ACM第二题

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Time limit1000 msMemory limit262144 kBSourceCodeforces Round #267 (Div. 2)Tagsimplementation *800EditorialAnnouncement

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.

Sample Input
3
1 1
2 2
3 3

Sample Output
0

问题链接：https://vjudge.net/problem/CodeForces-467A

问题简述：在现有的房间中，选择能同时铸两个人的房间，求出符合房间的数目。

问题分析：要能循环输入每个房间的状况，并用判断语句统计符合的数目。

程序说明：用for语句循环输入，用两个数相减大于或者等于2作为判断条件，循环内自增来统计。

AC通过的C语言程序如下：

#include "pch.h"
#include <iostream>
using namespace std;
int main()
{
	int room, pi, qi,a,b=0;
	cin >> a;
	for (int i = 0; i < a; i++) {
		cin >> pi;
		cin >> qi;
		if ((qi - pi) >= 2)
			b++;
	}
	cout << b;
}