How Many Calls? UVA - 10518

How Many Calls? UVA - 10518
题意：求f(n)的递归调用次数，用输出base进制的最后一个数。
思路：f(0)=1,f(1)=1，f(n)=f(n-1)+f(n-2)+1;

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll mod;
struct  matrix{
    ll x[3][3];
};
matrix multi(matrix a,matrix b){
    matrix temp;
    memset(temp.x,0,sizeof(temp.x));
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
            for(int k=0;k<3;k++)
            {
                temp.x[i][j]+=a.x[i][k]*b.x[k][j];
                temp.x[i][j]%=mod;
                //if(temp.x[i][j]<0) temp.x[i][j]=(temp.x[i][j]+mod)%mod;
            }
    return temp;
}
matrix quick_multi(matrix a,ll n)//矩阵快速幂
{
    matrix temp=a;
    n--;
    while(n){
        if(n&1)
            temp=multi(temp,a);
        a=multi(a,a);
        n>>=1;
    }
    return temp;
}
/*void p(matrix a)
{
    for(int i=0;i<5;i++)
    {
        for(int j=0;j<5;j++)
            printf("%lld ",a.x[i][j]);
        printf("\n");
    }
}*/
int main()
{
    ll n,kcase=0;
    while(scanf("%lld%lld",&n,&mod)!=EOF&&(n||mod))
    {
        printf("Case %lld: %lld %lld ",++kcase,n,mod);
        matrix A,B;
        if(n<=1) n==1?printf("1\n"):printf("1\n");
        else
        {
            memset(A.x,0,sizeof(A.x));
            memset(B.x,0,sizeof(B.x));
            A.x[0][0]=A.x[0][1]=A.x[0][2]=1;
            A.x[1][0]=A.x[2][2]=1;
            B.x[0][0]=B.x[1][0]=B.x[2][0]=1;
            A=quick_multi(A,n-1);
            B=multi(A,B);
            printf("%lld\n",B.x[0][0]%mod);
        }
    }
    return 0;
}