A_B_Good Bye 2018_cf

Posted 2021-01-31 tws-yifei

A. New Year and the Christmas Ornament

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Alice and Bob are decorating a Christmas Tree.

Alice wants only 33 types of ornaments to be used on the Christmas Tree: yellow, blue and red. They have yy yellow ornaments, bb blue ornaments and rr red ornaments.

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

• the number of blue ornaments used is greater by exactly 11 than the number of yellow ornaments, and
• the number of red ornaments used is greater by exactly 11 than the number of blue ornaments.

That is, if they have 88 yellow ornaments, 1313 blue ornaments and 99 red ornaments, we can choose 44 yellow, 55 blue and 66 red ornaments (5=4+15=4+1 and 6=5+16=5+1).

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 77 yellow, 88 blue and 99 red ornaments. If we do it, we will use 7+8+9=247+8+9=24ornaments. That is the maximum number.

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 66 (1+2+3=61+2+3=6) ornaments.

Input

The only line contains three integers yy, bb, rr (1≤y≤1001≤y≤100, 2≤b≤1002≤b≤100, 3≤r≤1003≤r≤100) — the number of yellow, blue and red ornaments.

It is guaranteed that it is possible to choose at least 66 (1+2+3=61+2+3=6) ornaments.

Output

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

Examples

input

Copy

8 13 9

output

Copy

24

input

Copy

13 3 6

output

Copy

9

Note

In the first example, the answer is 7+8+9=247+8+9=24.

In the second example, the answer is 2+3+4=92+3+4=9.

我的代码:

#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;
}

B. New Year and the Treasure Geolocation

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bob is a pirate looking for the greatest treasure the world has ever seen. The treasure is located at the point TT, which coordinates to be found out.

Bob travelled around the world and collected clues of the treasure location at nn obelisks. These clues were in an ancient language, and he has only decrypted them at home. Since he does not know which clue belongs to which obelisk, finding the treasure might pose a challenge. Can you help him?

As everyone knows, the world is a two-dimensional plane. The ii-th obelisk is at integer coordinates (xi,yi)(xi,yi). The jj-th clue consists of 22integers (aj,bj)(aj,bj) and belongs to the obelisk pjpj, where pp is some (unknown) permutation on nn elements. It means that the treasure is located at T=(xpj+aj,ypj+bj)T=(xpj+aj,ypj+bj). This point TT is the same for all clues.

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 nn (1≤n≤10001≤n≤1000) — the number of obelisks, that is also equal to the number of clues.

Each of the next nn lines contains two integers xixi, yiyi (?106≤xi,yi≤106?106≤xi,yi≤106) — the coordinates of the ii-th obelisk. All coordinates are distinct, that is xi≠xjxi≠xj or yi≠yjyi≠yj will be satisfied for every (i,j)(i,j) such that i≠ji≠j.

Each of the next nn lines contains two integers aiai, bibi (?2?106≤ai,bi≤2?106?2?106≤ai,bi≤2?106) — the direction of the ii-th clue. All coordinates are distinct, that is ai≠ajai≠aj or bi≠bjbi≠bj will be satisfied for every (i,j)(i,j) such that i≠ji≠j.

It is guaranteed that there exists a permutation pp, such that for all i,ji,j it holds (xpi+ai,ypi+bi)=(xpj+aj,ypj+bj)(xpi+ai,ypi+bi)=(xpj+aj,ypj+bj).

Output

Output a single line containing two integers Tx,TyTx,Ty — the coordinates of the treasure.

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

Examples

input

Copy

2
2 5
-6 4
7 -2
-1 -3

output

Copy

1 2

input

Copy

4
2 2
8 2
-7 0
-2 6
1 -14
16 -12
11 -18
7 -14

output

Copy

9 -12

Note

As n=2n=2, we can consider all permutations on two elements.

If p=[1,2]p=[1,2], then the obelisk (2,5)(2,5) holds the clue (7,?2)(7,?2), which means that the treasure is hidden at (9,3)(9,3). The second obelisk (?6,4)(?6,4) would give the clue (?1,?3)(?1,?3) and the treasure at (?7,1)(?7,1). However, both obelisks must give the same location, hence this is clearly not the correct permutation.

If the hidden permutation is [2,1][2,1], then the first clue belongs to the second obelisk and the second clue belongs to the first obelisk. Hence (?6,4)+(7,?2)=(2,5)+(?1,?3)=(1,2)(?6,4)+(7,?2)=(2,5)+(?1,?3)=(1,2), so T=(1,2)T=(1,2) is the location of the treasure.

In the second sample, the hidden permutation is [2,3,4,1][2,3,4,1].

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...)