1、N个分数求和

博客链接 https://blog.csdn.net/weixin_73142957/article/details/128798429

public class Demo3_01 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int sumA = 0;//分子和
        int sumB = 0;//分母和
        for (int i = 0; i < n; i++) {
            String[] split = scanner.next().split("/");
            int a = Integer.parseInt(split[0]);//分子
            int b = Integer.parseInt(split[1]);//分母
            if (sumA == 0) {//只有一个算式时
                sumA = a;
                sumB = b;
            } else {//通分
                sumA = a * sumB + b * sumA;
                sumB = sumB * b;
            }
            int max = resGcd(sumA, sumB);
            //分子和分母化为最简
            sumA /= max;
            sumB /= max;
        }
        if (sumA != 0 && (sumA / sumB == 0)) {//只有分数
            System.out.println(sumA + "/" + sumB);
        } else if (sumA % sumB == 0) {//只有整数部分
            System.out.println(sumA / sumB);
        } else {//包含整数和分数
            System.out.println(sumA / sumB + " " + (sumA % sumB) + "/" + sumB);
        }

    }
    //计算最大公约数
    private static int resGcd(int a, int b) {
        if (b == 0) return a;
        return resGcd(b, a % b);
    }
}

时间：2023-07-01