2013長沙網路賽H題Hypersphere (蛋疼的題目 神似邀請賽A題)

In the world of k-dimension, there's a large hypersphere made by mysterious metal. People in the world of k-dimension are performing a ceremony to worship the goddess of dimension. They are melting the large hypersphere into metal flow, and then they will cast the metal flow into unit hyperspheres. An unit hypersphere is a hypersphere which radius is 1.

The mysterious metal in k-dimension world has a miraculous property: if k unit hyperspheres made by the mysterious metal are constructed, all of these k unit hyperspheres will be sacrificed to goddess of dimension and disappear from the k-dimension world immediately.

After the ceremony, there will be some unit hyperspheres and a little metal flow left, and people in the world of k-dimension want to know the number of unit hyperspheres left.

You might want to know that how the large hypersphere was constructed. At first, people in the world created a long ruler which length is l. And then, they drew a rectangle with lengthl - 1 and width l. Using some mysterious method, they changed the rectangle into a square but didn't change the area. After that, they extended the ruler's length by the length of the square's side. After successfully made the ruler, people started using magic to construct the large hypersphere. The magic could make a hypersphere become more and more larger. Started with a hypersphere of zero radius, the magic will be continuously used until the radius reached the ruler's length.

Input

There will be several test cases. Each test case contains two integers k (3 ≤ k ≤ 1000000000) and l (2 ≤ l ≤ 1000000000), which are the same meanings as in the description part.

Output

For each test case, please output the number of unit hyperspheres people will get in one line.

Sample Input

3 3
5 6

Sample Output

2
1

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
long long l,mo;
long long tmp[2][2],p[2][2],ret[2][2];

void init()
{
    ret[0][0]=2*l,ret[0][1]=-l,ret[1][0]=1,ret[1][1]=0;
    p[0][0]=2*l,p[0][1]=-l,p[1][0]=1,p[1][1]=0;
}

void cal1()  //矩陣的平方
{
    int i,j,k;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            tmp[i][j]=p[i][j];
            p[i][j]=0;
        }
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
            for(k=0;k<2;k++)
               p[i][j]=(p[i][j]+tmp[i][k]*tmp[k][j])%mo;
}

void cal2()  //矩陣的乘法
{
    int i,j,k;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            tmp[i][j]=ret[i][j];
            ret[i][j]=0;
        }
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
            for(k=0;k<2;k++)
               ret[i][j]=(ret[i][j]+tmp[i][k]*p[k][j])%mo;
}

int main()
{
    long long t;
    while(~scanf("%lld%lld",&mo,&l))
    {
        init();
        t=mo-1;
        while(t)
        {
            if(t&1) cal2();
            cal1();
            t>>=1;
        }
        printf("%lld\n",((ret[1][0]*2*l+ret[1][1]*2)%mo-1+mo)%mo);
    }
}