8 条题解

  • 0
    @ 2025-7-19 9:30:18

    #include using namespace std;

    // 计算最大公约数 int gcd(int a, int b) { while (b != 0) { int temp = b; b = a % b; a = temp; } return a; }

    int main() { int a, b, c, d; cin >> a >> b >> c >> d;

    // 计算和的分子和分母
    int numerator = a * d + c * b;
    int denominator = b * d;
    
    // 约分
    int common = gcd(numerator, denominator);
    numerator /= common;
    denominator /= common;
    
    // 输出结果
    cout << a << "/" << b << "+" << c << "/" << d << "=";
    if (numerator >= denominator) {
        int integer = numerator / denominator;
        numerator = numerator % denominator;
        cout << integer;
        if (numerator != 0) {
            cout << "+" << numerator << "/" << denominator;
        }
    } else {
        cout << numerator << "/" << denominator;
    }
    
    return 0;
    

    }

    信息

    ID
    907
    时间
    1000ms
    内存
    128MiB
    难度
    5
    标签
    递交数
    396
    已通过
    157
    上传者