问题描述：
Write a program that will accept a fraction of the form N/D, where N is the numerator and D is the denominator and print the decimal representation. If the decimal representation has a repeating sequence of digits, indicate the sequence by enclosing it in brackets. For example, 1/3 = .33333333...is denoted as 0.(3), and 41/333 = 0.123123123...is denoted as 0.(123). Use xxx.0 to denote an integer. Typical conversions are:
1/3 = 0.(3) //将循环部分用( )扩起来。
22/5 = 4.4
1/7 = 0.(142857)
2/2 = 1.0
3/8 = 0.375
45/56 = 0.803(571428)
Input ：
A single line with two space separated integers,N and D,1<=N,D<=100000.
Output ：
The decimal expansion, as detailed above. If the expansion exceeds 76 characters in length, print it on multiple lines with 76 characters per line.
例子：
输入：45 56
输出：0.803(571428)
程序代码：
#include <stdio.h>
#include <stdlib.h>
#include <windows.h>
#define LEN 100 //最大小数位数
typedef struct _div
{
long int num; //分子
long int den; //分母
long int quot[LEN]; //商
long int resi[LEN]; //余数
unsigned long int cycle_point; //循环点
unsigned long int length; //商的个数
}DIV;
void divide(DIV &tmp)
{
unsigned int len=0, i=0;
ldiv_t result;
if(!tmp.den)
{
printf("Denominator Error!\n");
exit(1);
}
while(len<LEN)
{
result=ldiv(tmp.num ,tmp.den);
tmp.quot[len]=result.quot;
tmp.length+=1;
if(result.rem!=0)
{
tmp.resi[len]=result.rem ;
tmp.num=result.rem*10;
while(i!=len)
{
if(tmp.resi[i]==result.rem)
{
tmp.cycle_point=i;
return ;
}
i++;
}
}
else
return ;
len++;
i=0;
}
}
int main()
{
DIV oper;
unsigned int i=0, len=0 ,repair=0, flag=0;
oper.length=0;
printf(" 请输入分子(nume):");
scanf("%ld",&oper.num);
printf(" 请输入分母(demo):");
scanf("%ld",&oper.den);
divide(oper);
printf("%ld%c",oper.quot[i++],’.’);
if(oper.length==1)
printf("%d",0);
else
{
while(i<oper.length)
{
if(i==oper.cycle_point+1)
{
printf("%c%ld",’(’,oper.quot[i]);
flag=1;
}
else
printf("%ld",oper.quot[i]);
i++;
}
if(flag)
printf("%c",’)’);
}
printf("\n");
return 0;
}
责任编辑：小草
