Hastiness（1727）

Hastiness

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1441    Accepted Submission(s): 571

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

2034 1234 123 24 0

 

Sample Output

two thousand and thirty-four one thousand and two hundred and thirty-four one hundred and twenty-three twenty-four

zero

#include 
#include 
#include 
#include 
#include 
int main()
{	int t;
	int a,b,c,d,flag,i,l;
	char name[10];
	char s[11][10]={"zero","one","two","three","four","five","six","seven","eight","nine"}; 
	char shi[11][10]={"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
	char ty[10][10]={"ten","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"};
	while(scanf("%d",&t)!=EOF)
	{	flag=0;
		sprintf(name,"%d",t);
		l=strlen(name);
		if(l==1)
		{
			printf("%s\n",s[name[0]-'0']);
			continue;
		}
		if(l==4)
		{
			a=name[0]-'0';
			b=name[1]-'0';
			c=name[2]-'0';
			d=name[3]-'0';
		}
		else if(l==3)
		{	a=0;
			b=name[0]-'0';
			c=name[1]-'0';
			d=name[2]-'0';
		}
		else if(l==2)
		{	a=0,b=0;
			c=name[0]-'0';
			d=name[1]-'0';
		}
		//printf("%d %d %d %d",a,b,c,d);
		if(a>0)
		{
			printf("%s thousand",s[a]);
			flag++;
		}
		if(b>0)
		{
			if(flag==0)
			{
				printf("%s hundred",s[b]);
				flag++;
			}
			else
			printf(" and %s hundred",s[b]);
		}
		if(c>0)
		{
			if(c==1)
			{
				if(flag==0)
				{
					printf("%s",shi[d]);
					flag++;
				}
				else
				printf(" and %s",shi[d]);				
			}
			else
			{
				if(flag==0)
				{
					if(d==0)
					{printf("%s",ty[c-1]);	flag++;}
					else
					{printf("%s-%s",ty[c-1],s[d]);	flag++;}									
				}
				else
				{
					if(d==0)
					printf(" and %s",ty[c-1]);	
					else
					printf(" and %s-%s",ty[c-1],s[d]);				
				}
			}
		}
		if(c==0 && (a!=0 || b!=0) && d!=0)
		printf(" and %s",s[d]);
		printf("\n");	
	}

	return 0;
}