main()
{double r,y;
int x, m,n,j,i,k,flag=0;
int str1[10]={"零","壹","贰","叁","肆","伍","陆","柒","捌","玖"};
int str2[14]={"仟","佰","拾","亿","仟","佰","拾","萬","仟","佰","拾","元","角","分"};
char str3[30];
printf("请输入金额(范围小于千亿):\n");
scanf("%lf",&r);
if(r>=1000000000000)
printf("输入的数超出范围，请重新输入\n");
y=r-(int)r; /*y为小数部分*/
x=(int)r; /*x为整数部分*/
for(i=0;i<=11;i++) /*整数部分*/
{
m=(int)((int)x%10);
str3[11-i]=m;
x=(int)(x/10);
}
str3[12]=(int)(y*10); /*小数部分*/
str3[13]=(int)(y*100)%10;
for(k=0;k<=13&&flag==0;k++) /*判断最大位是多少*/
{if(str3[k]>0)
{flag=1;n=k;
}
}
for(j=n;j<=13;j++)
printf("%s%s",str1[str3[j]],str2[j]);
printf("\n");
}
2．(拓展题，分值25)编程实现，输入一个人民币小写金额值，转化为大写金额值输出。要求实现完善的功能，如输入1002300.90，应该输出“壹佰万贰仟三佰元零玖角整”。
# includemain()
{double r,y;
int x, m,n,j,i,k,flag=0;
int str1[10]={"","壹","贰","叁","肆","伍","陆","柒","捌","玖"};
int str2[14]={"仟","佰","拾","亿","仟","佰","拾","萬","仟","佰","拾","元","角","分"};
char str3[30];
printf("请输入金额(范围小于千亿):\n");
scanf("%lf",&r);
if(r>=1000000000000)
printf("输入的数超出范围，请重新输入\n");
y=r-(int)r; /*y为小数部分*/
x=(int)r; /*x为整数部分*/
for(i=0;i<=11;i++)
{
m=(int)((int)x%10);
str3[11-i]=m;
x=(int)(x/10);
}
str3[12]=(int)(y*10);
str3[13]=(int)(y*100)%10;
if(y==0) /*当没有小数时的输出*/
{for(k=0;k<=13&&flag==0;k++) /*判断最大位是多少*/
{if(str3[k]>0)
{flag=1;n=k;
}
}
for(j=n;j<=11;j++)
{printf("%s",str1[str3[j]]);
if(str3[j]!=0||j==3||j==7)
printf("%s",str2[j]);
}
printf("整\n");
printf("\n");
}
if(y!=0) /*当有小数时的输出*/
{for(k=0;k<=13&&flag==0;k++) /*先输出整数*/
{if(str3[k]>0)
{flag=1;n=k;
}
}
for(j=n;j<=11;j++)
{printf("%s",str1[str3[j]]);
if(str3[j]!=0||j==3||j==7)
printf("%s",str2[j]);
}
printf("零");
if(str3[12]==0&&str3[13]!=0) /*输出小数*/
printf("%s%s整",str1[str3[13]],str2[13]);
if(str3[12]!=0&&str3[13]==0)
printf("%s%s整",str1[str3[12]],str2[12]);
if(str3[12]!=0&&str3[13]!=0)
printf("%s角%s分整",str1[str3[12]],str1[str3[13]]);
printf("\n");
}
}
