Problem:
Range Sum Query

Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:

add(i, x): add x to ai.

getSum(s, t): print the sum of as, as+1,...,at.

Note that the initial values of ai (i = 1, 2, . . . , n) are 0.

Input

n q
com1 x1 y1
com2 x2 y2
...
comq xq yq

In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).

Output

For each getSum operation, print the sum in a line.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 100000

If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.

If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.

Sample Input 1

3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2

Sample Output 1

3
2

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

void getSum(int a[],int x,int y){
    int i;
    int sum=0;
    for(i=x;i<=y;i++){
        sum+=a[i];
    }
    printf("%d\n",sum);
}

int main(void){
    int i;
    
    int n;
    scanf("%d ",&n);
    
    int a[n];
    for(i=0;i<n;i++)a[i]=0;

    int q;
    scanf("%d\n",&q);
    
    int com,x,y;
    while(q--){
        scanf("%d %d %d\n",&com,&x,&y);
        switch(com){
            case 0:
                a[x-1]+=y;
                break;
            case 1:
                getSum(a,x-1,y-1);
                break;
        }
    }

    return 0;
}