= 1

I got AC in 0.125s and 284KB!!!
If you want my AC code,post me:happyzeyu@163.com

Edited by author 25.12.2006 17:30

Edited by author 25.12.2006 17:30

Time limit test 20 failed on python. Rewrited on C# - everything OK.

Edited by author 04.10.2019 02:08

Edited by author 04.10.2019 02:08

if you have TLE #20. Use BufferedReader and Printwriter in Java. Good luck...

1. Segment tree.
2. Fenwick tree.
3. Partial sums (b[i] = SUM(a[0], ..., a[i])).
4. Sqrt decomposition.
Hope, it helps to understand how to solve this task.

Edited by author 28.02.2017 15:25

How to do my program more simply? Who khow?
When I send this program - Time limit Exceed on 20 test.
Help, who can!

Program zadacha;
var a:array[1..10000] of integer; b,c,i,j,n:integer; d:array[0..100000] of longint; m:longint;
begin
readln(n);
for i:=1 to n do
 read(a[i]);

readln(m);
for i:=1 to m do begin
 read(b,c);
  for j:=b to c do
   d[i]:=d[i]+a[j];
 end;
for i:=1 to m do
writeln(d[i]);
end.

Use scanf and printf instead of cin,cout.
I had a really hard time to figure it out))

[code deleted]

Edited by moderator 01.12.2019 21:31

Hello

My python code is as follows:

[code deleted]

I get TLE in # 20
Where can I save time?

Edited by moderator 01.12.2019 21:35

Who wants solution, e-mail me:
nevertheless_94@hotmail.com
:D

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class N1330 {



    public static void main(String[] args) {

        try {
            BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
            int n = Integer.parseInt(in.readLine());
            int list[] = new int[n];
            for (int i = 0; i < n; i++)
                list[i] = Integer.parseInt(in.readLine());
            int q = Integer.parseInt(in.readLine());
            for (int i = 0; i < q; i++) {
                int sum = 0;
                StringTokenizer tok = new StringTokenizer(in.readLine());
                int start = Integer.parseInt(tok.nextToken())-1;
                int end = Integer.parseInt(tok.nextToken())-1;
                for (int j = start; j <= end; j++)
                    sum += list[j];
                System.out.println(sum);
            }

        }
        catch(IOException e) {

        }

    }

}

Same algorithm (sum[b]-sum[a-1]) got AC using C/C++, but got time exceeded using Java

[code deleted]

Edited by moderator 01.12.2019 21:45

I recommend you to use the simplest input you have.
1. For a line with one number just use

int.Parse(Console.ReadLine())

2. For a line with two numbers use the simplest parser

string query = Console.ReadLine();
int separator = query.IndexOf(' ');
int a = int.Parse(query.Substring(0, separator)) - 1,
b = int.Parse(query.Substring(separator + 1)) - 1;

Otherwise you 'll get TL.

P.S. Does anybody know a faster input?

5
1
2
3
-1
4
2
1 4
4 1

Resul will be:
5
5

???

My solve O(N), but output too slow. How fast print answer?

AC this problem is very simple!
Try to solve the problem with O(n).

Navchin Crash barib duribdi?

here is my  C code
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
    int n,x,i,l,m;
    long *a=0,*b=0;
    scanf("%d",&n);
    a=(long*)malloc((n+1)*n);
    a[0]=0;
    for(i=1;i<n+1;++i)
    {
        scanf("%d",&x);
        a[i]=a[i-1]+ x;
    }
    scanf("%d",&x);
    b=(long*)malloc(x*sizeof(long));
    for(i=0;i<x;++i)
    {
        scanf("%d %d",&l,&m);
        b[i]=a[m]-a[l-1];
    }
    for(i=0;i<x-1;++i)
        printf("%ld\n",b[i]);
    printf("%ld",b[i]);
    return 0;
}