Question

所以现在计算平均值，我试图打印出用户输入，只在其自己的行中严格低于平均值。我尝试这样做，但是，错误“二元运算符的坏操作数”＆lt;“。

const std::array<int, 3> array1 = {{1, 2, 3}};
const std::array<int, 2> array2 = {{4, 5}};
const std::array<int, 4> array3 = {{6, 7, 8, 9}};

const auto result = concatenate(array1, array2, array3);

Answer 1

for (int i = howMany - 1; i >= 0; i--){
     sum+=myArray[i];//sum up the array
}
avg = sum / howMany;

要在一行中打印每个double：

System.out.println(avg);

for (int i = 0; i < howMany; i++) {
    if(myArray[i] < avg){
        System.out.println(myArray[i]);//single line for every double
    }
}

在一行中打印所有double：

String oneLine = "["; 
int lowerAvg = 0;
for (int i = 0; i < howMany; i++) {
if(myArray[i] < avg){
        oneLine += (i > 0 && ++lowerAvg > 0)?", ":""; // add separator
        oneLine += (myArray[i]);
    }
}
oneLine+="]";
System.out.println(oneLine);//prints like [1.0, 2.0, ...]

使用Java 8：

avg = Arrays.stream(myArray, 0, howMany).average().orElse(0);
final double finalAvg = avg;//final variable to use in filter condition
System.out.println(finalAvg);
Arrays.stream(myArray, 0, howMany).filter(x -> x < finalAvg).forEach(System.out::println);//will print every double in a single line
System.out.println(Arrays.toString(Arrays.stream(myArray, 0, howMany).filter(x -> x < finalAvg).toArray()));//Will print the array like [1.0, 2.0, ...]

仅打印用户输入的数字，严格低于平均值

1 个答案: