[算法系列]-算法刷题及关于引用的总结-2

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1027 Colors in Mars (20 分)

People in Mars represent the colors in their computers in a similar way as the Earth people. That is, a color is represented by a 6-digit number, where the first 2 digits are for Red, the middle 2 digits for Green, and the last 2 digits for Blue. The only difference is that they use radix 13 (0-9 and A-C) instead of 16. Now given a color in three decimal numbers (each between 0 and 168), you are supposed to output their Mars RGB values.

Input Specification:

Each input file contains one test case which occupies a line containing the three decimal color values.

Output Specification:

For each test case you should output the Mars RGB value in the following format: first output #, then followed by a 6-digit number where all the English characters must be upper-cased. If a single color is only 1-digit long, you must print a 0 to its left.

Sample Input:

15 43 71

Sample Output:

#123456

#include<iostream>
using namespace std;
char radix[13] = {
    '0','1','2','3','4','5','6','7','8','9','A','B','C'
};
int main(){
    int r,g,b;
    cin>>r>>g>>b;
    cout<<("#");
    cout<<radix[r/13]<<radix[r%13];
    cout<<radix[g/13]<<radix[g%13];
    cout<<radix[b/13]<<radix[b%13];
    return 0;
}

算法笔记：引用

虽然在定义上，说是别名，不占用内存空间，但是再实际实现的时候，还是很难做到，所以大多数编译器在处理引用的时候，还是将其翻译为const 指针, 即保存原变量的指针，当然，如果编译器继续优化，则可能出现很多情况

指针变量其实是unsigned int类型的整数。

所以

void swap(int* &p1, int* &p2)

的意思是，将 p1 , p2 看作两个变量的别名，直接对存放源地址的变量进行操作。

void swap(int* p1, int* p2)

如果不带引用符号，那就可以理解传入的是两个unsigned int的整数，所谓对原变量的操作，其实是通过这连个整数，找到在对应内存中所存在的位置，然后对内存进行操作。

还有，实际上对于计算机而言，并不存在指针或者引用，你可以将它们看作是是对 地址的封装，还加上了一些额外信息。