【HDU - 3524】Perfect Squares（找规律，推公式，除法取模）

A number x is called a perfect square if there exists an integer b 
satisfying x=b^2. There are many beautiful theorems about perfect squares in mathematics. Among which, Pythagoras Theorem is the most famous. It says that if the length of three sides of a right triangle is a, b and c respectively(a < b In this problem, we also propose an interesting question about perfect squares. For a given n, we want you to calculate the number of different perfect squares mod 2^n. We call such number f(n) for brevity. For example, when n=2, the sequence of {i^2 mod 2^n} is 0, 1, 0, 1, 0……, so f(2)=2. Since f(n) may be quite large, you only need to output f(n) mod 10007. 

Input

The first line contains a number T<=200, which indicates the number of test case. 
Then it follows T lines, each line is a positive number n(0

Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is f(x).

Sample Input

2
1
2

Sample Output

Case #1: 2
Case #2: 2

题意：求完全平方数摸2^n有多少不同的结果。

思路：

尝试了打表，最后还是看的别人的博客找到的规律（博客）。

这道题目的规律是奇数项之间的增量有规律，偶数项之间的增量有规律，发现这个后，能写出通项公式。这里要注意除法取余问题。

自己推出的公式：

这里需要解决的只有除法取余:（转自该博客）
    1.可以用乘法的逆来解决，当然可知当成定理来用(a/b)%mod=(a*b^(mod-2))%mod,mod为素数
原理是费马小定理：a^(mod-1)%mod=1,又a^0%mod=1,所以a^(-1)=a^(mod-2),推出a/b=a*b^(-1)=a*b^(mod-2)
    2.上面的方法适合b比较大的时候，这里的b=3.很小，可知直接(a/b)%mod=(a%(mod*b)/b)%mod

AC代码：

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