欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

1082. Read Number in Chinese (25)【字符串处理】——PAT (Advanced Level) Practise

程序员文章站 2022-06-24 20:33:57
题目信息 1082. Read Number in Chinese (25) 时间限制400 ms 内存限制65536 kB 代码长度限制16000 B Given an...

题目信息

1082. Read Number in Chinese (25)

时间限制400 ms
内存限制65536 kB
代码长度限制16000 B
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

解题思路

注意细节

AC代码

#include 
#include 
#include 
#include 
using namespace std;
char num[][5] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
char step[][5] = {"", "Shi", "Bai", "Qian"};
char bigStep[][5] = {"", "Wan", "Yi"};
vector ans, qans;
void show(string s, bool first)
{
    int i = 0, t = -1;
    while (i < s.size() && '0' == s[i]) ++i;
    t = i - 1;
    if (i != 0 && (ans.empty() || ans[ans.size() - 1] != "ling")) ans.push_back("ling");
    for (; i < s.size(); ++i){
        if (t + 1 != i) ans.push_back("ling");
        t = i;
        ans.push_back(num[s[i] - '0']);
        ans.push_back(step[s.size() - 1 - i]);
        while (i + 1 < s.size() && '0' == s[i + 1]) ++i;
        first = false;
    }
}
int main()
{
    char s[15];
    char *p = s;
    gets(s);
    int len = strlen(s);
    if (s[0] == '-'){
        ans.push_back("Fu");
        --len;
        ++p;
    }
    for (int i = 8; i >= 0; i -= 4){
        if (len > i){
            show(string(p, p + len - i), p == s);
            if (ans[ans.size() - 1] != "ling") ans.push_back(bigStep[i/4]);
            p += len - i;
            len -= len - i;
            if (strspn(p, "0") == len) break;
        }
    }
    for (int i = 0; i < ans.size(); ++i){
        if (ans[i] != "") qans.push_back(ans[i]);
    }
    for (int i = 0; i < qans.size() - 1; ++i){
        printf("%s ", qans[i].c_str());
    }
    printf("%s\n", qans[qans.size() - 1].c_str());
    return 0;
}