Meeting of Old Friends(CodeForces 714A)

最新推荐文章于 2024-10-30 02:03:01 发布

Baiyi_destroyer

最新推荐文章于 2024-10-30 02:03:01 发布

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分类专栏：水题集

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本文描述了一场发生在森林中的特殊聚会，刺猬菲利亚将拜访他的老朋友猫头鹰索尼娅。索尼娅在白天睡觉，晚上从特定时刻开始到结束保持清醒，但在某个时刻她会打盹。菲利亚计划在特定时间段内访问索尼娅。本文通过计算他们能够共度的时间，展示了如何找出两个时间段的交集。

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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.

题解：水题，找范围，注意使用 long long int .

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std;



int main()
{
    ll l1,r1,l2,r2,k;
    cin>>l1>>r1>>l2>>r2>>k;
    ll s=min(r1,r2)-max(l1,l2);
    if(s<0)
        cout<<"0"<<endl;
    else
    {
        if(k<=min(r1,r2)&&k>=max(l1,l2))
            cout<<s<<endl;
        else
            cout<<s+1<<endl;
    }
    return 0;
}