Hdu 3524 Perfect Squares inverse element

Hdu 3524 Perfect Squares inverse element

 

Perfect Squares Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 501 Accepted Submission (s): 272


Problem Description A number x is called a perfect square if there exists an integer B
Satisfying x = B ^ 2. there are too 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 brespon. 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

212

Sample Output

Case #1: 2Case #2: 2

 

Tell a number n, and find the result of the exact number 2 ^ n. I didn't say the range of the full number of partitions. I thought it would be a loop. If I had a look at the table, I would find that there were regular conditions when n was an odd number.

N is an odd number of 2 + (4 ^ N-1)/3

N is an even number, 2 + 2/3*(4 ^ (n/2-1)-1)

There are a large number of molecules in Division calculation, so we need to use reverse elements. For reverse elements, see this click to open the link.

 

 

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include typedef long long ll;const int mod=10007;using namespace std;ll fun(ll n,ll m){    ll b=1;    while(m)    {        if(m&1) b=b*n%mod;        n=n*n%mod;        m>>=1;    }    return b;}int main(){    int T;    ll n;    ll tmp=fun(3,mod-2);    scanf(%d,&T);    for(int ca=1;ca<=T;ca++)    {        scanf(%lld,&n);        ll ans;        if(n%2)        {            ll t=(fun(4,n/2)-1)%mod;            ans=(2+t*tmp%mod)%mod;        }        else        {            ll t=(fun(4,n/2-1)-1)%mod;             ans=(2+2*t*tmp%mod)%mod;        }        printf(Case #%d: %lld,ca,ans);    }    return 0;}