Codeforces Round #267 (Div. 2) A

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本文探讨了如何高效地为大学生提供宿舍分配解决方案，通过算法计算出能够容纳乔治和亚历克斯的空闲房间数量。利用输入数据（房间总数、当前入住人数及房间容量），该算法能快速找出满足条件的可用房间，简化了宿舍入住过程，体现了算法在实际生活问题中的应用价值。

题目：

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 test(s)
  

      input
    
3
1 1
2 2
3 3

      output
    
0

      input
    
3
1 10
0 10
10 10

      output
    
2

题意分析：

有N个寝室，每个寝室已经有p个人住了，总共能住下q个人。现在还有两个人能不能住下。直接写写写吧。大水题，看到一堆人1，2分钟就过了.

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>


using namespace std;


int main()
{
    int n,ans,q,p;
    while(scanf("%d",&n)!=EOF)
    {
       ans=0;
       for(int i=0;i<n;i++)
       {
           scanf("%d%d",&q,&p);
           if(p-q>=2)
            ans++;
       }
       printf("%d\n",ans);
    }
}