A. Apple

###Problem Description

Apple is Taotao’s favourite fruit. In his backyard, there are three apple trees with coordinates (x1,y1), (x2,y2), and (x3,y3). Now Taotao is planning to plant a new one, but he is not willing to take these trees too close. He believes that the new apple tree should be outside the circle which the three apple trees that already exist is on. Taotao picked a potential position (x,y) of the new tree. Could you tell him if it is outside the circle or not?

###Input

The first line contains an integer T, indicating that there are T(T≤30) cases.
In the first line of each case, there are eight integers x1,y1,x2,y2,x3,y3,x,y, as described above.
The absolute values of integers in input are less than or equal to 1,000,000,000,000.
It is guaranteed that, any three of the four positions do not lie on a straight line.

###Output

For each case, output “Accepted” if the position is outside the circle, or “Rejected” if the position is on or inside the circle.

###Sample Input

3
-2 0 0 -2 2 0 2 -2
-2 0 0 -2 2 0 0 2
-2 0 0 -2 2 0 1 1

###Sample Output

Accepted
Rejected
Rejected

###Source

輸入輸出測試

程式碼塊

####錯誤程式碼
在做這一個題目時，我首先考慮到的是找三角形的外接圓，然後確定外心，確定外接圓的半徑。
在確定了外接圓的半徑和圓心後，我們就很容易判斷第四個點是否在圓內了。
步驟：
1.計算外接圓圓心
2.計算外接圓半徑
3.判斷點是否在圓內
4.在圓內輸出“Rejected”，不在圓內輸出“Accepted”
但是最終的結果是錯誤的！！

//錯誤程式碼
#include<stdio.h>//runtime error 除零錯誤   
#include<iostream>  
#include<math.h>  
using namespace std;  
int main()  
{  
  
        int x0,y0,x1,y1,x2,y2,x3,y3;  
        int x4,y4;  
        double dis,r;  
        int n;  
        cin>>n;  
        for(int i=0;i<n;i++)  
        {  
            cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;              
            x0 = ((y3-y1)*(y2*y2-y1*y1+x2*x2-x1*x1)+(y2-y1)*(y1*y1-y3*y3+x1*x1-x3*x3))/(2*(x2-x1)*(y3-y1)-2*(x3-x1)*(y2-y1));  
            y0 = ((x3-x1)*(x2*x2-x1*x1+y2*y2-y1*y1)+(x2-x1)*(x1*x1-x3*x3+y1*y1-y3*y3))/(2*(y2-y1)*(x3-x1)-2*(y3-y1)*(x2-x1));  
            r  = sqrt((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0));  
            dis= sqrt((x4-x0)*(x4-x0)+(y4-y0)*(y4-y0));  
            if(dis <= r)  
            {   
            cout<<"Rejected"<<endl;  
            }  
             
            else  
            { 
			 cout<<"Accepted"<<endl;  
            }   
    }  
    return 0;  
} 

下面看這樣一個程式碼

public class ssss {  
  
    public static void main(String[] ages){  
        double d1=2.07;  
        double d2=1.03;  
        System.out.println(d1+d2);  
    }  
} 

其輸出結果為
3.0999999999999996

雖然計算結果離精確值誤差很小，但其不是精確的！這在像如金融計算一樣計算精確度要求很高的領域是無法接受的，但這是二進位制本身的問題，而計算機普遍採用二進位制表示，使用基本資料型別無法解決。

為了解決基本資料型別浮點數不能進行精確計算的問題，Java中專門提供了java.math.BigDecimal類，其提供浮點數的精確計算功能。與BigInteger類相同，其運算操作均使用方法呼叫完成

以下是java.math.BigDecimal.multiply()方法宣告
public BigDecimal multiply(BigDecimal multiplicand)
BigDecimal.valueOf(qty)是把qty這個數轉為BigDecimal型別的，BigDecimal是可以處理任意長度的浮點數運算的。

Ⅰ基本函式：
1.valueOf(parament); 將引數轉換為制定的型別
String s=”12345”; BigInteger c=BigInteger.valueOf(s); 則c=12345；
BigInteger a=new BigInteger(“23”); BigInteger b=new BigInteger(“34”); a. add(b); 3.subtract(); 相減
5.divide(); 相除取整
6.remainder(); 取餘
8.gcd(); 最大公約數
10.negate(); 取反數
11.mod(); a.mod(b)=a%b=a.remainder(b);
13.punlic int comareTo();
14.boolean equals(); 是否相等
15.BigInteger建構函式： 一般用到以下兩種： BigInteger(String val); 將指定字串轉換為十進位制表示形式； BigInteger(String val,int radix); 將指定基數的 BigInteger 的字串表示形式轉換為 BigInteger
Ⅱ.基本常量： A=BigInteger.ONE 1
C=BigInteger.ZERO 0
Ⅲ.基本操作

1. 讀入： 用Scanner類定義物件進行控制檯讀入,Scanner類在java.util.*包中
   Scanner cin=new Scanner(System.in);// 讀入 while(cin.hasNext()) //等同於!=EOF { int n; BigInteger m; n=cin.nextInt(); //讀入一個int;
   m=cin.BigInteger();//讀入一個BigInteger; System.out.print(m.toString()); }

package 大數問題;

