幼儿园数学题I

ssl 2513

题目大意

定义$f_n={\left(\frac{\sqrt{5}+1}{2}\right)}^{n-1}$，求前n项的和,（对$10^9+7$取模）（题目貌似出了些问题，实际上$f$等价于斐波那契数列）

输入样例#1

1

输出样例#1

1

输入样例#2

2

输出样例#2

2

数据范围

$1⩽n⩽2^{31}-1$

解题思路

通过打表可以发现$f$等价于斐波那契数列，那题目就是求斐波那契数列前$n$项的和
n十分大，不能直接递推
考虑矩阵乘法
设矩阵$\left[\begin{array}{ccc}{f}_{i-2}& f_{i-1}& s_{i-2}\end{array}\right]$
在通过乘矩阵$\left[\begin{array}{ccc}0& 1& 0\\ 1& 1& 1\\ 0& 0& 1\end{array}\right]$来递推
然后快速幂即可

代码

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define wyc 1000000007
using namespace std;
ll n;
struct matrix
{ll n, m, a[10][10];matrix operator *(const matrix &b) const{matrix c;c.n = n;c.m = b.m;for (int i = 1; i <= c.n; ++i)for (int j = 1; j <= c.m; ++j)c.a[i][j] = 0;for (int i = 1; i <= c.n; ++i)for (int k = 1; k <= m; ++k)for (int j = 1; j <= c.m; ++j)c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j] % wyc) % wyc;return c;}
}A, B;
void Counting(ll x)
{while(x){if (x&1) A = A * B;B = B * B;x>>=1;}
}
int main()
{scanf("%lld", &n);A.n = 1;A.m = B.n = B.m = 3;A.a[1][1] = 1, A.a[1][2] = 1, A.a[1][3] = 1;//初始状态B.a[1][1] = 0, B.a[1][2] = 1, B.a[1][3] = 0;//相乘的矩阵B.a[2][1] = 1, B.a[2][2] = 1, B.a[2][3] = 1;B.a[3][1] = 0, B.a[3][2] = 0, B.a[3][3] = 1;Counting(n - 1);printf("%lld", A.a[1][3]);return 0;
}