Codeforces 714A Meeting of Old Friends

简介: A. Meeting of Old Friends time limit per test:1 second memory limit per test:256 megabytes input:standard input output:s...
A. Meeting of Old Friends
time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!

Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.

Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.

Calculate the number of minutes they will be able to spend together.

Input

The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.

Output

Print one integer — the number of minutes Sonya and Filya will be able to spend together.

Examples
Input
1 10 9 20 1
Output
2
Input
1 100 50 200 75
Output
50
Note

In the first sample, they will be together during minutes 9 and 10.

In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.

 题目链接:http://codeforces.com/problemset/problem/714/A

解题思路:

【题意】

Sonya每天只有[l1,r1]时间段内空闲,且在k时刻,她要打扮而不能够见Filya

Filya每天[l2,r2]时间段内空闲

问他们俩每天有多少时间能够在一起

【类型】
区间交

【分析】

显然,要求他们俩每天有多少时间在一起

其实就是求两区间的交集

那无外乎就是对两区间的位置关系进行分析




,。当然还有几个图这里就不一一列举了,主要就是找到两个的

相交区间,然后判断k是否在这个区间中,在的话减一;

这道题要用long long,否则会超!

下面给出AC代码:

 

 1 #include <bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4 int main()
 5 {
 6     ll l1,l2,r1,r2,k;
 7     while(cin>>l1>>r1>>l2>>r2>>k)
 8     {
 9         ll minn = min(r1,r2);
10         ll maxn = max(l1,l2);
11         ll ans = minn-maxn+1;
12         if(maxn > minn)
13         {
14             cout<<0<<endl;
15             continue;
16         }
17         if(k >= maxn && k <= minn)
18             ans--;
19         cout<<ans<<endl;
20     }
21     return 0;
22 }

 

 

 

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