CodeForces-467A George and Accommodation(语法练习题)_george and accommodation time limit per test1 seco-优快云博客

乔治和他的朋友亚历克斯正寻找可以共同居住的大学宿舍。他们需要找到哪些房间有足够的空间供两人同时入住。本文章详细介绍了如何通过编程解决这一问题，包括输入输出格式、算法思路及代码实现。

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George and Accommodation

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
inputCopy
3
1 1
2 2
3 3
outputCopy
0
inputCopy
3
1 10
0 10
10 10
outputCopy
2

问题简述：

George和他朋友要住旅馆，而且要同一个房间。第一行给出房间总数，后面给出每个房间的已入住人数和可容纳的总人数。请问有几间房间可以选择？

问题分析：

对于每个房间，判断其空位数是否大于等于2即可。将符合条件的房间数加起来。

程序说明：

得到房间数n后，申请空间创建一个长度为2n的数组存储房间信息。用一个for循环判断遍历房间信息，统计符合条件的房间数。

程序实现：

#include<iostream>
using namespace std;

int main()
{
    int n,sum=0;
    cin>>n;
    int *a=new int[2*n];
    for(int i=0;i<2*n;i+=2)
    {
        cin>>a[i]>>a[i+1];
        if(a[i]<a[i+1]-1) sum++;
    }
    cout<<sum;
}