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>
#define MAX 2147483647
#define NMAX 100001


int getsum(int A[],int x,int y){
int i;
int sum=0;

for(i=x;i<=y;i++){
 sum+=A[i];
}

return sum;
}



int main(){
int n,q,i;
int com,x,y;
int A[NMAX]={0};
int C[NMAX];
scanf("%d %d",&n,&q);

for(i=0;i<q;i++)C[i]=-1;

for(i=0;i<q;i++){
 scanf("%d %d %d",&com,&x,&y);
 switch(com){
  case 0:A[x]+=y;
         break;
  case 1:C[i]=getsum(A,x,y);
         break;
 }
}


for(i=0;i<q;i++){if(C[i]!=-1){printf("%d\n",C[i]);}}

return 0;
}