Problem Description

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

此題可以用兩種解法求解：
方法一：

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;
const int Mod=10007;
typedef long long ll;
int t;
ll n;
ll fastpow (ll a,ll b)
{
    ll sum=1;
    while (b>0)
    {
        if(b&1)
            sum=sum*a%Mod;
        a=a*a%Mod;
        b>>=1;
    }
    return sum;
}
int main()
{
    scanf("%d",&t);
    ll inv3=fastpow(3,Mod-2);
    for (int i=1;i<=t;i++)
    {
        scanf("%lld",&n);
        ll ans;
        if(n&1)
           ans=(fastpow(2,n-1)+5)%Mod*inv3%Mod;
        else
           ans=(fastpow(2,n-1)+4)%Mod*inv3%Mod;
        printf("Case #%d: %lld\n",i,ans);
    }
    return 0;
}
 

方法二：
用的迴圈節...

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;
const int Mod=10007;
typedef long long ll;
int t;
int n;
int f[Mod+3];
int main()
{
    f[0]=2;
    f[1]=2;
    for (int i=2;i<Mod;i++)
    {
        if(i%2==0)
            f[i]=(2*f[i-1]-1+Mod)%Mod;
        else
            f[i]=(2*f[i-1]-2+Mod)%Mod;
       // printf("%d\n",f[i]);
    }
    scanf("%d",&t);
    for (int i=1;i<=t;i++)
    {
        scanf("%d",&n);
            printf("Case #%d: %d\n",i,f[n%(Mod-1)-1]);
    }
    return 0;
}