2020牛客暑期多校训练营（第六场）（线性代数+找规律）

2020牛客暑期多校训练营（第六场）（线性代数+找规律）

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld
judge：牛客

题目描述

Roundgod is obsessive about linear algebra. Let $A=\{0,1\}$, everyday she will generate a binary vector randomly in $A^n$. Now she wonders the probability of generating $n$ linearly independent vectors in the next $n$ days modulo $10^9+7$. Formally, it can be proved that the answer has the form of $\frac{P}{Q}$, where $P$ and $Q$ are coprime and $Q$ is not a multiple of $10^9+7$. The answer modulo $10^9+7$ thus means $P\cdot Q^{-1}(\text{mod}10^9+7)$, where $Q^{-1}$ is the multiplicative inverse of $10^9+7$.

Wcy thinks the problem too easy. Let the answer of $n$ be $f_n$, she wants to know $f_1\oplus f_2\oplus ...\oplus f_N$, where $\oplus$ denotes bitwise exclusive or operation.

Note that when adding up two vectors, the components are modulo $2$.

输入描述:

The input contains multiple test cases. The first line of input contains one integer $T(1\le T\le 1000)$, denoting the number of test cases.

In the following $T$ lines, each line contains an integer $N(1\le N\le 2\ast 10^7)$ describing one test case.

输出描述:

For each test case, output one integer indicating the answer.

输入

3
1
2
3

输出

500000004
194473671
861464136

说明

$$\begin{gathered} f(1)=\frac{1}{2} \\ f(2)=\frac{3}{8} \\ f(3)=\frac{21}{64} \end{gathered}$$

题目大意

设$A$={0,1}，每天Roundgod从$A^n$（即维度为n，每一位由01组成的所有向量的集合）中随机选择一个二进制向量。现在他想知道n天中选取n个线性独立向量的概率。
设$f_n$​表示n的答案，最后输出 ，表示异或。

线性独立是什么？（其实就是任意一个向量不能通过其他两个的“加减”运算得到）

题解

由于这$n$个向量线性无关所以它们张成了满秩空间($N$维)
每一个向量都不属于之前的空间总共$2^N$ 个向量

当$N=1$时，设取到向量 $a$共有2个向量与它线性相关： $a$

当$N=2$时，设取到向量$ab$共有4个向量与它们线性相关：$a$, $b$,$a+b$,$0$

当$N=3$时，设取到向量 $abc$共有8个向量与它们线性相关： $a$, $b$, $c$, $a+b$, $b+c$, $a+c$, $a+b+c$, $0$

…

以此类推，当$N=i$时，有$2^i$个向量线性相关就有$2^N-2^i$个向量线性无关

$P(linearindependent)=\frac{2^n-2^i}{2^n}$

$f(n)=\prod _{i=1}^n\frac{2^n-2^i}{2^n}$

$=\prod _{i=1}^n\frac{2^{n-i}-1}{2^{n-i}}$

$=\prod _{i=1}^n\frac{2^i-1}{2^i}$

然后就可以递推了

#include 
const int N = 2e7, mod = 1e9 + 7;
int T, n, mul[N], inv[N], ans[N];
int ksm(int a, int b) {
    int r = 1;
    while (b) {
        if (b & 1) r = 1ll * r * a % mod;
        a = 1ll * a * a % mod, b >>= 1;
    }
    return r;
}
int main() {
    mul[0] = 1;
    for (int i = 1; i <= N; i++) mul[i] = 2ll * mul[i - 1] % mod;
    inv[N] = ksm(mul[N], mod - 2);
    for (int i = N - 1; i; i--) inv[i] = 2ll * inv[i + 1] % mod;
    int x = inv[1];
    ans[1] = inv[1];
    for (int i = 2; i <= N; i++)
        x = 1ll * x * (mul[i] - 1) % mod * inv[i] % mod,
        ans[i] = ans[i - 1] ^ x;
    scanf("%d", &T);
    while (T--) scanf("%d", &n), printf("%d\n", ans[n]);
}