Codeforces Round #267 (Div. 2) A

时间:2023-03-08 22:25:59
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\n",ans);
}
}