Problem description

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

Input

3
1 1
2 2
3 3

Output

0

Input

3
1 10
0 10
10 10

Output

2
解题思路:统计房间内剩余人数不小于2的房间总数(两个人住在同一间),水过!
AC代码:
 #include<bits/stdc++.h>
using namespace std;
int main(){
int n,p,q,num=;
cin>>n;
while(n--){
cin>>p>>q;
if(q-p>=)num++;
}
cout<<num<<endl;
return ;
}

A - George and Accommodation的更多相关文章

  1. cf467A George and Accommodation

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

  2. CF467 AB 水题

    Codeforces Round #267 (Div. 2) (C和D的题解单独写:CF467C George and Job (DP) CF467D Fedor and Essay 建图DFS) C ...

  3. Codeforces Round #267 (Div. 2) A

    题目: A. George and Accommodation time limit per test 1 second memory limit per test 256 megabytes inp ...

  4. codeforces467-A水题

    题目链接:http://codeforces.com/problemset/problem/467/A A. George and Accommodation time limit per test ...

  5. CF467C George and Job (DP)

    Codeforces Round #267 (Div. 2) C. George and Job time limit per test 1 second memory limit per test ...

  6. Codeforces Round #227 (Div. 2) E. George and Cards set内二分+树状数组

    E. George and Cards   George is a cat, so he loves playing very much. Vitaly put n cards in a row in ...

  7. Codeforces 467C George and Job(DP)

    题目 Source http://codeforces.com/contest/467/problem/C Description The new ITone 6 has been released ...

  8. Codeforces Round #267 (Div. 2) C. George and Job DP

                                                  C. George and Job   The new ITone 6 has been released ...

  9. HDU 4118 Holiday's Accommodation

    Holiday's Accommodation Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 200000/200000 K (Jav ...

随机推荐

  1. MSCRM4 在过滤后的LOOKUP框中实现查找

    在MSCRM中让Lookup根据一定的条件实现过滤功能, 这个需求很常见, 在我接触的诸多项目中似乎都需要有这个功能. 但非常遗憾是, MSCRM 的SDK并没有提供实现这个功能的方法. 不过我们应该 ...

  2. 【sqli-labs】 less41 GET -Blind based -Intiger -Stacked(GET型基于盲注的堆叠查询整型注入)

    整型的不用闭合引号 http://192.168.136.128/sqli-labs-master/Less-41/?id=1;insert into users(id,username,passwo ...

  3. 我所理解的monad(4):函子(functor)是什么--可把范畴简单的看成高阶类型

    大致介绍了幺半群(monoid)后,我们重新回顾最初引用wadler(haskell委员会成员,把monad引入haskell的家伙)的那句话: 现在我们来解读这句话中包含的另一个概念:自函子(End ...

  4. windows 下安装mysql 成功版

    mysql 下载地址 http://dev.mysql.com/downloads/ zip版下载 解压到本地 假设文件保存在C:\mysql-5.7.17-winx64 1.以管理员身份运行cmd. ...

  5. Java 8 函数接口详细教程

    ay = new byte[array.length]; for (int i = 0; i < array.length; i++) { transformedArray[i] = funct ...

  6. matlab 读取输入数组

    In an assignment A(I) = B, the number of elements in B and I must be the same MATLAB:index_assign_el ...

  7. POJ 2229 Sumsets(找规律,预处理)

    题目 参考了别人找的规律再理解 /* 8=1+1+1+1+1+1+1+1+1 1 8=1+1+1+1+1+1+1+2 2 3 8=1+1+1+1+2+2 8=1+1+1+1+4 4 5 8=1+1+2 ...

  8. 爬虫写法进阶:普通函数--->函数类--->Scrapy框架

    本文转载自以下网站: 从 Class 类到 Scrapy https://www.makcyun.top/web_scraping_withpython12.html 普通函数爬虫: https:// ...

  9. vue.js 中 data, prop, computed, method,watch 介绍

    vue.js 中 data, prop, computed, method,watch 介绍 data, prop, computed, method 的区别 类型 加载顺序 加载时间 写法 作用 备 ...

  10. Git 基础教程 之 分支管理及策略

    创建一个属于自己的分支,别人看不到,你在你自己的分支上干活, 想提交就提交,直至开发完毕后,再一次性合并到原来分支上.这样,既安全,又不影响他人工作.          在实际的开发过程中,应照几个基 ...