2020牛客暑期多校训练营（第六场）B.Binary Vector

2020牛客暑期多校训练营（第六场）B.Binary Vector

题目链接

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 nn linearly independent vectors in the next nn 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.

示例1

输入

3
1
2
3

输出

500000004
194473671
861464136

这种毒瘤题建议大家别读题(如果不是数学大佬就别浪费时间阅读理解，我们队就是一直死磕题目意思浪费了很多时间)，直接找规律得了，事实证明这题就是规律题，我们通过 $oeis$ 可以发现分子就是 $\prod 2^i-1,i\in [1,n]$，分母就是 $2^{i\ast (i+1)/2}$，如果直接预处理你就会 T 飞，所以我们考虑优化，因为这个 n 非常大，所以我们要保证每次循环里的计算尽可能接近 $O(1)$，下面观察答案，这题其实和杭电多校那个斐波那契求和很像，就是每次循环更新一下答案，避免重复计算，可以发现答案每次就是乘上一个 $\frac{2^i-1}{2^i}$，那么我们可以设一个初值 $p1=1$，每次乘二，表示分子，设一个初值 $p2=(500000004\equiv \frac{1}{2}\%1e9+7)$，每次乘 $500000004$ 表示分母，那么更新答案其实就是乘上 $(p1-1)\ast p2$ 就行了，这样能保证循环内的运算都是 $O(1)$ 的，AC代码如下：

#include
using namespace std;
typedef long long ll;
const int N=2e7+5;
const ll mod=1e9+7;
const ll inv2=500000004;
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
ll f[N],ans[N];
void init(){
    f[0]=1;
    ll p1=1,p2=1;
    for(ll i=1;i<=2e7;i++){
        p1=2*p1%mod;
        p2=inv2*p2%mod;
        f[i]=f[i-1]*p2%mod*(p1-1)%mod;
        ans[i]=f[i]^ans[i-1];
    }
}
int main(){
    init();
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        printf("%lld\n",ans[n]);
    }
    return 0;
}