啦啦啦

这道题是找可以组成矩形的四根木棍，使组成的满足那个公式最小，用memset可能会超时，化简公式之后，要让x=y最小，那我们就要尽可能的接近

C. Minimum Value Rectangle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have $\mathrm{nn sticks\; of\; the\; given\; lengths.}$

Your task is to choose exactly four of them in such a way that they can form a rectangle. No sticks can be cut to pieces, each side of the rectangle must be formed by a single stick. No stick can be chosen multiple times. It is guaranteed that it is always possible to choose such sticks.

Let $\mathrm{SS be\; the\; area\; of\; the\; rectangle\; and PP be\; the\; perimeter\; of\; the\; rectangle.}$

The chosen rectangle should have the value $\frac{{\mathrm{P2SP2S minimal\; possible.\; The\; value\; is\; taken\; without\; any\; rounding.}}^{}}{}$

If there are multiple answers, print any of them.

Each testcase contains several lists of sticks, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer $\mathrm{TT (T\ge 1T\ge 1)\; —\; the\; number\; of\; lists\; of\; sticks\; in\; the\; testcase.}$

Then $\mathrm{2T2T lines\; follow\; —\; lines (2i-1)(2i-1) and 2i2i of\; them\; describe\; the ii-th\; list.\; The\; first\; line\; of\; the\; pair\; contains\; a\; single\; integer nn (4\le n\le 1064\le n\le 106)\; —\; the\; number\; of\; sticks\; in\; the ii-th\; list.\; The\; second\; line\; of\; the\; pair\; contains nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le aj\le 1041\le aj\le 104)\; —\; lengths\; of\; the\; sticks\; in\; the ii-th\; list.}$

It is guaranteed that for each list there exists a way to choose four sticks so that they form a rectangle.

The total number of sticks in all $\mathrm{TT lists\; doesn\text{'}t\; exceed 106106 in\; each\; testcase.}$

Output

Print $\mathrm{TT lines.\; The ii-th\; line\; should\; contain\; the\; answer\; to\; the ii-th\; list\; of\; the\; input.\; That\; is\; the\; lengths\; of\; the\; four\; sticks\; you\; choose\; from\; theii-th\; list,\; so\; that\; they\; form\; a\; rectangle\; and\; the\; value P2SP2S of\; this\; rectangle\; is\; minimal\; possible.\; You\; can\; print\; these\; four\; lengths\; in\; arbitrary\; order.}$

If there are multiple answers, print any of them.

Example

input

Copy

3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5

output

Copy

2 7 7 2
2 2 1 1
5 5 5 5

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int a[1000001],j,b[1000001];
int main()
{
    int t,n,i,m,x,y;
    cin>>t;
    while(t--)
    {
        for(i=0; i<10001; i++)
            a[i]=0;
        int p=0;
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
           scanf("%d",&m);
            a[m]++;
            if(a[m]==2||a[m]==4)
            {
                b[p++]=m;
            }
        }
        sort(b,b+p);
        double minn=inf;
        for(i=0;i<p-1;i++)
        {
            if(b[i]==b[i+1])
            {
                x=b[i];
                y=b[i+1];
                break;
            }
            double temp=(double)b[i+1]/b[i];
            if(temp<minn)
            {
                minn=temp;
                x=b[i];
                y=b[i+1];
            }
        }return 0;

}

这道题就是找规律我们可以得到答案2k+1∗n−(2k−1)，但是为了避免超时，要使用快乘和快速幂解决，还有要注意的一点是取模之后公式前面可能比后面小，所以我们要加上mod再取余一次，不然就是负的

C. Nastya and a Wardrobe

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Nastya received a gift on New Year — a magic wardrobe. It is magic because in the end of each month the number of dresses in it doubles (i.e. the number of dresses becomes twice as large as it is in the beginning of the month).

Unfortunately, right after the doubling the wardrobe eats one of the dresses (if any) with the 50% probability. It happens every month except the last one in the year.

Nastya owns x dresses now, so she became interested in the expected number of dresses she will have in one year. Nastya lives in Byteland, so the year lasts for k + 1 months.

Nastya is really busy, so she wants you to solve this problem. You are the programmer, after all. Also, you should find the answer modulo 109 + 7, because it is easy to see that it is always integer.

Input

The only line contains two integers x and k (0 ≤ x, k ≤ 1018), where x is the initial number of dresses and k + 1 is the number of months in a year in Byteland.

Output

In the only line print a single integer — the expected number of dresses Nastya will own one year later modulo 109 + 7.

Examples

input

Copy

2 0

output

Copy

4

input

Copy

2 1

output

Copy

7

input

Copy

3 2

output

Copy

21

#include <iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<cstdio>
using namespace std;
const int mod=1e9+7;
long long quickmul(long long a,long long b)
{
    long long ans=0;
    while(b)
    {
    if(b&1)
    {
        ans=(ans+a)%mod;
    }
    a=(a+a)%mod;
    b>>=1;
    }
    return  ans;
}
long long quickpow(long long a,long long b)
{
    long long ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=(ans*a)%mod;
        }
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}
int main()
{
  long long n,k;
  cin>>n>>k;
  if(n==0)
  {
      cout<<0<<endl;
      return 0;
  }
  long long res=(quickmul(quickpow(2, k + 1), n) % mod) - (quickpow(2, k) - 1 + mod) % mod;
  res=(res+mod)%mod;
  cout<<res<<endl;
  return 0;
}

还有一道是那天打比赛的题，一定要记住多边形内角和公式：（n-2）*180，然后计算出多边形面积，减去圆的面积，每层叠加就行了。

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#define pi acos(-1.0)
using namespace std;
int main()
{
    freopen("glorious.in","r",stdin);
    int t,n,k,cnt=0;
    double r;
    scanf("%d",&t);
    while(t--)
    {
        cnt++;
        scanf("%d%lf%d",&n,&r,&k);
         double sum=0;
        while(n--)
        {
            double r2=(double)r/(sin(pi*(k-2)/k/2));
            double kepa=2.0*sqrt(r2*r2-(double)r*r);
            sum+=(double)k/2.0*kepa*r-pi*r*r;