C++實(shí)例:循環(huán)小數(shù)的處理

問題描述：
    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) //將循環(huán)部分用( )擴(kuò)起來。
    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：
    輸入：45 56
    輸出：0.803(571428)
    程序代碼：
    #include
    #include
    #include
    #define LEN 100 //小數(shù)位數(shù)
    typedef struct _div
    {
    long int num; //分子
    long int den; //分母
    long int quot[LEN]; //商
    long int resi[LEN]; //余數(shù)
    unsigned long int cycle_point; //循環(huán)點(diǎn)
    unsigned long int length; //商的個(gè)數(shù)
    }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    {
    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(" 請(qǐng)輸入分子(nume):");
    scanf("%ld",&oper.num);
    printf(" 請(qǐng)輸入分母(demo):");
    scanf("%ld",&oper.den);
    divide(oper);
    printf("%ld%c",oper.quot[i++],’.’);
    if(oper.length==1)
    printf("%d",0);
    else
    {
    while(i    {
    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;
    }