Shuffling Machine
程序员文章站
2024-02-17 13:51:16
...
解题思路:分两步分析:
第一解决替换的问题,从序号为1开始索引,找到所在的位置得到对应序号,此时的序号即为对应字符串的序号,将该字符串与a数组
(就是原数组)的第一位进行替换,此时,必须也对数值数组进行类似替换,然后从序号为2开始索引,再到3,以此类推。
第二个,解决第二个问题,就是捆绑字母与数字,采用二维数组的方式可以做到,一行即为一个字符串 。
#include <iostream>
#include <stdio.h>
using namespace std;
void swap(char a[][4], int b[], int e[]);
void output(char a[][4]);
int main()
{//对a数值 一共是54个元素 占有54行, 这里涉及到二维数组的初始化的知识, 假如赋值没有填满空间的话,系统会自动分配为 '\0'
char a [54][4] = {"S1", "S2", "S3","S4", "S5","S6","S7","S8","S9","S10","S11","S12","S13",
"H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13",
"C1", "C2", "C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13",
"D1", "D2", "D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13","J1","J2"};
//定义b数组, 赋值需要掉换的顺序(即数字) 定义e数组的目的是保证下一次调用swap函数时 b数组的值不变,详见下
int b[55], e[55];
int n;
cin >> n;
//吸收回车符,保证数组b能够赋值
getchar();
for (int g=0; g<54; g++)
cin >> b[g];
//复制 b数组
for (int i=0; i<54; i++)
e[i] = b[i];
//调用已经定义的swap函数
for (int c=0; c<n; c++)
swap(a, b, e);
output(a);
return 0;
}
void swap(char a[][4], int b[], int e[])
{
// 保证每一次调用b数组的值都不变的关键步骤
for (int v=0; v<54; v++)
b[v] = e[v];
int i, j=1, p, q=0;
//交换的中介
char t;
int d;
//元素替换
for (p=0; p<54; p++)
{
//得到数字为j的下标
for (i=p; i<54; i++)
if (b[i] == j) q=i;
//对字符数组进行替换 ,f的意思是使得每次交换 3 次,保证最长的行都能够交换
int k=0, f=3;
for ( ;k<f; k++)
{
t = a[p][k];
a[p][k] = a[q][k];
a[q][k] = t;
}
// 对数值数组进行替换,如果不进行该操作的话会导致上一步的索引出现问题
d = b[p];
b[p] = b[q];
b[q] = d;
j++;
}
}
void output(char a[][4])
{//保证最后一个字符后面没有空格
int i;
for (i=0; i<53; i++)
cout << a[i] << " ";
cout << a[i];
}
上一篇: git 配置 命令行别名
推荐阅读
-
Docker Machine
-
Shuffling Machine
-
Docker Machine
-
Machine Learning
-
《Hands-On Machine Learning with Scikit-Learn & TensorFlow》读书笔记 第三章 分类
-
《Hands-On Machine Learning with Scikit-Learn & TensorFlow》读书笔记 第十四章 循环神经网络
-
Machine Learning With Spark--读书笔记
-
《Hands-On Machine Learning with Scikit-Learn & TensorFlow》读书笔记 第二章 机器学习项目
-
Machine Learning In Action 学习笔记之 KNN算法
-
KNN算法 第二章 Pandas & sklearn 机器学习实战 Machine Learning in action