hdoj-1727Hastiness

Hastiness

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1801    Accepted Submission(s): 699

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. 

 

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.

 

Sample Input

20341234123240

 

Sample Output

two thousand and thirty-fourone thousand and two hundred and thirty-fourone hundred and twenty-threetwenty-fourzero

 

题目链接

 

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int main(){    char a[21][10],b[10][10];    strcpy(a[0],"zero");    strcpy(a[1],"one");    strcpy(a[2],"two");    strcpy(a[3],"three");    strcpy(a[4],"four");    strcpy(a[5],"five");    strcpy(a[6],"six");    strcpy(a[7],"seven");    strcpy(a[8],"eight");    strcpy(a[9],"nine");    strcpy(a[10],"ten");    strcpy(a[11],"eleven");    strcpy(a[12],"twelve");    strcpy(a[13],"thirteen");    strcpy(a[14],"fourteen");    strcpy(a[15],"fifteen");    strcpy(a[16],"sixteen");    strcpy(a[17],"seventeen");    strcpy(a[18],"eighteen");    strcpy(a[19],"nineteen");    strcpy(a[20],"twenty");    strcpy(b[0],"twenty");    strcpy(b[1],"thirty");    strcpy(b[2],"forty");    strcpy(b[3],"fifty");    strcpy(b[4],"sixty");    strcpy(b[5],"seventy");    strcpy(b[6],"eighty");    strcpy(b[7],"ninety");    int n,m;    while(~scanf("%d",&n))    {        if(!n){            printf("zero\n");            continue;        }        if(n>=1000)        {            m=n/1000;            printf("%s thousand",a[m]);            n-=m*1000;            if(!n)            {                printf("\n");                continue;            }            printf(" and ");        }        if(n>=100)        {            m=n/100;            printf("%s hundred",a[m]);            n-=m*100;            if(!n){                printf("\n");                continue;            }            printf(" and ");        }        if(n)        {            if(n<=20)                printf("%s\n",a[n]);            else            {                m=n/10;                printf("%s",b[m-2]);                n-=m*10;                if(!n)                {                    printf("\n");                    continue;                }                printf ("-%s\n", a[n]);            }        }    }    return 0;}