题目例如以下：
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:-123456789Sample Output 1:
Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiuSample Input 2:
100800Sample Output 2:
yi Shi Wan ling ba Bai
题目要求依照中国人的习惯阅读一个不超过9位的数字。这道题的坑比較多，一定要考虑到全部的情况。
首先要抓住规律。我们能够发现，一个数字的读法在每一个4位是一致的。
比如12341234。我们读作“一千二百三十四万一千二百三十四”，我们能够看到除去万字以外读法全然一致。
因此我们集中精力解决四位数的读法。然后加上亿、万等就可以。
在解决的时候，注意0的读法，比如1000读作一千。而1050读作一千零五十，1005读作一千零五。另外依据题目要求，10读作一十而不是十。
四位的读法分析：一次性传入4位，高位同意全0，由于要处理不同部位的4位。
①一位数直接读。读作零到九。
②两位数推断十位是否是0，假设是，而且个位不是0，设个位为x，应该读作零x，比如10,0005，传入0005，整个数应该读作十万零五。假设十位为0而且个位为0，则不读。比如10,0000。万已经由高4位处理完成，读作10万，低位不必读。
③三位数、四位数和两位数的思路一致，首先推断是否为0，假设发现当前位为0而且该位后面有不为0的，应该读一个零。注意不要读多了。
④普通情况仅仅须要依照不同的位先输出数字，然后输出Qian Bai Shi就可以，注意缩进处理。
把处理四位数的方法封装成一个函数。
整个数的处理方法为≤4位的直接用上面的函数。>4的截取不同的字符段来得到不同的位，每4位一截取。
代码例如以下：
#include <iostream>
#include <string>
#include <string.h>
#include <sstream>
#include <stdio.h>
using namespace std;
char* values[] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
int char2int(char c){
return c - '0';
}
void handleNum(string num){
int bits = num.length();
bool printZero = false;
switch(bits){
case 1:{
int ge = char2int(num[0]);
printf("%s",values[ge]);
break;
}
case 2:{
int shi = char2int(num[0]);
int ge = char2int(num[1]);
if(shi != 0) printf("%s Shi",values[shi]);
else if(!printZero){
printf("ling");
printZero = true;
}
if(ge != 0) printf(" %s",values[ge]);
break;
}
case 3:{
int bai = char2int(num[0]);
int shi = char2int(num[1]);
int ge = char2int(num[2]);
if(bai != 0) {printf("%s Bai",values[bai]); printZero = false;}
else if(!printZero && (shi !=0 || ge != 0)){
printf("ling");
printZero = true;
}
if(shi != 0) { printf(" %s Shi",values[shi]); printZero = false; }
else if(!printZero && ge!=0){
printf(" ling");
printZero = true;
}
if(ge != 0) printf(" %s",values[ge]);
break;
}
case 4:{
int qian = char2int(num[0]);
int bai = char2int(num[1]);
int shi = char2int(num[2]);
int ge = char2int(num[3]);
if(qian != 0) {printf("%s Qian",values[qian]); printZero = false;}
else if(!printZero && (bai!=0 || shi!=0 || ge!=0)){
printf("ling");
printZero = true;
}
if(bai != 0) {printf(" %s Bai",values[bai]); printZero = false;}
else if(!printZero && (shi != 0 || ge != 0)){
printf(" ling");
printZero = true;
}
if(shi != 0) {printf(" %s Shi",values[shi]); printZero = false;}
else if(!printZero && ge != 0){
printf(" ling");
printZero = true;
}
if(ge != 0) printf(" %s",values[ge]);
}
}
}
int main()
{
string num;
cin >> num;
if(num[0] == '-'){
num = num.substr(1);
cout << "Fu ";
}
int bits = num.length();
switch(bits){
case 1:
case 2:
case 3:
case 4:
handleNum(num);
break;
case 5:{
int wan = char2int(num[0]);
printf("%s Wan ",values[wan]);
handleNum(num.substr(1));
break;
}
case 6:{
handleNum(num.substr(0,2));
printf(" Wan ");
handleNum(num.substr(2));
break;
}
case 7:{
handleNum(num.substr(0,3));
printf(" Wan ");
handleNum(num.substr(3));
break;
}
case 8:{
handleNum(num.substr(0,4));
printf(" Wan ");
handleNum(num.substr(4));
break;
}
case 9:{
handleNum(num.substr(0,1));
printf(" Yi ");
handleNum(num.substr(1,4));
printf(" Wan ");
handleNum(num.substr(5));
break;
}
}
return 0;
}