import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        BigDecimal x1,x2,x3,y1,y2,y3,x,y,x0,y0,r;
        BigDecimal a,b,c,d,e,f,g,tt,dis;
        Scanner cin = new Scanner(System.in);
        int ncase;
        ncase = cin.nextInt();
        while(ncase-->0)
        {
            x1  = cin.nextBigDecimal();
            y1  = cin.nextBigDecimal();
            x2  = cin.nextBigDecimal();
            y2  = cin.nextBigDecimal();
            x3  = cin.nextBigDecimal();
            y3  = cin.nextBigDecimal();
            x  = cin.nextBigDecimal();
            y  = cin.nextBigDecimal();
            a = x3.subtract(x2).multiply(BigDecimal.valueOf(2));// 2*(x3-x2)
            b = y3.subtract(y2).multiply(BigDecimal.valueOf(2));// 2*(y3-y2)
            c = x3.pow(2).subtract(x2.pow(2)).add(y3.pow(2).subtract(y2.pow(2)));
              //x3^2-x2^2+(y3^2-y2^2)
            e = x2.subtract(x1).multiply(BigDecimal.valueOf(2));//2*(x2-x1)
            f = y2.subtract(y1).multiply(BigDecimal.valueOf(2));//2*(y2-y1)
            g = x2.pow(2).subtract(x1.pow(2)).add(y2.pow(2).subtract(y1.pow(2)));
          x0=g.multiply(b).subtract(c.multiply(f)).divide(e.multiply(b).subtract(a.multiply(f)));
               y0=a.multiply(g).subtract(c.multiply(e)).divide(a.multiply(f).subtract(b.multiply(e)));
            tt = (x1.subtract(x0)).pow(2).add((y1.subtract(y0)).pow(2));//(x1-x0)^2+(y1-y0)^2
            dis = (x.subtract(x0)).pow(2).add((y.subtract(y0)).pow(2));
            if (dis.compareTo(tt)>0) {//dis.compareTo(tt) 相當於 dis與tt相比較 類比strcmp()
                System.out.println("Accepted");
            }
            else
                System.out.println("Rejected");
        }
    }

}

H. Chinese Zodiac

The Chinese Zodiac, known as Sheng Xiao, is based on a twelve-year cycle, each year in the cycle related to an animal sign. These signs are the rat, ox, tiger, rabbit, dragon, snake, horse, sheep, monkey, rooster, dog and pig.
Victoria is married to a younger man, but no one knows the real age difference between the couple. The good news is that she told us their Chinese Zodiac signs. Their years of birth in luner calendar is not the same. Here we can guess a very rough estimate of the minimum age difference between them.
If, for instance, the signs of Victoria and her husband are ox and rabbit respectively, the estimate should be 22 years. But if the signs of the couple is the same, the answer should be 1212 years.
Input
The first line of input contains an integer T (1≤T≤1000)T (1≤T≤1000) indicating the number of test cases.
For each test case a line of two strings describes the signs of Victoria and her husband.
Output
For each test case output an integer in a line.

Sample Input
3
ox rooster
rooster ox
dragon dragon
Sample Output
8
4
12

#include<stdio.h> 
#include<stdlib.h> 
#include<iostream> 
#include<algorithm> 
#include<string.h>
using namespace std; 
int num1,num2,num=12;
//rat, ox, tiger, rabbit, dragon, snake, horse, sheep, monkey, rooster, dog and pig
int s(char *str)
{
	if(strcmp(str,"rat")==0)
	 return 1;
	if(strcmp(str,"ox")==0)
	 return 2;
	if(strcmp(str,"tiger")==0)
	 return 3;
	if(strcmp(str,"rabbit")==0)
	 return 4;
	if(strcmp(str,"dragon")==0)
	 return 5;
	if(strcmp(str,"snake")==0)
	 return 6;
	if(strcmp(str,"horse")==0)
	 return 7;
	if(strcmp(str,"sheep")==0)
	 return 8;
	if(strcmp(str,"monkey")==0)
	 return 9;
	if(strcmp(str,"rooster")==0)
	 return 10;
	if(strcmp(str,"dog")==0)
	 return 11;
	if(strcmp(str,"pig")==0)
	 return 12;	 

}
int main(){
  int T;
  char str1[20],str2[20];
  scanf("%d",&T);
  while(T--)
  {
     scanf("%s%s",str1,str2);
	  num1=s(str1); 
	  num2=s(str2);   
	 int ans=num2-num1;
	 if(ans>0)
	 {
	 	printf("%d\n",ans);
	 }else
	 {
	 	ans=ans+12;
	 	printf("%d\n",ans);
	 }
  }
  return 0;	
}

K.A Cubic number and A Cubic Number

A cubic number is the result of using a whole number in a multiplication three times. For example, 3×3×3=273×3×3=27 so 2727 is a cubic number. The first few cubic numbers are 1,8,27,641,8,27,64 and 125125. Given an prime number pp. Check that if pp is a difference of two cubic numbers.
Input
The first of input contains an integer T (1≤T≤100)T (1≤T≤100) which is the total number of test cases.
For each test case, a line contains a prime number p (2≤p≤1012)p (2≤p≤1012).
Output
For each test case, output ‘YES’ if given pp is a difference of two cubic numbers, or ‘NO’ if not.

Sample Input
10
2
3
5
7
11
13
17
19
23
29
Sample Output
NO
NO
NO
YES
NO
NO
NO
YES
NO
NO

公式推導：

#include<stdio.h> 
#include<stdlib.h> 
#include<iostream> 
#include<algorithm> 
#include<string.h>
using namespace std;
typedef long long int ll;
bool flag=0;
int main()
{
    ll T,p;
    double n;
    scanf("%lld",&T);
    while(T--)
    {
    	flag=0;
    	scanf("%lld",&p);
    	ll ans;
        for(ll i=1;(3*i*i+3*i+1)<=p;i++)
		{
		  if((3*i*i+3*i+1)==p)
		  {
		    	flag=1;
		    	printf("YES\n");
		    	break;
     	  }      	
		}
		if(flag==0)
		{
		   printf("NO\n");
		}	
	}
    return 0;
}