问题 C: 【例题3Fibonaccin项和

时间限制: 1 Sec  内存限制: 128 MB
提交: 12  解决: 8
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题目链接:http://acm.ocrosoft.com/problem.php?cid=1704&pid=2

题目描述

大家都知道Fibonacci数列吧,f[1]=1,f[2]=1,f[3]=2,f[4]=3...也就是f[n]=f[n-1]+f[n-2]。现在,问题很简单,输入nm,求前n项和取模m

输入

输入nm

1<=n<=2 000 000 000

1<=m<=1 000 000 010

输出

输出前n项和取模m

样例输入

5 1000

样例输出

12

思路:矩阵快速幂,加推导斐波那契数列的矩阵递推式,另外,我们知道菲波那切数列的前n项和等于菲波那切数列的第n-2项值-1

即:S(n)=F(n+2)-1

代码:

#include<bits/stdc++.h>

using namespace std;

#define ll long long

struct MUL

{

    ll m[3][3];

}ans, res;

MUL mul(MUL a, MUL b, int n, int mod)//矩阵乘法,定义结构体函数,传入结构体,返回结构体,这样比较方便

{

    MUL tmp;

    for (int i = 1; i <= n; i++)

    {

         for (int j = 1; j <= n; j++)

         {

             tmp.m[i][j] = 0;

         }

    }

    for (int i = 1; i <= n; i++)

    {

         for (int j = 1; j <= n; j++)

         {

             for (int k = 1; k <= n; k++)

             {

                  tmp.m[i][j] += ((a.m[i][k] % mod) *(b.m[k][j] % mod)) % mod;

             }

         }

    }

    return tmp;

}

ll quickpower(int N, int n, int mod)//矩阵快速幂

{

    MUL C, res;//C是常数矩阵

    memset(res.m, 0, sizeof(res.m));

    C.m[1][1] = 1;

    C.m[1][2] = 1;

    C.m[2][1] = 1;

    C.m[2][2] = 0;

    for (int i = 1; i <= n; i++)res.m[i][i] = 1;//单位矩阵

    while (N)

    {

         if (N & 1)

             res = mul(res, C, n, mod);

         C = mul(C, C, n, mod);

         N = N >> 1;

    }

    return res.m[1][1];

}

int main()

{

    ll n, mod;

    cin >> n >> mod;

    cout << quickpower(n + 1, 2, mod) % mod - 1 << endl;//S(n)=F(n+2)-1

}