F Takio与Blue的人生赢家之战

Time Limit:1000MS  Memory Limit:65535K

题型: 编程题   语言: 无限制

描述

在那个风起云涌的SCAU ACM里，有两位人生赢家，他们分别是大洲Takio神和Blue神。      (尤其是blue神。)
由于这两位人生赢家代码能力强，才高八斗，学富五车，英俊潇洒，玉树临风，独步江湖，呼风唤雨，妹子纷至沓来。
而小邪由于太渣了，只能默默地帮他们记录下他们换了多少个妹子。
以上背景纯属题目需要，其实两位大神是很专情的。
终于有一天，小邪计算出他们身边妹子的总数n，想要给Takio神和Blue神。
但是Takio神和Blue神的邮箱是使用英文的，而小邪的英语又很渣，于是无法将n翻译成英语发过去。
但是，小邪想到了你——聪明的14级新生，向你寻求答案。

出题人:K·小邪

输入格式

第一行是一个整数t(t <= 100)，代表样例个数
对于每个样例有一个整数n(0<=n<=2000000000)

输出格式

对于每个n，输出其英文表现形式，具体格式见样例输出

输入样例

4
5
121
1010
1008611

输出样例

five
one hundred and twenty-one
one thousand and ten
one million, eight thousand, six hundred and eleven

Hint

输出不一定符合英语规范，但是要符合Sample的规范
对于一个n>1000,若n%1000 >= 100(%代表取余操作)且不为0，且在n%1000对应的英文输出前（如果存在）用","相连而不是"and"

需要用到的英文单词为(不包括引号)：
"zero","one","two","three","four","five","six","seven","eight","nine"
"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"
"twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"
"hundred","thousand","million","billion"
分别代表
0,1,2,3,4,5,6,7,8,9
10,11,12,13,14,15,16,17,18,19
20,30,40,50,60,70,80,90
100,1000,1000000,1000000000

#include<stdio.h>
#include<string.h>
void go(void);
int a[],k,num;
int leap,leap1,leap2,leap3,leap4,leap5,leap6,leap7,leap8,leap9,leap10;
char str[][]=
{
    "zero","one","two","three","four","five","six",
    "seven","eight","nine","ten","eleven","twelve",
    "thirteen","fourteen","fifteen","sixteen",
    "seventeen","eighteen","nineteen","twenty"
};

int main()
{
    int i,k,T;

    strcpy(str[],"thirty");
    strcpy(str[],"forty");
    strcpy(str[],"fifty");
    strcpy(str[],"sixty");
    strcpy(str[],"seventy");
    strcpy(str[],"eighty");
    strcpy(str[],"ninety");
    scanf("%d",&T);
    while(T--)
    {
        memset(a,,sizeof(a));
        leap=leap1=leap2=leap3=leap4=leap5=leap6=leap7=leap8=leap9=leap10=;
        scanf("%d",&num);
        i=;
        k=num;
        if(num>&&num%>=) leap=;
        while(k!=)
        {
            a[i++]=k%;
            k=k/;
        }
        k=i-;
        if(k==||k==) {printf("%s\n",str[num]);continue;}
        if(k==) {if(num>&&num<=||num%==) printf("%s\n",str[num]);
        else printf("%s-%s\n",str[num-num%],str[num%]);continue;}
      go();
      printf("\n");

    }
    return ;
}
void go(void)
{
    int temp1,temp2,temp3;

    if(a[]) {printf("%s billion",str[a[]]);leap10=;}
    if((leap&&k>=)||(leap10&&a[]!=)||(leap10&&a[]!=)||(leap10&&a[]!=)||(leap10&&a[]!=)||(leap10&&a[]!=)||(leap10&&a[]!=))
    printf(", ");

    if(a[]) {printf("%s hundred",str[a[]]);leap9=;}
    temp1=a[]*+a[];
    if(leap9&&temp1!=) printf(" and ");
    if(temp1!=)
    {
        if((temp1>&&temp1<=)||temp1%==) printf("%s",str[temp1]);
        else printf("%s-%s",str[temp1-temp1%],str[temp1%]);
        leap7=;
    }
    if(leap9||leap7) printf(" million");
    if((leap&&k>=)||((leap9&&a[]!=)||(leap7&&a[]!=))||((leap9&&a[]!=)||(leap7&&a[]!=))||((leap9&&a[]!=)||(leap7&&a[]!=))) printf(", ");

    if(a[]) {printf("%s hundred",str[a[]]);leap6=;}
    temp2=a[]*+a[];
    if(leap6&&temp2!=) printf(" and ");
    if(temp2!=)
    {
        if((temp2>&&temp2<=)||temp2%==) printf("%s",str[temp2]);
        else printf("%s-%s",str[temp2-temp2%],str[temp2%]);
        leap4=;
    }
    if(leap6||leap4) printf(" thousand");
    if(leap) printf(", ");
    if(a[]) {printf("%s hundred",str[a[]]);leap3=;}
    temp3=a[]*+a[];
    if((leap3&&temp3!=)||(a[]==&&temp3!=)) printf(" and ");
    if(temp3!=)
    {
        if((temp3>&&temp3<=)||temp3%==) printf("%s",str[temp3]);
        else printf("%s-%s",str[temp3-temp3%],str[temp3%]);
    }
}