PAT乙级 1034

思路：是个水题，但是有坑。不能被题目忽悠了，题目保证正确的输出中没有超过整型范围的整数。 它只是保证结果不超出int，但是我们在运算过程中的乘法可能会超出int，直接把所有int改成long long

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AC代码

#include <stdio.h>
#include <algorithm>
using namespace std;
typedef long long LL;

LL a, b, c, d;
LL e, f, g, h;

LL gcd(LL a, LL b) {
    return b == 0 ? a : gcd(b, a%b);
}

void kill(LL &a, LL &b, LL &e, LL &f) {
    if(b == 0) {
        printf("Inf");
        return;
    }
    if(a == 0) {
        printf("0");
        e = 0, f = 1;
        return;
    } else {
        LL r = gcd(a, b);
        e = a / r, f = b / r;
    }
    LL l = abs(e) / abs(f);
    LL p = abs(e) - l * abs(f);
    if(e < 0 || f < 0) {
        printf("(");
    }

    if(e < 0 || f < 0) {
        printf("-");
    }
    //output
    if(l != 0) printf("%d", l);
    if(l != 0 && p != 0) printf(" ");
    if(p != 0) {
        printf("%d/%d", p, abs(f));
    }

    if(e < 0 || f < 0) {
        printf(")");
    }
}

void add() {
    kill(a, b, e, f);
    printf(" + ");
    kill(c, d, g, h);
    printf(" = ");
    LL p = f*h / gcd(f, h);
    LL res = e * (p/f) + g * (p/h);
    kill(res, p, res, p);
    printf("\n");
}

void sub() {
    kill(a, b, e, f);
    printf(" - ");
    kill(c, d, g, h);
    printf(" = ");
    LL p = f*h / gcd(f, h);
    LL res = e * (p/f) - g * (p/h);
    kill(res, p, res, p);
    printf("\n");
} 

void mul() {
    kill(a, b, e, f);
    printf(" * ");
    kill(c, d, g, h);
    printf(" = ");
    LL res = e * g;
    LL p = f * h;
    kill(res, p, res, p);
    printf("\n");
}

void div() {
    kill(a, b, e, f);
    printf(" / ");
    kill(c, d, g, h);
    printf(" = ");
    LL res = e * h;
    LL p = f * g;
    kill(res, p, res, p);
    printf("\n");
}

int main() {

    while(scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d) != EOF) {
        add();
        sub();
        mul();
        div();
    }
    return 0;
}

如有不当之处欢迎指出！