103.1(3)0.(3)1.(25)

47/151/3124/99

#include <stdio.h>#include <string.h>//按照题目条件，设小数是a.b(c)的，比如3.1(3)但是有可能b是空串，比如0.(10)//思路，拿3.1(3)举例：x = 3.1(3) ; 10x = 31.(3) ; 100x = 313.(3) ; => 90x = 313.(3) - 31.(3) ，循环节将就这样消去了void extract(char *str, int *pb, int *pc){    char *p = str;    *pb = *pc = 0;    if (*p != ‘(‘) //能得到b    {        do        {            *pb = *pb * 10 + *p++ - ‘0‘;        } while (*p != ‘(‘);//直到遇到左括号，说明b读完了    }    else //得不到b，只能得到c    {        *pb = -1;//用-1表示b读取失败    }    p++;//跳过左括号    do    {        *pc = *pc * 10 + *p++ - ‘0‘;    } while (*p != ‘)‘);//同理取出c，到右括号为止}int gcd(int a, int b)//最大公约数{    if (!b)        return a;    else return gcd(b, a%b);}int pow10(int x)//10的x次方{    int result = 1;    while (x--)        result *= 10;    return result;}int main(void){    int n;    int a, b, c;    int M, N, gcd_m_n; //最后结果表示为N/M，因为要化成最简分数所以用gcd_m_n约分    char buf[16];    scanf("%d", &n);    while (n--)    {        scanf("%d.", &a);//一定有a，把a和小数点取出，注意是"%d."不是"%d"        scanf("%s", buf);//剩下的放入缓冲区        extract(buf, &b, &c);        // /*测试输入3.1(4) 3.(4)是否识别*/ printf("%d %d %d", a, b, c);        if (b == -1)        {            int c_mul;            char c_buf[16];            sprintf(c_buf, "%d", c);            c_mul = pow10(strlen(c_buf));            N = c_mul - 1;            M = a * c_mul + c - a;            gcd_m_n = gcd(M, N);            M /= gcd_m_n;            N /= gcd_m_n;            printf("%d/%d\n", M, N);        }        else        {            int b_mul, c_mul;            char b_buf[16];            char c_buf[16];            sprintf(b_buf, "%d", b);            sprintf(c_buf, "%d", c);            b_mul = pow10(strlen(b_buf));            c_mul = pow10(strlen(c_buf));            N = c_mul * b_mul - b_mul;            M = a * b_mul * c_mul + b * c_mul + c - a * b_mul - b;            gcd_m_n = gcd(M, N);            M /= gcd_m_n;            N /= gcd_m_n;            printf("%d/%d\n", M, N);        }    }    return 0;}