P - Meeting of Old Friends

ゞ浴缸里的玫瑰 2022-03-20 08:12 101阅读 0赞

### 题目描述： ###

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.  
题意：就是一个求交集的问题。  
样例输入是l1,r1,l2,r2,k;  
求出l1,r1,l2,r2的交集不能包括点k。  
分析：这里注意几种情况：  
1，l1大于l2;  
2, r2<r1。  
注意这几点情况，就是很简单的贪心问题了。

#include"stdio.h"
    #include"string.h"
    typedef long long ll;
    int main()
    {
        ll l1,r1,l2,r2,k;
        while(~scanf("%lld%lld%lld%lld%lld",&l1,&r1,&l2,&r2,&k))
        {
            ll ans=0;
            if(l2<l1)
            {
                ll t=l2;l2=l1;l1=t;
                t=r2;r2=r1;r1=t;
            }
            if(l2>r1)
            {
                printf("%d\n",0);
            }
            else
            {
                ll Min=r1;
                if(Min>r2)
                    Min=r2;
                if(k>=l2&&k<=Min)
                    printf("%lld\n",Min-l2);
                else
                    printf("%lld\n",Min-l2+1);
            }
        }
    }