HDOJ-1713-相遇周期

1. 题目中的Input描述是错的，正确解释为：输入n/t，表示转t圈需要n天。相遇周期T = 1 / fabs(t1/n1 - t2/n2)，即：周期 = 路程差 / 速度差

2. WA若干次，在Discuss中了解到，64-bit int在windows系统和Linux系统中输入输出格式是 I64d 还是 lld 的问题。在我的系统上用Dev C++测试后两种格式都正确。说明与编译器的版本和系统版本有关。这里提交选择G++/GCC则需要用 I64d 格式，否则，会WA。

3. 如果计算过程中尽量避免连续相乘操作，可用32-bit int版本AC。[见此题Discuss]

 

#include <stdio.h>
#include <string.h>
#define INT64 long long
// to calculate the greatest common divisor
INT64 GCD(INT64 a, INT64 b)
{
    INT64 t;
    while (b) {
        t = a%b;
        a = b;
        b = t;
    }
    return a;
}
// to calculate the lowest common multiple
INT64 LCM(INT64 t1, INT64 t2)
{
    return t1/GCD(t1,t2)*t2;
}

int main()
{
    int cs;
    scanf("%d", &cs);
    while (cs--) {
        INT64 n1, n2, t1, t2, f1, f2;
        scanf("%I64d/%I64d %I64d/%I64d", &n1,&t1,&n2,&t2);
        INT64 lcm1, lcm2, gcd, gcd1, gcd2;
        // reduce the input data to the simplest form
        gcd1 = GCD(n1,t1);
        gcd2 = GCD(n2,t2);
        n1 /= gcd1;  
        t1 /= gcd1;
        n2 /= gcd2;
        t2 /= gcd2;
        // reduction of fractions to a common denominator
        lcm1 = LCM(t1, t2);
        f1 = lcm1/t1; // f1 = t2/GCD(t1,t2);
        f2 = lcm1/t2; // f2 = t1/GCD(t1,t2);
        lcm2 = LCM(n1*f1, n2*f2);
        gcd = GCD(lcm1, lcm2);
        if (lcm1/gcd == 1) {
            printf("%I64d\n", lcm2/gcd);
        }else {
            printf("%I64d/%I64d\n", lcm2/gcd, lcm1/gcd);
        }
    }
    return 0;
}