PTA甲级1082 Read Number in Chinese (25分)

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0
        bool 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;
}