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:-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; }