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");

}

}