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
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <vector> 5 #include <cmath> 6 #include <algorithm> 7 using namespace std; 8 #define N 100005//一开始是10005,一直超时/ 9 #define mod 1000000007 10 #define gep(i,a,b) for(int i=a;i<=b;i++) 11 #define mem(a,b) memset(a,b,sizeof(a)) 12 #define ll long long 13 /* 14 a(n)=(5+sqrt(6))^n,b(n)=(5-sqrt(6))^n 15 c(n)=a(n)+b(n)//是个整数 16 a(n)-(1-b(n))=a(n)+b(n)-1=c(n)-1为a(n)向下取整的结果。 17 c(n)*((5+sqrt(6))+(5-qrt(6))) 18 因此10*c(n)=c(n+1)+c(n-1) 19 c(n+1)=10*c(n)-c(n-1) 20 */ 21 int t; 22 ll x,m; 23 ll c[N]; 24 ll len; 25 ll solve(){ 26 c[1]=10%m; 27 c[2]=98%m; 28 //cout<<m<<endl; 29 for(int i=3;;i++) 30 { 31 c[i]=(10*c[i-1]-c[i-2]+m)%m; 32 //cout<<c[i]<<endl; 33 if(c[i-1]==c[1]&&c[i]==c[2]){ 34 //cout<<len<<endl; 35 return len=i-2; 36 } 37 } 38 } 39 ll poww(ll a,ll b,ll p){ 40 ll ans=1%p; 41 a%=p; 42 while(b){ 43 if(b&1) ans=(ans*a)%p; 44 b>>=1; 45 a=(a*a)%p; 46 } 47 return ans%p; 48 } 49 int main() 50 { 51 scanf("%d",&t); 52 gep(i,1,t){ 53 scanf("%lld%lld",&x,&m);//1+2^x(x很大)一定有循环节。 54 //cout<<x<<" "<<m<<endl; 55 solve();//因为%m的m不大，因此每次都去找循环节。 56 ll id=poww(2,x,len); 57 id=(id+1)%len; 58 printf("Case #%d: %lld ",i,((c[id]-1)%m+m)%m); 59 } 60 return 0; 61 }