UVA10518 How Many Calls? —— 矩阵快速幂

题目链接：https://vjudge.net/problem/UVA-10518

题解：

问：求斐波那契数f[n]的时候调用了多少次f[n] = f[n-1] + f[n-2]，没有记忆化，一直递归到f[0]、f[1]，其中f[0]、f[1]也算调用了一次。

设求f[n]调用了S[n]次，则可知： S[n] = S[n-1] + S[n-2] + 1。构造矩阵求解即可。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
//const int MOD = 1e9+7;
const int MAXN = 1e6+100;

int MOD;
const int Size = 3;
struct MA
{
    LL mat[Size][Size];
    void init()
    {
        for(int i = 0; i<Size; i++)
        for(int j = 0; j<Size; j++)
            mat[i][j] = (i==j);
    }
};

MA mul(MA x, MA y)
{
    MA ret;
    memset(ret.mat, 0, sizeof(ret.mat));
    for(int i = 0; i<Size; i++)
    for(int j = 0; j<Size; j++)
    for(int k = 0; k<Size; k++)
        ret.mat[i][j] += 1LL*x.mat[i][k]*y.mat[k][j]%MOD, ret.mat[i][j] %= MOD;;
    return ret;
}

MA qpow(MA x, LL y)
{
    MA s;
    s.init();
    while(y)
    {
        if(y&1) s = mul(s, x);
        x = mul(x, x);
        y >>= 1;
    }
    return s;
}

MA tmp ={
    1, 1, 1,
    1, 0, 0,
    0, 0, 1
};

int main()
{
    LL n, b, f[2] = {1,1}, kase = 0;
    while(scanf("%lld%lld",&n,&b)&&(n||b))
    {
        MOD = b;
        if(n<=1)
        {
            printf("Case %lld: %lld %lld %lld
", ++kase, n, b, f[n]%MOD);
            continue;
        }

        MA s = tmp;
        s = qpow(s, n-1);
        LL ans = ((s.mat[0][0]+s.mat[0][1])%MOD+s.mat[0][2])%MOD;
        printf("Case %lld: %lld %lld %lld
", ++kase, n, b, ans);
    }
}