[programming training 4] calculate candy + hexadecimal conversion

1. Calculate candy --》 link

【 describe 】
A,B,C Three people are good friends , Everyone has some candy in his hand , We don't know how many sweets each of them has , But we know the following information ：

A - B, B - C, A + B, B + C.

These four values . Each letter represents the number of sweets each person has .

Now we need to calculate how many sweets each person has in his hand through these four values , namely A,B,C. There is guaranteed to be at most one set of integers A,B,C Meet all problem setting conditions .

Input description ：
Enter as one line , altogether 4 It's an integer , Respectively A - B,B - C,A + B,B + C, Space off . The range is -30 To 30 Between ( Closed interval ).

Output description ：
Output as a line , If there is an integer that satisfies A,B,C Then output... In sequence A,B,C, Space off , There is no space at the end of the line . If such an integer does not exist A,B,C, The output No

Example 1

Input ： 1 -2 3 4

Output ： 2 1 3

【 title 】：
A,B,C It's the amount of candy in three people's hands , We don't know A,B,C How much is the ？ But we know that A - B, B - C, A + B, B + C Result , The result title is given to us by inputting test cases . So this problem is essentially an expression solving problem .

【 Their thinking 】：

1、A - B = a
2、B - C = b
3、A + B = c
4、B + C = d

The essence of this topic is ： Judge whether there is a solution to the ternary system of first-order equations and solve it
1+3 You can get A=(a+c)/2;
4-2 You can get C=(d-b)/2;
2+4 You can get B2=(b+d)/2,
3-1 You can get B1=(c-a)/2;
If B1 Unequal B2 Then the expression has no solution

Code implementation ：

#include<iostream>
using namespace std;
int main()
{
    
int a,b,c,d;
cin>>a>>b>>c>>d;
int A=(a+c)/2;
int C=(d-b)/2;
int B1=(c-a)/2;
int B2=(b+d)/2;
if(B1!=B2)
cout<<"No";
else
cout<<A<<" "<<B1<<" "<<C;
return 0;
}

2. Hexadecimal conversion --》 link

【 describe 】
Given a decimal number M, And the base number to be converted N. Will decimal number M Turn into N Hexadecimal number

Input description ：
Enter as one line ,M(32 An integer )、N(2 ≤ N ≤ 16), Space off .

Output description ：
Output the converted number for each test instance , One line per output . If N Greater than 9, The corresponding number rules refer to 16 Base number （ such as ,10 use A Express , wait ）

Example

Input ： 7 2

Output ： 111

【 Their thinking 】：
The idea is very simple , Take the mold （M%N）, The remainder is the value of the current low base digit , Then by removing the hexadecimal number （M/=N）, Enter the calculation of the next hexadecimal digit .

Code implementation ：

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;
int main()
{
    
    string s,v="0123456789ABCDEF";
    int m,n;
    cin>>m>>n;
    
    //  If it's a negative number , Then it turns into a positive number , And mark it 
    bool flag=false;
    if(m<0)
    {
    
        m=-m;
        flag=true;
    }
    if(m==0)
        cout<<'0'<<endl;

    //  Convert into corresponding characters by base and add them to s
    while(m)
    {
    
        s+=v[m%n];
        m/=n;
    }
   // Indicating starting m It's a negative number , Add the symbol to s in 
    if(flag)
        s+='-';
    reverse(s.begin(),s.end());// Inverted string 
    
    cout<<s<<endl;
    
    
    return 0;
}