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: Time Limit Exceeded)

#include <stdio.h>

#define MAX 100002

int A[MAX];
int B[MAX];

int main(void) {
	int i, j, n, q, m;
	int com, x, y;
	int *p, *b, *a;
	int pos = MAX;
	int last_y = 0;

	scanf("%d %d\n", &n, &q);
	for(i=0;i<=n;i++) {
		A[i] = 0;
		B[i] = 0;
	}
	
	for(i=0;i<q;i++) {
		scanf("%d %d %d\n", &com, &x, &y);
		if(com == 0) {
			A[x] += y;
			if(pos > x)	pos = x;
		} else {
			if(pos != MAX) {
				for(j=pos,a=&A[j],b=&B[j];j<=n;j++,a++,b++) *b = *a + *(b-1);
				pos = MAX;
				last_y = y;
			}
			
			printf("%d\n", B[y]-B[x-1]);
		}
	}
	
	return 0;
}