2020牛客多校 第六场B

规律题，在OEIS查了一下，找到了这个，所以说$f(i) = frac{(2^i - 2^{i-1}) (2 ^ i - 2 ^ {i-2}) ..... (2 ^ i - 2 ^ 0)}{(2 ^ i) ^ i}$

列出：

$f(i) = frac{(2 ^ i - 2 ^ 0) (2 ^ i - 2 ^ 1).......(2^i - 2^{i-1})}{(2 ^ i) ^ i} = frac{(2 ^ {i + 1} - 2 ^ 1) (2 ^ {i + 1} - 2 ^ 2).......(2^{i + 1} - 2^ i)}{(2 ^ i) ^ {i + 1}}$

即$f(i) = frac{(2 ^ {i + 1} - 2 ^ 1) (2 ^ {i + 1} - 2 ^ 2).......(2^{i + 1} - 2^ i)}{(2 ^ {i + 1}) ^ i}$

$f(i + 1) = frac{(2 ^ {i + 1} - 2 ^ 0) (2 ^ {i + 1} - 2 ^ 1)(2 ^ {i + 1} - 2 ^ 2).......(2^{i + 1} - 2^ i)}{(2 ^ {i + 1}) ^ {i + 1}}$

可推出$f(i + 1) = f(i) * frac{2 ^ {i + 1} - 2 ^0 }{2 ^ {i + 1}}$

Code:

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

using namespace std;

typedef long long ll;

const int N = 2e7 + 10;
const int mod = 1e9 + 7;

ll qmi(ll a, ll b, ll p){
    ll res = 1;
    while(b){
        if(b & 1) res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}

ll ans[N], res[N];

void init(){
    res[0] = 1;
    ll k = 1, infact = 1, mid = qmi(2, mod - 2, mod);
    for(int i = 1; i <= N - 10; i ++){
        k = k * 2 % mod;
        infact = infact * mid % mod;
        res[i] = res[i - 1] * (k - 1 + mod) % mod * infact % mod;
    }
    ans[1] = res[1];
    for(int i = 2; i <= N - 10; i ++){
        ans[i] = res[i] ^ ans[i - 1];
    }
}

void solve(){
    int n;
    scanf("%d", &n);
    printf("%lld
",ans[n]);
}

int main(){
    int t = 1;
    scanf("%d", &t);
    
    init();
    while(t --){
        solve();
    }
    return 0;
}