Day4 计算糖果、进制转换

#include <stdio.h>
int cnt = 0;
int fib(int n) 
{
    cnt++;
    if (n == 0)
        return 1;
    else if (n == 1)
        return 2;
    else
        return fib(n - 1) + fib(n - 2);
}
void main() 
{
    fib(8);
    printf("%d", cnt);
}

#include <iostream>
using namespace std;
//A = x1 + B
//B = (x3 - x1) / 2
//C = x4 - B
int main() 
{
    int x1, x2, x3, x4;
    cin >> x1 >> x2 >> x3 >> x4;
    int A, B, C;
    B = (x3 - x1) / 2;
    A = x1 + B;
    C = x4 - B;
    //需要进行合法性检测
    if(x1 == A - B && x2 == B - C && x3 == A + B && x4 == B + C)
        cout << A << " " << B << " " << C << endl;
    else
     cout << "No" << endl;
    return 0;
}

class Solution {
public:
    string solve(int M, int N) 
    {
        //所求值数组
        vector<int> solveVal = {'0', '1', '2', '3', '4', '5', '6', '7', '8',
                                '9', 'A', 'B', 'C', 'D', 'E', 'F'};
        int val = M;
        int base = N;   //进制数
        //需要先判断是否为负数
        bool flag = false;
        if(val < 0)
        {
            val *= -1;  //转为正数计算
            flag = true;
        }
        string target;  //所求目标值
        while(val)
        {
            target += solveVal[val % base];
            val /= base;
        }
        if(flag)
            target += '-';  //需要注意负号
        reverse(target.begin(), target.end());  //逆置
        return target;
    }
};