CSP.ac #61乘积求和

题面大意：

#61乘积求和
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solution

帕斯卡法则：

(dbinom{n - 1}{k} + dbinom{n - 1}{k - 1} = dbinom{n}{k})

证明

(dbinom{n}{k})是从(n)中选(k)个，有多少种组合
从(n)个数中拿出一个数(x)，有两种情况
(1.)选(x)的话，那就为(dbinom{n - 1}{k - 1})
(2.)不选(x)的话，那就为(dbinom{n}{k - 1})
两者加和就为(dbinom{n}{k})

二项式定理

对于一个二项式，((a+b)^n＝sumlimits_{k = 0}^{n}dbinom{n}{k}a^{n - k}b^k)

证明1：

由数学归纳法，当(n=1),左边(=a+b),右边(=dbinom{0}{1}a+dbinom{1}{1}b)
设(n=k)时该式成立，则(n=k+1)时，
(=(a+b)^n∗(a+b))
(=a∗sumlimits_{k = 0}^{ n}dbinom{n}{k}a^{n - k}b^k+b∗sumlimits_{j = 0}^{ n}dbinom{n}{j}a^{n - j}b^j)
(=sumlimits_{k = 0}^{n}dbinom{n}{k}a^{n - k + 1}b^k+sumlimits_{j = 0}^{n}dbinom{n}{j}a^{n - j}b^{j + 1})
(=a^{n + 1}sumlimits_{k = 1}^{n}dbinom{n}{k}a^{n - k + 1}b^k+sumlimits_{j = 0}^{n}dbinom{n}{j}a^{n - j}b^{j + 1})(把(k = 0)单独取出来)
(=a^{n + 1}+sumlimits_{k = 1}^{n}dbinom{n}{k}a^{n - k + 1}b^k+sumlimits_{k = 1}^{n + 1}dbinom{n}{k - 1}a^{n - k + 1}b^{k})(令(j = k - 1))
(=a^{n + 1}+sumlimits_{k = 1}^{n}dbinom{n}{k}a^{n - k + 1}b^k+b^{n + 1}+sumlimits_{k = 1}^{n}dbinom{n}{k - 1}a^{n - k + 1}b^{k})(取出第(k = n + 1)项)
(=a^{n + 1}+b^{n + 1}+sumlimits_{k = 1}^{n}left(dbinom{n}{k} + dbinom{n}{k - 1} ight)a^{n - k + 1}b^{k})(加和)
(=a^{n + 1}+b^{n + 1}+sumlimits_{k = 1}^{n}dbinom{n + 1}{k}a^{n - k + 1}b^{k})(套用帕卡斯定理)
(=sumlimits_{k = 0}^{ n + 1}dbinom{n + 1}{k}a^{n - k + 1}b^{k + 1})(再把k = 0加回去)

证明2:

((a+b)^n) 是由(n)个((a+b))相乘，对于其中(a^k)中的这一项，我们肯定是从中选择了(k)个(a)相乘，剩下的(b)相乘就是(b^{n−k})，这样的选法共有(dbinom{n}{k})个，因此该项为(dbinom{n}{k}a^{n - k}b^k) ,那么((a+b)^n=sumlimits_{k = 0}^{n}dbinom{n}{k}a^{n - k}b^k)
来自Zsf大佬整理的数学组提供的详细证明2

变形

当(a = 1) 时就变成了((1+b)^n＝sumlimits_{k = 0}^{n}dbinom{n}{k}b^k),这里的(b)可以为一个式子，既(一会用得到)
(left(1 + dfrac{1 + sqrt{5}}{2} ight)^n = sumlimits_{i = 0}^{n}dbinom{n}{i}left(dfrac{1 + sqrt{5}}{2} ight)^i)
(left(1 + dfrac{1 - sqrt{5}}{2} ight)^n = sumlimits_{i = 0}^{n}dbinom{n}{i}left(dfrac{1 - sqrt{5}}{2} ight)^i)

