#P835. Shifty Sum
Shifty Sum
題目描述
Suppose we have a number like 12. Let’s define shifting a number to mean adding a zero at the end. For example, if we shift that number once, we get the number 120. If we shift the number again we get the number 1200. We can shift the number as many times as we want.
In this problem you will be calculating a shifty sum, which is the sum of a number and the numbers we get by shifting. Specifically, you will be given the starting number N and a non-negative integer k. You must add together N and all the numbers you get by shifting a total of k times.
For example, the shifty sum when N is 12 and k is 1 is: 12 + 120 = 132. As another example, the shifty sum when N is 12 and k is 3 is 12 + 120 + 1200 + 12000 = 13332.
輸入格式
The first line of input contains the number N (1 ≤ N ≤ 10000). The second line of input contains k, the number of times to shift N (0 ≤ k ≤ 5).
輸出格式
Output the integer which is the shifty sum of N by k.
Samples
12
3
13332
原始資料
- Zero1 題號:
a835 - Hydro 題號:
Z0835 - Locale:
zh_TW - Display:
open