ACM新手训练题（CodeForces - 467A）

George and Accommodation
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has n rooms in total. At the moment the i-th room has pi people living in it and the room can accommodate qi people in total (pi ≤ qi). Your task is to count how many rooms has free place for both George and Alex.

    Input
    The first line contains a single integer n (1 ≤ n ≤ 100) — the number of rooms.

The i-th of the next n lines contains two integers pi and qi (0 ≤ pi ≤ qi ≤ 100) — the number of people who already live in the i-th room and the room’s capacity.

    Output
    Print a single integer — the number of rooms where George and Alex can move in.

    Examples

Input

3
1 1
2 2
3 3

Output

0

Input

3
1 10
0 10
10 10

Output

2
问题链接：https://vjudge.net/problem/53354/origin
问题简述：
输入一个数作为房间数，接着输入这些房间现住人数与该房间最大容纳量，一个房间占一行。判断这个房间能否再住下两个人，并且统计所有能再住下两个人的房间数量。
问题分析：
用两个数组分别接收现住人数与该房间的最大容纳量。用if结构判断该房间能否再住下两人，循环结构可进行计数。

AC代码如下：

#include
using namespace std;
int main()
{
 int n, pi[101], qi[101], i,count=0;
 cin >> n;
 for (i = 0; i < n; i++)
 {
  cin >> pi[i] >> qi[i];
  if (qi[i] - pi[i] >= 2)
   count++;
 }
 cout << count << endl;
 return 0;
}