【31.42%】【CF 714A】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.

【题解】

求两个区间的交集。

很明显的线段树问题

(滑稽)

把两个区间的左端点的位置确定一下就好。

然后再看两个区间的右端点。

再判断k在不在交集中。

吧嗒吧嗒。。

【代码】

#include <cstdio>
 
long long left1, right1, left2, right2, k,ans;
 
void input_data()
{
	scanf("%I64d%I64d%I64d%I64d%I64d", &left1, &right1, &left2, &right2, &k);
}
 
void get_ans()
{
	if (left1 > left2) //固定两个区间的左端点为从左到右
	{
		long long t = left1;
		left1 = left2;
		left2 = t;
		t = right1;
		right1 = right2;
		right2 = t;
	}
	if (right1 < left2) //这是没有交集的情况。
		printf("0\n");
	else
		if (right2 <= right1)
		{
			ans = right2 - left2 + 1;
			if (left2 <= k && k <= right2)
				ans--;
			printf("%I64d\n", ans);
		}
		else
		{
			ans = right1 - left2 + 1;
			if (left2 <= k && k <= right1)
				ans--;
			printf("%I64d\n", ans);
		}
 
}
 
int main()
{
	//freopen("F:\\rush.txt", "r", stdin);
	input_data();
	get_ans();
	return 0;
}