pat-1082 Read Number in Chinese(25)(字符串处理)
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 ...
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 asFu 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 isyi 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
作者: CHEN, Yue
单位: 浙江大学
时间限制: 400 ms
内存限制: 64 MB
代码长度限制: 16 KB
注意输出的大小写和空格处理。
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <set>
using namespace std;
char num[10][5] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
char wei[5][5] = {"Shi","Bai","Qian","Wan","Yi"};
int main()
{
string s;
char str[15];
getline(cin,s);
strcpy(str,s.c_str());
int len = strlen(str);
int left = 0,right = len - 1;
if(str[0] == '-')
{
printf("Fu");
left ++;
}
while(left +4 <= right)
{
right -= 4;
}
while(left < len)
{
bool flag = false;
bool isprint = false;
while(left <= right)
{
if(left > 0 && str[left] == '0')
flag = true;
else
{
if(flag == true)
{
printf(" ling");
flag = false;
}
if(left > 0)
printf(" ");
printf("%s",num[str[left] - '0']);
isprint = true;
if(left != right)
printf(" %s",wei[right - left - 1]);
}
left ++;
}
if(isprint == true && right != len - 1)
printf(" %s",wei[(len - 1 - right) / 4 + 2]);
right += 4;
}
return 0;
}
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