題干：

Polish mathematician Wac?aw Sierpiński?(1882-1969) described the 2D geometric figure known as the Sierpiński?triangle as part of his work on set theory in 1915. The triangle, which is really an infinite collection of points, can be constructed by the following algorithm:

The initial shape is a solid triangle.

Shrink the current shape to half its dimensions (both height and width), and make two more copies of it (giving three copies total).

Arrange the three copies so that each touches the two others at their corners. Set the current shape to be the union of these three.

Repeat from step 2.

Here is an illustration of the first few iterations:

?
iteration	00	11	22	33	44

As the iterations go to infinity, this process creates an infinite number of connected points. However, consider the case of a finite number of iterations. If the initial triangle has a circumference of?33, what is the sum of the circumferences of all (black) triangles at a given iteration? Write a program to find out not the exact circumference, but the number of decimal digits required to represent its integer portion. That is, find the number of decimal digits required to represent the largest integer that is at most as large as the circumference.

Input

Each test case is a line containing a non-negative integer?0≤n≤100000≤n≤10000?indicating the number of iterations.

Output

For each case, display the case number followed by the number of decimal digits required to represent the integer portion of the circumference for the given number of iterations. Follow the format of the sample output.

Sample Input 1Sample Output 1

0 1 5 10 100	Case 1: 1 Case 2: 1 Case 3: 2 Case 4: 3 Case 5: 19

?解題報告：

? ?推出公式發現需要求3* (1.5)^n的位數，處理方法很多，可以double存答案，然后每一次進到1e9的時候，就答案+=9，數字/=1e9，最后肯定得到了一個1e9范圍以內的數字，再單獨處理就行。也可以用Java的大整數類先算出分母和分子，再做除法（向下取整），然后按位處理。

? ?也可以直接用C++中取log的操作得到答案。整數n的位數就是log10(n) + 1.

AC代碼：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; typedef long long ll; const double eps=1e-8; int main() {int t,q,i,j,n,cnt=0,ans;while(scanf("%d",&n)!=EOF){ans=log10(3)+n*log10(3.0/2);ans++;printf("Case %d: %d\n",++cnt,ans);}return 0; }

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