【题解】csp.ac #61. 乘积求和(T2-20)

题目

csp.ac #61. 乘积求和(T2-20)

思路

求 \(\sum_{i=0}^{n} C_{n}^{i} \times F_i\)，其中 \(C_n^i\) 为组合数，\(F_i\) 为斐波那契数列。

\(\begin{aligned} \sum_{i=0}^{n} C_{n}^{i} \times F_i &= \sum_{i=0}^{n}C_{n}^{i}\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^i-(\frac{1-\sqrt{5}}{2})^i] \ \ \ \ \ \ \ \ \ \ \ \ \ \text{（斐波那契通项公式）}\\\\ &=\frac{1}{\sqrt{5}}\sum_{i=0}^{n}C_n^i[(\frac{1+\sqrt{5}}{2})^i-(\frac{1-\sqrt{5}}{2})^i]\ \ \ \ \ \ \ \ \ \ \ \ \ \text{（提出一个常数）} \\\\ &=\frac{1}{\sqrt{5}}\sum_{i=0}^{n}C_n^i(\frac{1+\sqrt{5}}{2})^i-\frac{1}{\sqrt{5}}\sum_{i=0}^{n}C_n^i(\frac{1-\sqrt{5}}{2})^i \\\\ &=\frac{1}{\sqrt{5}}[(1+\frac{1+\sqrt{5}}{2})^n- (1+\frac{1-\sqrt{5}}{2})^n] \ \ \ \ \ \ \ \ \ \ \ \ \ \text{（二项式定理）}\\\\ &=\frac{1}{\sqrt{5}}[(\frac{3+\sqrt{5}}{2})^n- (\frac{3-\sqrt{5}}{2})^n] \\\\ &=\frac{1}{\sqrt{5}}[[(\frac{1+\sqrt{5}}{2})^2]^n- (\frac{1-\sqrt{5}}{2})^2]^n] \\\\ &=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{2n}- (\frac{1-\sqrt{5}}{2})^{2n}] \\\\ &=F_{2n} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{（斐波那契通项公式）}\end{aligned}\)

二项式定理，不会的可以康康
\((1+b)^n = \sum\limits_{i=0}^{n}C_{n}^{i}b^i\)

Code

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>

inline void read(int &T) {
	int x=0;bool f=0;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=!f;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	T=f?-x:x;
}

const int mod=1e9+7;
int t,n,f[2000001];

int main() {
	read(t);f[1]=f[0]=1;
	for(int i=2;i<=2000000;++i) {
		f[i]=(f[i-1]+f[i-2])%mod;
	}
	while(t--) {
		read(n);
		std::cout<<f[2*n]<<'\n';;
	}
	return 0;
}