HDU1727数字转换成英文
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http://acm.hdu.edu.cn/showproblem.php?pid=1727
Hastiness
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1297 Accepted Submission(s): 495
Problem Description
How many problems did you AC?
When you read this problem, don’t hasty and careless, this is also simple, haha, I didn’t cheat you.
The game over soon, WisKey starts using English begin countdown. He not only have no gene in math, but also bad in English. Fortunately, He met you who have gift in programming. So please help him to translate.
When you read this problem, don’t hasty and careless, this is also simple, haha, I didn’t cheat you.
The game over soon, WisKey starts using English begin countdown. He not only have no gene in math, but also bad in English. Fortunately, He met you who have gift in programming. So please help him to translate.
Input
Give you an integer T, output T in English, and note that all of words are lower case. (0<=T<=9999)
Output
One answer One line.
Details see sample.
Details see sample.
Sample Input
20341234123240
Sample Output
two thousand and thirty-fourone thousand and two hundred and thirty-fourone hundred and twenty-threetwenty-fourzero
题目:题目意思就是将一个数字按照英文习惯输出,比较麻烦,关键是细心。
#include <stdio.h>char tab[][20] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty"};char dec[][20] = { "zero", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety", "hundred"};int main (){ int n; while (scanf ("%d", &n)!=EOF) { if (n == 0) { puts ("zero") ; continue ; } if (n >= 1000){ printf ("%s thousand", tab[n/1000]);//n -= n/1000*1000 ;n = n % 1000; if (n != 0) printf (" and ") ; } if (n >= 100){ printf ("%s hundred", tab[n/100]);//n -= n/100*100 ;n = n % 100; if (n != 0) printf (" and ") ; } if (n != 0) { if (n <= 20) printf ("%s", tab[n]) ; else { printf ("%s", dec[n/10]) ; if (n%10!=0) printf ("-%s", tab[n%10]) ; } } printf ("\n") ; } return 0 ;}
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