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  • Codeforces Round #267 (Div. 2) A

    题目:

    A. George and Accommodation
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard 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.

    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,每一个寝室已经有p个人住了,总共能住下q个人。如今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
    ",ans);
        }
    }
    


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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5078206.html
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