题解 | #数列求和#

数列为：9，99，999，...，9999999999

9 = 10 -1 = 10的1次幂-1；
99 = 100-1 = 10的2次幂-1；
999 = 1000-1 = 10的3次幂-1；
...
9999999999 = 10000000000-1 = 10的10次幂-1；

public class Main {
    public static void main(String[] args) {

        //write your code here........
        long[] arr = new long[10];
        long sum = 0L;
        for(int i =0; i < 10; i++){
            arr[i] = (long)(Math.pow(10,i+1)-1);
            sum += arr[i];
        }
        System.out.println(sum);

    }
}