HDU3524 数论

Perfect Squares

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 582    Accepted Submission(s): 323

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

 

Source

2010 ACM-ICPC Multi-University Training Contest（9）——Host by HNU

题意：

求i^2%2^n有多少个不同的结果（i>=0,n是给出的数），最后总数%10007；

代码：

/*
就是打表+找规律 
    n为偶数,f[n]=(2^(n-1)-2)/3+2; 
    n为奇数,f[n]=(2^(n-1)-1)/3+2; 
    这里需要解决的只有除法取余: 
    
    费马小定理(Fermat Theory)是数论中的一个重要定理，其内容为： 假如p是质数，且gcd(a,p)=1，那么 
    a(p-1)≡1（mod p）。即：假如a是整数，p是质数，且a,p互质(即两者只有一个公约数1)，那么a的
    (p-1)次方除以p的余数恒等于1。
    
    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) 
*/
#include<iostream>
#include<cmath>
using namespace std;
const int mod=10007;
int solve(int a,int n)
{
    if(n==0) return 1;
    int x=solve(a,n/2);
    long long ans=(long long)x*x%mod;
    if(n&1) ans=ans*a%mod;
    return (int)ans;
}
int main()
{
    int t,n;
    cin>>t;
    for(int i=1;i<=t;i++){
        cin>>n;
        cout<<"Case #"<<i<<": ";
        int ans;
        if(n&1) ans=((solve(2,n-1)-1)*solve(3,mod-2))%mod+2;
        else ans=((solve(2,n-1)-2)*solve(3,mod-2))%mod+2;
        cout<<ans<<endl;
    }
    return 0;
}