![Codeforces Round #267 (Div. 2) A Codeforces Round #267 (Div. 2) A](https://image.shishitao.com:8440/aHR0cHM6Ly9ia3FzaW1nLmlrYWZhbi5jb20vdXBsb2FkL2NoYXRncHQtcy5wbmc%2FIQ%3D%3D.png?!?w=700&webp=1)
题目:
1 second
256 megabytes
standard input
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.
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.
Print a single integer — the number of rooms where George and Alex can move in.
3
1 1
2 2
3 3
0
3
1 10
0 10
10 10
2
题意分析:
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);
}
}