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Codeforces Round #267 (Div. 2) A

时间:2014-09-19 10:16:05      阅读:233      评论:0      收藏:0      [点我收藏+]

标签:codeforces   算法   acm   algorithm   

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);
    }
}


Codeforces Round #267 (Div. 2) A

标签:codeforces   算法   acm   algorithm   

原文地址:http://blog.csdn.net/notdeep__acm/article/details/39388303

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