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首先,先贴柳神的博客

https://www.liuchuo.net/ 这是地址

想要刷好PTA,强烈推荐柳神的博客,和算法笔记

题目原文

Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way. Output Fu first if it is negative. For example, -123456789 is read as Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu. Note: zero (ling) must be handled correctly according to the Chinese tradition. For example, 100800 is yi Shi Wan ling ba Bai.

Input Specification:

Each input file contains one test case, which gives an integer with no more than 9 digits.

Output Specification:

For each test case, print in a line the Chinese way of reading the number. The characters are separated by a space and there must be no extra space at the end of the line.

Sample Input 1:

-123456789

Sample Output 1:

Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu  

Sample Input 2:

100800

Sample Output 2:

yi Shi Wan ling ba Bai

题目大意

就是给你一个数字,让你用中国的读法输出出来

解题思路

① 找到这个九位数字的特点,可以把它4个一分为一节

② 用两个bool类型的量

一个用来判断一节里面有没有零

一个用来判断一节里面有没有已经输出的数

具体的我也不是很懂,这题是抄的算法笔记的

代码如下

#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
using namespace std;
char num[10][5] = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };
char wei[5][5] = { "Shi","Bai","Qian","Wan","Yi" };
int main() {char str[15];cin >> str;int len = strlen(str);	//字符串的长度int left = 0, right = len - 1;	//left与right分别指向字符串的首尾元素if (str[0] == '-') {printf("Fu");		//如果是负数left++;}while (left + 4 <= right) {right -= 4;			//将right每次左移4位,直到left与right在同一节}while (left < len) {	//循环每次处理数字的一节(4位或小于4位)bool flag = false;		//flag==flase	表示没有积累的0bool isPrint = false;	//isPrint==false	表示该节没有输出过其中的位while (left <= right) {if (left > 0 && str[left] == '0') {flag = true;		//令标记flag为true}else {if (flag == true) {printf(" ling");flag = false;}//只要不是首位(包括负号),后面的每一位前都要输出空格if (left > 0)printf(" ");printf("%s", num[str[left] - '0']);isPrint = true;		//该节至少有一位被输出if (left != right) {printf(" %s", wei[right - left - 1]);}}left++;	//left右移1位}if (isPrint == true && right != len - 1) {printf(" %s", wei[(len - 1 - right) / 4 + 2]);}right += 4;			//right右移4位,输出下一节}return 0;
}