PAT乙级1034. 有理数四则运算(20)

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题目:

本题要求编写程序,计算2个有理数的和、差、积、商。

输入格式:

输入在一行中按照“a1/b1 a2/b2”的格式给出两个分数形式的有理数,其中分子和分母全是整型范围内的整数,负号只可能出现在分子前,分母不为0。

输出格式:

分别在4行中按照“有理数1 运算符 有理数2 = 结果”的格式顺序输出2个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式“k a/b”,其中k是整数部分,a/b是最简分数部分;若为负数,则须加括号;若除法分母为0,则输出“Inf”。题目保证正确的输出中没有超过整型范围的整数。

输入样例1:
2/3 -4/2
输出样例1:
2/3 + (-2) = (-1 1/3)2/3 - (-2) = 2 2/32/3 * (-2) = (-1 1/3)2/3 / (-2) = (-1/3)
输入样例2:
5/3 0/6
输出样例2:
1 2/3 + 0 = 1 2/31 2/3 - 0 = 1 2/31 2/3 * 0 = 01 2/3 / 0 = Inf
思路:
1. 本题主要任务是对输入的数字化简,在化简过程中需要约分、判断分子分母大小关系和正负,考虑用函数来实现;
2. 首先定义函数判断两数是否互质,求出最大公因数并约分;
3. 之后定义化简函数和加减乘除四则运算;
4. 最后是主函数;
5. 需要注意的是数字可能较大,要用到long long int,并且要注意在输入输出时在scanf和printf函数里要用%lld,不能用%d,没注意会卡很长时间。
#include<cstdio>using namespace std;int gca (long long int m, long long int n) {    long long r;    if (m < n) {        m=m+n;        n=m-n;        m=m-n;    }    r=m%n;    while (r) {        m = n;        n = r;        r = m%n;    }    return n;}void huajian (long long int x, long long int y) {    long long int p, q, flag = 1;    if (x < 0) {        x = -x;        flag = -flag;    }    if (y < 0) {        y = -y;        flag = -flag;    }    if (y == 0)        printf ("Inf");    else if (x == 0)        printf ("0");    else if (x < y) {        p = gca (y, x);        if (p != 1) {            y /= p;            x /= p;        }        if (flag == -1)            printf ("(");        if (x != 0)            printf ("%lld/%lld", x*flag, y);        if (x == 0)            printf ("0");        if (flag == -1)            printf (")");    }    else if (x >= y) {        q = x/y;        x %= y;        if (x != 0) {            p = gca(y, x);            y /= p;            x /= p;        }        if (flag == -1)            printf ("(");        if (x != 0)            printf ("%lld %lld/%lld", q*flag, x, y);        if (x == 0)            printf ("%lld", q*flag);        if (flag == -1)            printf (")");    }    return;}void add (long long int a, long long int b, long long int c, long long int d) {    huajian (a, b);    printf (" + ");    huajian (c, d);    printf (" = ");    huajian (a*d+b*c, b*d);    printf ("\n");}void minu (long long int a, long long int b, long long int c, long long int d) {    huajian (a, b);    printf (" - ");    huajian (c, d);    printf (" = ");    huajian (a*d-b*c, b*d);    printf ("\n");}void multi (long long int a, long long int b, long long int c, long long int d) {    huajian (a, b);    printf (" * ");    huajian (c, d);    printf (" = ");    huajian (a*c, b*d);    printf ("\n");}void divide (long long int a, long long int b, long long int c, long long int d) {    huajian (a, b);    printf (" / ");    huajian (c, d);    printf (" = ");    huajian (a*d, b*c);}int main() {    long long int a, b, c, d;    scanf ("%lld/%lld %lld/%lld", &a, &b, &c, &d);    add (a, b, c, d);    minu (a, b, c, d);    multi (a, b, c, d);    divide (a, b, c, d);    return 0;}