1082. Read Number in Chinese (25)
题目如下:
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:
- -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
题目要求按照中国人的习惯阅读一个不超过9位的数字,这道题的坑比较多,一定要考虑到所有的情况。
首先要抓住规律,我们可以发现,一个数字的读法在每个4位是一致的。
例如12341234,我们读作“一千二百三十四万一千二百三十四”,我们可以看到除去万字以外读法完全一致。因此我们集中精力解决四位数的读法,然后加上亿、万等即可。
在解决的时候,注意0的读法,例如1000读作一千,而1050读作一千零五十,1005读作一千零五,另外根据题目要求,10读作一十而不是十。
四位的读法分析:一次性传入4位,高位允许全0,因为要处理不同部位的4位。
①一位数直接读,读作零到九。
②两位数判断十位是否是0,如果是,并且个位不是0,设个位为x,应该读作零x,例如10,0005,传入0005,整个数应该读作十万零五;如果十位为0并且个位为0,则不读。例如10,0000,万已经由高4位处理完毕,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,首先判断是否为0,如果发现当前位为0并且该位后面有不为0的,应该读一个零,注意不要读多了。
④一般情况只需要按照不同的位先输出数字,然后输出Qian Bai Shi即可,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。
代码如下:
- #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;
- }
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