Problem:
Score : 200 points

Problem Statement

In a factory, there are N robots placed on a number line.
Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative.

We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect.
Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints.

Find the maximum number of robots that we can keep.

Constraints

1 \leq N \leq 100,000

0 \leq X_i \leq 10^9 (1 \leq i \leq N)

1 \leq L_i \leq 10^9 (1 \leq i \leq N)

If i \neq j, X_i \neq X_j.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 L_1
X_2 L_2
\vdots
X_N L_N

Output

Print the maximum number of robots that we can keep.

Sample Input 1

4
2 4
4 3
9 3
100 5

Sample Output 1

3

By removing Robot 2, we can keep the other three robots.

Sample Input 2

2
8 20
1 10

Sample Output 2

1

Sample Input 3

5
10 1
2 1
4 1
6 1
8 1

Sample Output 3

5

Solution in C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

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

int *X, *L, n;

int culc_max(int left, int right, int num);

int main(void)
{
  int i, j, temp_x, temp_l, answer;

  /*ロボットの個数の入力*/
  scanf("%d", &n);

  /*動的配列の作成*/
  X = (int*)malloc(sizeof(int) * n);
  if (X == NULL) printf("MEMORY ERROR OF X\n");
  L = (int*)malloc(sizeof(int) * n);
  if (L == NULL) printf("MEMORY ERROR OF L\n");

  /*ロボット毎の位置とアームの長さの入力*/
  for (i=0;i<n;i++) {
    scanf("%d %d", &X[i], &L[i]);
  }

  /*ロボットの位置でソートする*/
  for (i=0;i<n;i++) {
    for (j=i+1;j<n;j++) {
      if (X[i] > X[j]) {
        temp_x = X[i];
        temp_l = L[i];
        X[i] = X[j];
        L[i] = L[j];
        X[j] = temp_x;
        L[j] = temp_l;
      } else {
        break;
      }
    }
  }

  /*処理*/
  answer = culc_max(0, 1, 1);

  /*出力*/
  printf("%d\n", answer);
  return 0;
}

/*再帰関数*/
int culc_max(int left, int right, int num)
{
  /*ベースケース*/
  if (right == n) {
    return num;
  }

  if (X[left] + L[left] <= X[right] - L[right]) {
    /*かぶらないとき*/
    return culc_max(right, right + 1, num + 1);
  } else {
    /*かぶるとき*/
    if (culc_max(left, right + 1, num) < culc_max(right, right + 1, num)) {
      /*右残しの方がいいとき*/
      return culc_max(right, right + 1, num);
    } else {
      return culc_max(left, right + 1, num);
    }
  }
}