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
題意：求完全平方數摸2^n有多少不同的結果。
思路：
嘗試了打表，最後還是看的別人的博客找到的規律（博客）。
這道題目的規律是奇數項之間的增量有規律，偶數項之間的增量有規律，發現這個後，能寫出通項公式。這裏要注意除法取餘問題。
自己推出的公式：
這裏需要解決的只有除法取餘:（轉自該博客）
1.可以用乘法的逆來解決，當然可知當成定理來用(a/b)%mod=(a*b^(mod-2))%mod,mod爲素數
原理是費馬小定理：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.上面的方法適合b比較大的時候，這裏的b=3.很小，可知直接(a/b)%mod=(a%(mod*b)/b)%mod
AC代碼：
#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);
}
}