HDU

Best Solver

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

It is known that y=(5+26–√)1+2xy=(5+26)1+2x. 
For a given integer x (0≤x<232)x (0≤x<232) and a given prime number M (M≤46337)M (M≤46337), print [y]%M[y]%M. ([y][y] means the integer part of yy) 

InputAn integer T (1<T≤1000)T (1<T≤1000), indicating there are TT test cases. 
Following are TT lines, each containing two integers xx and MM, as introduced above.OutputThe output contains exactly TT lines. 
Each line contains an integer representing [y]%M[y]%M.Sample Input

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

Sample Output

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



循环节题目常见的有两种情况：
1.MOD-1
2.MOD^2-1
通过推导或暴力可求出。

本题循环节MOD^2-1。

#include<bits/stdc++.h>
#define MAX 3
using namespace std;
typedef long long ll;

ll n,MOD;
struct mat{
    ll a[MAX][MAX];
};

mat operator *(mat x,mat y)
{
    mat ans;
    memset(ans.a,0,sizeof(ans.a));
    for(int i=1;i<=2;i++){
        for(int j=1;j<=2;j++){
            for(int k=1;k<=2;k++){
                ans.a[i][j]+=(x.a[i][k]*y.a[k][j]+MOD)%MOD;
                ans.a[i][j]%=MOD;
            }
        }
    }
    return ans;
}
mat qMod(mat a,ll n)
{
    mat t;
    t.a[1][1]=10;t.a[1][2]=-1;
    t.a[2][1]=1;t.a[2][2]=0;
    while(n){
        if(n&1) a=t*a;
        n>>=1;
        t=t*t;
    }
    return a;
}
ll qsortMod(ll a,ll b)
{
    ll ans=1;
    a%=((MOD+1)*(MOD-1));
    while(b){
        if(b&1) ans=ans*a%((MOD+1)*(MOD-1));
        b>>=1;
        a=a*a%((MOD+1)*(MOD-1));
    }
    return ans;
}
int main()
{
    int tt=0,t,i;
    scanf("%d",&t);
    while(t--){
        scanf("%I64d%I64d",&n,&MOD);
        mat a;
        a.a[1][1]=10;a.a[1][2]=0;
        a.a[2][1]=2;a.a[2][2]=0;
        a=qMod(a,qsortMod(2,n));
        printf("Case #%d: %I64d
",++tt,(a.a[1][1]-1+MOD)%MOD);
        
    }
    return 0;
}