瀋陽網絡賽

E .

You promised your girlfriend a rounded cake with at least SSS strawberries.

But something goes wrong, you don't have the final cake but an infinite plane with NNN strawberries on it. And you need to draw a circle which contains strawberries completely to make the cake by yourself.

To simplify the problem, the strawberries are represented as circles in 222D plane. Every strawberry has its center (Xi,Yi)(X_i, Y_i)(Xi​,Yi​) and the same radius RRR.

The cost of your cake is the radius of the circle you draw. So you want to know the minimal cost of the cake, or if you can't give your girlfriend such a cake.

Input

The input file contains several test cases.

The first line of the file is an integer TTT, giving the number of test cases. (T≤20)(T\leq20)(T≤20)

For each test case, there are two integers NNN and SSS in the first line, denoting the number of total strawberries and the number of strawberries on cake as you promised. (1≤N,S≤300)(1\leq N, S \leq 300)(1≤N,S≤300)

In the following NNN lines, each line contains two integers Xi,YiX_i, Y_iXi​,Yi​, denoting the coordinates of the strawberry center. (0≤Xi,Yi≤10000)(0\leq X_i, Y_i\leq 10000)(0≤Xi​,Yi​≤10000)

The next line contains one integer RRR, denoting the radius of all strawberries. (1≤R≤10000)(1\leq R \leq 10000)(1≤R≤10000)

It's guaranteed that sum of NNN is smaller than 100010001000.

Output

Output the minimal cost correct to four decimal places if you can make a rounded cake with at least SSS strawberries, or "The cake is a lie." if not.

樣例輸入

2
5 3
0 0
0 2
2 0
0 0
1 1
1
1 2
0 0
1

樣例輸出

1.7071
The cake is a lie.

題意 ： 給你平面上的 n 個點，問你最少要用半徑爲多少的圓能夠使園內包含 m 個點
思路分析 ： 最小圓覆蓋板子題
代碼示例：

#include <bits/stdc++.h>
using namespace std;
 
#define Mn 305
const double eps = 1e-7;
const double pi = acos(-1.0);
struct Point
{
	double x,y;
	Point() {}
	Point(double tx,double ty)
	{
		x=tx;
		y=ty;
	}
} p[Mn+5];
struct Node
{
	double angle;
	bool in;
} arc[Mn * Mn + 5];
int n;
double dist(Point p1,Point p2)
{
	return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
bool cmp(Node &n1,Node &n2)
{
	return n1.angle<n2.angle;
}
double R;
 
int MaxCircleCover()
{
    int ans=1;
    for(int i=0;i<n;i++)
	{
        int cnt=0;
        for(int j=0;j<n;j++)
		{
            if(i==j) continue;
            double D = dist(p[i],p[j]);
            if(D>2.0*R) continue;
            
            double angle=atan2(p[i].y-p[j].y,p[i].x-p[j].x);
            if(angle < 0)
                angle += 2 * pi;
                
            double phi=acos(D/(2.0*R));
            arc[cnt].angle=angle-phi+ 2 * pi;arc[cnt++].in=true;
            arc[cnt].angle=angle+phi+ 2 * pi;arc[cnt++].in=false;
        }
        sort(arc,arc+cnt,cmp);
        int tmp=1;
        for(int j=0;j<cnt;j++)
		{
            if(arc[j].in)  tmp++;
            else tmp--;
            ans=max(ans,tmp);
        }
    }
    return ans;
}
 
int main()
{
	int t;scanf("%d",&t);
	while(t--)
	{
		int s;
		scanf("%d%d",&n,&s);
		for(int i=0;i<n;i++)
			scanf("%lf%lf",&p[i].x,&p[i].y);
		double r;
		scanf("%lf",&r);
		double l1=0,r1=100000;
		if(n<s)
		{
			printf("The cake is a lie.\n");continue;
		}
		while(abs(r1-l1)>eps)
		{
			double mid=(l1+r1)*0.5;
			R=mid;
			int p=MaxCircleCover();
			if(p>=s)
			{
				r1=mid;
			}
			else
			{
				l1=mid;
			}
		}
		printf("%.4lf\n",l1+r);
	}
	return 0;
}

 

G .

 

思路分析 ：

　　一個長整型數字他的質因子最多隻有 10 個，且分解的次數不會超過 40 次

　　再計算 1-n  中與 m 互質的狀況下，能夠用經典的容斥原理，這裏有一個二進制的優化

代碼示例 ：

#define ll long long
const ll maxn = 1e6+5;
const ll mod = 1e9+7;
const ll inv6 = 166666668;
const ll inv2 = 500000004;

ll n, m;
ll prime[100];
ll cnt;
void getprime(){
    cnt = 0; ll num = m;
    for(ll i = 2; i*i <= m; i++){
        if (num%i == 0) {
            prime[cnt++] = i;
            while(num%i == 0){
                num /= i;
            }
        }
        if (num == 1) break;
    }
    if (num > 1) prime[cnt++] = num; 
}

ll cal1(ll tem, ll x){
    ll ans = tem*tem%mod*x%mod*(x+1)%mod*(2*x+1)%mod*inv6%mod;
    return ans;
}
ll cal2(ll tem, ll x){
    ll ans = tem*x%mod*(x+1)%mod*inv2%mod;
    return ans;
}

void solve() {
    ll ans = 0;
    for(ll i = 1; i < (1<<cnt); i++){
        ll f = 0; ll tem = 1;
        for(ll j = 0; j < cnt; j++){
            if (i&(1<<j)) {
                f++;
                tem *= prime[j];
            }
        }
        ll time = n/tem;
        if (f&1) ans = (ans+cal1(tem, time)+cal2(tem, time))%mod;
        else ans = (ans-cal1(tem, time)-cal2(tem, time))%mod;
        ans = (ans+mod)%mod;
    }
    ll sum = (cal1(1, n)+cal2(1, n))%mod; 
    ans = (sum-ans)%mod;
    printf("%lld\n", (ans+mod)%mod);
}

int main() {

    while(~scanf("%lld%lld", &n, &m)){
        getprime();
        solve();
    }
    return 0;
}