ACM-ICPC 2015 Shenyang Preliminary Contest B. Best Solver

The so-called best problem solver can easily solve this problem, with his/her childhood sweetheart.

It is known that y=(5+2 *sqrt(6))^(1+2^x)

For a given integer x(0≤x<2^32) and a given prime number M(M≤46337), print [y]%M. ([y]means the integer part of y)

Input Format

An integer T(1<T≤1000), indicating there are T test cases.

Following are T lines, each containing two integers x and M, as introduced above.

Output Format

The output contains exactly T lines.

Each line contains an integer representing [y]%M.

样例输入

7
0 46337
1 46337
3 46337
1 46337
21 46337
321 46337
4321 46337

样例输出

Case #1: 97
Case #2: 969
Case #3: 16537
Case #4: 969
Case #5: 40453
Case #6: 10211
Case #7: 17947

题目来源

ACM-ICPC 2015 Shenyang Preliminary Contest

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define N 100005//一开始是10005,一直超时/
 #define mod 1000000007
 #define gep(i,a,b) for(int i=a;i<=b;i++)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define ll  long long
 /*
 a(n)=(5+sqrt(6))^n,b(n)=(5-sqrt(6))^n
 c(n)=a(n)+b(n)//是个整数
 a(n)-(1-b(n))=a(n)+b(n)-1=c(n)-1为a(n)向下取整的结果。
 c(n)*((5+sqrt(6))+(5-qrt(6)))
 因此10*c(n)=c(n+1)+c(n-1)
 c(n+1)=10*c(n)-c(n-1)
 */
 int t;
 ll x,m;
 ll c[N];
 ll len;
 ll  solve(){
     c[]=%m;
     c[]=%m;
     //cout<<m<<endl;
     for(int i=;;i++)
     {
         c[i]=(*c[i-]-c[i-]+m)%m;
         //cout<<c[i]<<endl;
         if(c[i-]==c[]&&c[i]==c[]){
             //cout<<len<<endl;
             return len=i-;
         }
     }
 }
 ll poww(ll a,ll b,ll p){
     ll ans=%p;
     a%=p;
     while(b){
         if(b&)  ans=(ans*a)%p;
         b>>=;
         a=(a*a)%p;
     }
     return ans%p;
 }
 int main()
 {
     scanf("%d",&t);
     gep(i,,t){
         scanf("%lld%lld",&x,&m);//1+2^x(x很大)一定有循环节。
         //cout<<x<<" "<<m<<endl;
         solve();//因为%m的m不大，因此每次都去找循环节。
         ll id=poww(,x,len);
         id=(id+)%len;
         printf("Case #%d: %lld\n",i,((c[id]-)%m+m)%m);
     }
     return ;
 }