斐波那契通项公式

(F_n = dfrac{1}{sqrt{5}}left(left(dfrac{1 + sqrt{5}}{2} ight)^n -left(dfrac{1 - sqrt{5}}{2} ight)^n ight))

证明

见百度百科

题解：

(sumlimits_{i = 0}^{n}dbinom{n}{i}F_i)
(= sumlimits_{i = 0}^{n}dbinom{n}{i}dfrac{1}{sqrt{5}}left(left(dfrac{1 + sqrt{5}}{2} ight)^i -left(dfrac{1 - sqrt{5}}{2} ight)^i ight))
(= dfrac{1}{sqrt{5}}sumlimits_{i = 0}^{n}dbinom{n}{i}left(left(dfrac{1 + sqrt{5}}{2} ight)^i -left(dfrac{1 - sqrt{5}}{2} ight)^i ight))
(= dfrac{1}{sqrt{5}}sumlimits_{i = 0}^{n}dbinom{n}{i}left(dfrac{1 + sqrt{5}}{2} ight)^i -dfrac{1}{sqrt{5}}sumlimits_{i = 0}^{n}dbinom{n}{i}left(dfrac{1 - sqrt{5}}{2} ight)^i)
这里的变形上面讲过了
(= dfrac{1}{sqrt{5}}left(1 + dfrac{1 + sqrt{5}}{2} ight)^n -dfrac{1}{sqrt{5}}left(1 + dfrac{1 - sqrt{5}}{2} ight)^n)
(= dfrac{1}{sqrt{5}}left(dfrac{3 + sqrt{5}}{2} ight)^n -dfrac{1}{sqrt{5}}left(dfrac{3 - sqrt{5}}{2} ight)^n)
(= dfrac{1}{sqrt{5}}left(left(dfrac{3 + sqrt{5}}{2} ight)^n -left(dfrac{3 - sqrt{5}}{2} ight)^n ight))
(= dfrac{1}{sqrt{5}}left(left(dfrac{6 + 2sqrt{5}}{4} ight)^n -left(dfrac{6 - 2sqrt{5}}{4} ight)^n ight))
(= dfrac{1}{sqrt{5}}left(left(dfrac{1 + 2sqrt{5} + 5}{4} ight)^n -left(dfrac{1 - 2sqrt{5} + 5}{4} ight)^n ight))
(= dfrac{1}{sqrt{5}}left(left({left(dfrac{1 + sqrt{5}}{2} ight)^2} ight)^n-left({left(dfrac{1 + sqrt{5}}{2} ight)^2} ight)^n ight))
(= dfrac{1}{sqrt{5}}left(left(dfrac{1 + sqrt{5}}{2} ight)^{2n} - left(dfrac{1 - sqrt{5}}{2} ight)^{2n} ight))
(= F_{2n})

(Code)

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

#define int long long
#define rr register

#define MAXN 2000010
#define inf 1e9

using namespace std;

const int mod = 1e9 + 7;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) {f |= ch == '-'; ch = getchar();}
	while (isdigit(ch)) {s = s * 10 + (ch ^ 48); ch = getchar();}
	return f ? -s : s;
}

int T, n;

int g[MAXN], inv[MAXN], f[MAXN];

inline void init() {
	f[0] = f[1] = 1;
	g[0] = 1;
	inv[0] = inv[1] = 1;
	for (rr int i = 1; i <= 1000000; i++) 
		g[i] = (i * g[i - 1]) % mod;
	for (rr int i = 2; i <= 2000000; i++)
		f[i] = (f[i - 1] + f[i - 2]) % mod;
	for (rr int i = 2; i <= 1000000; i++) 
		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	for (rr int i = 2; i <= 1000000; i++)
		inv[i] = (inv[i - 1] * inv[i]) % mod;
}

inline int C(int n, int m) {
    return (g[n] * inv[m] % mod) * inv[n - m] % mod;
}

signed main() {
	init();
	T = read();
	while (T--) {
		n = read();
		cout << f[2 * n] << "
";
	}
}