A_B_Good Bye 2018

A_B_Good Bye 2018_cf

Alice and Bob are decorating a Christmas Tree.

Alice wants only

In Bob's opinion, a Christmas Tree will be beautiful if:

the number of blue ornaments used is greater by exactly
the number of red ornaments used is greater by exactly

That is, if they have

Alice wants to choose as many ornaments as possible, but she also wants the Christmas Tree to be beautiful according to Bob's opinion.

In the example two paragraphs above, we would choose

Since Alice and Bob are busy with preparing food to the New Year's Eve, they are asking you to find out the maximum number of ornaments that can be used in their beautiful Christmas Tree!

It is guaranteed that it is possible to choose at least

Input

The only line contains three integers

It is guaranteed that it is possible to choose at least

Output

Print one number — the maximum number of ornaments that can be used.

Examples

input

Copy

8 13 9

output

Copy

input

Copy

13 3 6

output

Copy

Note

In the first example, the answer is

In the second example, the answer is

#include <iostream>

using namespace std;

int main()
{
    int y,b,r;
    while(cin>>y>>b>>r){
        for(int i=r;i>=3;i--){
            if(b>=(i-1)&&y>=(i-2)){
                cout<<i+i-1+i-2<<endl;
                break;
            }
        }
    }
    return 0;
}

大佬的代码:

/**
 *    author:  tourist
 *    created: 30.12.2018 17:35:22       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int a, b, c;
  cin >> a >> b >> c;
  int x = min(b, min(a + 1, c - 1));
  cout << 3 * x << '
';
  return 0;
}

Bob is a pirate looking for the greatest treasure the world has ever seen. The treasure is located at the point

Bob travelled around the world and collected clues of the treasure location at

As everyone knows, the world is a two-dimensional plane. The

In other words, each clue belongs to exactly one of the obelisks, and each obelisk has exactly one clue that belongs to it. A clue represents the vector from the obelisk to the treasure. The clues must be distributed among the obelisks in such a way that they all point to the same position of the treasure.

Your task is to find the coordinates of the treasure. If there are multiple solutions, you may print any of them.

Note that you don't need to find the permutation. Permutations are used only in order to explain the problem.

Input

The first line contains an integer

Each of the next

It is guaranteed that there exists a permutation

Output

Output a single line containing two integers

If there are multiple answers, you may print any of them.

Examples

input

Copy

output

Copy

1 2

input

Copy

output

Copy

9 -12

Note

If the hidden permutation is

$\text{[math]}$

In the second sample, the hidden permutation is

我的代码,当然没通过了...在第33组数据超时了.

#include <iostream>
#include <map>
#include <string>
#include <sstream>
#include <cstdio>

using namespace std;

map<string,int> m1;
int obe[1005][2];


int main()
{
    int n,x,y,a,b;
    int resx,resy;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>obe[i][0]>>obe[i][1];
    }
    int t1,t2;std::stringstream ss;
    for(int i=0;i<n;i++){
        //scanf("%d %d",&t1,&t2);
        cin>>t1>>t2;
        int newx,newy;

        for(int j=0;j<n;j++){
            newx=obe[j][0]+t1;
            newy=obe[j][1]+t2;
            string tmp1,tmp2;
            ss.clear();
            ss<<newx;
            ss>>tmp1;
            ss.clear();
            ss<<newy;
            ss>>tmp2;
            tmp1=tmp1+",";
            tmp1=tmp1+tmp2;
            //cout<<tmp1;
            m1[tmp1]++;
        }
    }
    map<string,int>::iterator aa=m1.begin();
    for(;aa!=m1.end();aa++){
        if(aa->second==n){
            string tmp=aa->first;
            int i;
            for(i=0;tmp[i]!=',';i++){
                cout<<tmp[i];
            }
            printf(" ");

            for(i++;i<tmp.size();i++){
                //printf("%c",tmp[i]);
                cout<<tmp[i];
            }
            cout<<endl;
            break;
        }
    }
    return 0;
}

hencuo Code

还是看一下tourist的吧.

题意是n个起点,有n个向量,有个一一对应的关系使每个起点加上一个向量之后是同一个点,让输出那个点的坐标.

所以这些每个起点+向量都是得到相同的坐标,然后把这每个值再加起来就是坐标的n倍啊!

再就是我的基本操作都不会啊,连个int和string 的相互转换都不会...

/**
 *    author:  tourist
 *    created: 30.12.2018 17:36:58       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  long long x = 0, y = 0;
  for (int i = 0; i < 2 * n; i++) {
    int xx, yy;
    cin >> xx >> yy;
    x += xx;
    y += yy;
  }
  cout << (x / n) << " " << (y / n) << '
';
  return 0;
}

拖哥的代码中有两句看不懂的:

ios::sync_with_stdio(false);

cin.tie(0);

原来c++为了兼容c中的stdin stdout,防止混用cin scanf时出现问题,将输入输入流跟他们绑定到一块,然后就导致c++的cin cout速度慢了,
这两句代码可以解除两者的同步和cin与cout的绑定,加上之后再用cin cout 就与scanf printf的速度没什么差别了.
也有人对此做过测试 https://www.byvoid.com/zhs/blog/fast-readfile.
(我的代码方法太挫了,加上这个仍然超时...)