晚训 4.3

C. Watering Flowers

C. Watering Flowers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A flowerbed has many flowers and two fountains.

You can adjust the water pressure and set any values r1(r1 ≥ 0) and r2(r2 ≥ 0), giving the distances at which the water is spread from the first and second fountain respectively. You have to set such r1 and r2 that all the flowers are watered, that is, for each flower, the distance between the flower and the first fountain doesn't exceed r1, or the distance to the second fountain doesn't exceed r2. It's OK if some flowers are watered by both fountains.

You need to decrease the amount of water you need, that is set such r1 and r2 that all the flowers are watered and the r12 + r22 is minimum possible. Find this minimum value.

Input

The first line of the input contains integers n, x1, y1, x2, y2 (1 ≤ n ≤ 2000,  - 107 ≤ x1, y1, x2, y2 ≤ 107) — the number of flowers, the coordinates of the first and the second fountain.

Next follow n lines. The i-th of these lines contains integers xi and yi ( - 107 ≤ xi, yi ≤ 107) — the coordinates of the i-th flower.

It is guaranteed that all n + 2 points in the input are distinct.

Output

Print the minimum possible value r12 + r22. Note, that in this problem optimal answer is always integer.

Examples

input

Copy

2 -1 0 5 3
0 2
5 2

output

Copy

6

input

Copy

4 0 0 5 0
9 4
8 3
-1 0
1 4

output

Copy

33

Note

The first sample is (r12 = 5, r22 = 1):The second sample is (r12 = 1, r22 = 32):

显而易见，这是一个贪心，和导弹拦截差不多。

代码都是一样的。。。

思路呢,就是你把到两个点的距离全部算出来，注意用结构体存储，对到其中的一个点进行排序，大的排前面，然后再去贪心。

贪心的过程就是，首先先设置成全部到一号点，然后ans=max(ans,t+exa[i].s1)用这个来进行转移。

不算特别难吧。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <queue>
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 2020;
struct node
{
	int x, y;
	ll s1, s2;
}exa[maxn];
bool cmp(node a,node b)
{
	return a.s1 > b.s1;
}
ll dis(int x,int y,int a,int b)
{
	return 1ll * (x - a)*(x - a) + 1ll * (y - b)*(y - b);
}

int main()
{
	int n, x1, y1, x2, y2;
	cin >> n >> x1 >> y1 >> x2 >> y2;
	for(int i=1;i<=n;i++)
	{
		cin >> exa[i].x >> exa[i].y;
		exa[i].s1 = dis(exa[i].x, exa[i].y, x1, y1);
		exa[i].s2 = dis(exa[i].x, exa[i].y, x2, y2);
	}
	sort(exa + 1,exa + 1 + n, cmp);
	ll ans = inf,t=0;
	for(int i=1;i<=n;i++)
	{
		ans = min(ans, exa[i].s1 + t);
		t = max(t, exa[i].s2);
	}
	printf("%lld
", min(t, ans));
	return 0;
}

　　

C. Wet Shark and Flowers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks i and i + 1 are neighbours for all i from 1 to n - 1. Sharks n and 1 are neighbours too.

Each shark will grow some number of flowers si. For i-th shark value si is random integer equiprobably chosen in range from li to ri. Wet Shark has it's favourite prime number p, and he really likes it! If for any pair of neighbouring sharks i and j the product si·sj is divisible by p, then Wet Shark becomes happy and gives 1000 dollars to each of these sharks.

At the end of the day sharks sum all the money Wet Shark granted to them. Find the expectation of this value.

Input

The first line of the input contains two space-separated integers n and p (3 ≤ n ≤ 100 000, 2 ≤ p ≤ 109) — the number of sharks and Wet Shark's favourite prime number. It is guaranteed that p is prime.

The i-th of the following n lines contains information about i-th shark — two space-separated integers li and ri (1 ≤ li ≤ ri ≤ 109), the range of flowers shark i can produce. Remember that si is chosen equiprobably among all integers from li to ri, inclusive.

Output

Print a single real number — the expected number of dollars that the sharks receive in total. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

Examples

input

Copy

3 2
1 2
420 421
420420 420421

output

Copy

4500.0

input

Copy

3 5
1 4
2 3
11 14

output

Copy

0.0

Note

A prime number is a positive integer number that is divisible only by 1 and itself. 1 is not considered to be prime.

Consider the first sample. First shark grows some number of flowers from 1 to 2, second sharks grows from 420 to 421 flowers and third from 420420 to 420421. There are eight cases for the quantities of flowers (s0, s1, s2) each shark grows:

1. (1, 420, 420420): note that s0·s1 = 420, s1·s2 = 176576400, and s2·s0 = 420420. For each pair, 1000 dollars will be awarded to each shark. Therefore, each shark will be awarded 2000 dollars, for a total of 6000 dollars.
2. (1, 420, 420421): now, the product s2·s0 is not divisible by 2. Therefore, sharks s0 and s2 will receive 1000 dollars, while shark s1 will receive 2000. The total is 4000.
3. (1, 421, 420420): total is 4000
4. (1, 421, 420421): total is 0.
5. (2, 420, 420420): total is 6000.
6. (2, 420, 420421): total is 6000.
7. (2, 421, 420420): total is 6000.
8. (2, 421, 420421): total is 4000.

The expected value is .

In the second sample, no combination of quantities will garner the sharks any money.

借鉴博客

这个题目涉及到概率问题来求期望，我开始并不知道期望怎么求，后来看了一个题解，期望有加分原理，一个期望可以等于每一段的期望相加，这个就很简单了，

剩下的就是有一点点的数论问题。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <queue>
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 100;
ll len[maxn];
ll sum[maxn];
int main()
{
	ll n, p;
	cin >> n >> p;
	for (int i = 1; i <= n; i++)
	{
		int a, b;
		cin >> a >> b;
		len[i] = b - a + 1;
		sum[i] = b / p - (a - 1) / p;
	}
	sum[n + 1] = sum[1];
	len[n + 1] = len[1];
	double ans = 0;
	for(int i=2;i<=n+1;i++)
	{
		double s1 = (len[i] - sum[i])*1.0 / (len[i] * 1.0);
		double s2 = (len[i - 1] - sum[i - 1])*1.0 / (len[i-1] * 1.0);
		ans += 1 - s1 * s2;
	}
	printf("%.6lf
", ans*2000);
	return 0;
}