George and Accommoditon c++

题目

出处
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.

分析

题目的意思很简单，有两人想住同一间宿舍，现在有n间房，一间房间共能住q人，现在住了p人。不同房间可能不一样(人为输入的n,p,q)。要求计算能给两人住的房间有多少间，也就是q-p>=2的情况发生了多少次。用一个数i记录发生次数，初始化为0。输入n，后设置n次的循环，循环里输入p,q。如果q-p>=2，则i++；否则不操作。最后跳出循环后i就是合题意的房间数。

代码

#include
using namespace std;
int main()
{
	int n, p, q, i;
	cin >> n;
	i = 0;
	while (n--)
	{
		cin >> p >> q;
		if (q - p >= 2)i++;
	}
	cout << i << endl;
	return 0;
}