A1082. Read Number in Chinese

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

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int num[] = {}, pt;
void num2arr(int N){
    pt = ;
    do{
        num[pt++] = N % ;
        N = N / ;
    }while(N != );
}
char strNum[][] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
char duan[][] = {"", "Shi", "Bai", "Qian"};
char prt[][];
int main(){
    int N, neg = , len = ;
    scanf("%d", &N);
    if(N < ){
        neg = -;
        N = N * -;
    }
    num2arr(N);
    if(neg == -){
        strcpy(prt[len++], "Fu");
    }
    int zeroTag = , wanTag = ;
    if(N == ){
        printf("ling");
        return ;
    }
    for(int i = pt - ; i>= ; i--){
        if(i == ){
            strcpy(prt[len++], strNum[num[i]]);
            strcpy(prt[len++], "Yi");
        }else if(num[i] == ){
            zeroTag = ;
        }else if(num[i] != ){
            if(i >  && i < ){
                wanTag = ;
            }
            if(zeroTag == ){
                zeroTag = ;
                strcpy(prt[len++], strNum[]);
            }
            strcpy(prt[len++], strNum[num[i]]);
            if(i %  != )
                strcpy(prt[len++], duan[i % ]);
        }
        if(i ==  && wanTag == ){
                strcpy(prt[len++], "Wan");
        }
    }
    for(int i = ; i < len; i++){
        if(i == len - )
            printf("%s", prt[i]);
        else printf("%s ", prt[i]);
    }
    cin >> N;
    return ;
}

总结：

1、属于字符串处理问题，我写的不太好，写完后才发现问题，于是只能修修补补勉强过了测试点。由于题目所给的数字最多才到亿，可以分成三段（如果够分的话），亿、万、个，分别处理。对于每一段，分别读数（几千几百几十几），读完每一段再分别输出‘亿’、‘万’。

2、对于零的处理：由于多个零可能按一个零读（1001），也可能不读（1000），这个需要额外注意。可以采取这样的办法：遇到0时不要立马读出，而是设置一个zero标志，该符号表示是否已经有了未读的0，在遇到一个非0的数字时，再检查zero标志看看是否有0未读，如果有则先读0，再读这个非0数字，再修改zero标志。 另外，在读完一段数字时，需要把zero标志置为没有未读0（如 1,1230,1234）。

3、注意N = 0时的处理。注意万段全为0时不可读“万”，且要读一个0（1,0000,1234）。

4、不好直接输出的，可以先拷贝到prt数组中，之后统一输出。