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 <c), then a^2 + b^2=c^2.
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<n<2*10^9).
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：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
const int mod=10007;
ll qsm(ll a,ll b,ll mod)
{
ll t=1;
while(b)
{
if(b&1)
t=(t*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return t%mod;
}
int main()
{
int t,cas=1;
cin>>t;
ll n;
while(t--)
{
scanf("%lld",&n);
ll ans=0;
if(n&1)
ans=((qsm(4,(n+1)/2-1,3*mod)+5)%(3*mod))/3;
else
ans=((qsm(4,(n/2)-1,3*mod)*2+4)%(3*mod))/3;
printf("Case #%d: %lld\n",cas++,ans);
}
}