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

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 <cstdio>#include <string>#include <cstring>#include <vector>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<string> 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;}