Codeforces #371（Div.2）A.Meeting of Old Friends【模拟+思维】

最新推荐文章于 2018-10-09 15:39:55 发布

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本文介绍了一个计算两个时间段内有效共同时段的问题，考虑到其中一个时间段内的特定分钟不可用的情况，并提供了AC代码实现。

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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 minutek 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 andk (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

Input

1 100 50 200 75

Output

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.

题目大意：

有两个人，一个人会呆从l1到r1 这么长时间，另一个人会呆从l2，r2这么长时间，但是第一个人会在K这个时间点去化妆(一分钟)，问两个人同时在一起的时间为多少。

思路：

1、首先取l1，l2中大的，作为相同区间的左边界，然后取r1，r2中小的，作为相同区间的右边界，然后最后判一下k是否在区间内。

2、注意如果没有相同区间，千万不要输出负值。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define ll __int64
int main()
{
    ll l1,r1,l2,r2,k;
    while(~scanf("%I64d%I64d%I64d%I64d%I64d",&l1,&r1,&l2,&r2,&k))
    {
        ll output=0;
        if(l1<=l2)
        {
            ll you=min(r1,r2);
            output+=you-l2+1;
            if(k>=l2&&k<=you)output--;
        }
        else
        {
            ll you=min(r1,r2);
            output+=you-l1+1;
            if(k>=l1&&k<=you)output--;
        }
        if(output<0)output=0;
        printf("%I64d\n",output);
        fflush(stdout);
    }
}