CodeForces 714A

Description

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.

Sample Input

Input

1 10 9 20 1

Output

Input

1 100 50 200 75

Output

Hint

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.

算是一道非常非常简单的贪心做法题吧。

但我还是WA了无数发，只因为那个被我犯过好几次的错误，不是第一次犯这种错了。

只考虑到了两个区间相交和相离的情况，没有考虑到内含的情况。

这题应该注意的是k是在相交区间断点的问题。

 #include<stdio.h>

 #include<stdlib.h>

 #include<string.h>

 #include<math.h>

 #include<algorithm>

 #include<queue>

 #include<stack>

 #include<deque>

 #include<iostream>

 using namespace std;

 typedef long long  LL;

 struct con{

     LL head;

     LL tail;

 }con[];

 LL cmp(struct con a,struct con b)

 {

     if(a.head==b.head)

         return a.tail<b.tail;

     return a.head<b.head;

 }

 int main()

 {

     LL i,p,j,n;

     LL k,ans=;

 //    scanf("%lld",&p);

 //    printf("%lld\n",p);

     for(i=;i<=;i++)

     {

         scanf("%lld%lld",&con[i].head,&con[i].tail);

     }

     scanf("%lld",&k);

     sort(con,con+,cmp);

     if(con[].head<=con[].tail&&(k>con[].tail||k<con[].head))

     {

         ans=con[].tail-con[].head+;

     }

     if(con[].head<=con[].tail&&(k<=con[].tail&&k>=con[].head))

     {

         ans=con[].tail-con[].head;

     }

     if(con[].tail>=con[].tail&&(k>con[].tail||k<con[].head))

     {

         ans=con[].tail-con[].head+;

     }

     if(con[].tail>=con[].tail&&(k<=con[].tail&&k>=con[].head))

     {

         ans=con[].tail-con[].head;

     }

     printf("%lld\n",ans);

     return ;

 }

巴特西

CodeForces 714A

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