Visit of the Great

1.前言

真好玩，蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤蛤

2.题解

$$\begin{array}{rl}令G(x)=\{\begin{array}{l}1,2\mid x\\ 2,2\nmid x\end{array}& \\ gcd(k^{2^i}+1,k^{2^{i+1}}+1)& =gcd(k^{2^i}+1,k^{2^{i+1}}-k^{2^i})\\ & =gcd(k^{2^i}+1,k^{2^i}\times (k^{2^i}-1))\\ & =gcd(k^{2^i}+1,k^{2^i}-1)\\ & =gcd((k^{2^i}+1)-(k^{2^i}-1),k^{2^i}-1)\\ & =gcd(2,k^{2^i}-1)\\ & =G(k)\end{array}$$

$$\begin{array}{rl}{\mathrm{l}\mathrm{c}\mathrm{m}}_{i=l}^r(k^{2^i}+1)\mathrm{m}\mathrm{o}\mathrm{d}p& =lcm(lc{m}_{i=l}^{r-2}(k^{2^i}+1),\frac{(k^{2^{r-1}}+1)\times (k^{2^r}+1)}{gcd(k^{2^{r-1}}+1,k^{2^r}+1)})\\ & =lcm(lc{m}_{i=l}^{r-2}(k^{2^i}+1),\frac{(k^{2^{r-1}}+1)\times (k^{2^r}+1)}{G(k)})\\ & =lcm(lc{m}_{i=l}^{r-3}(k^{2^i}+1),\frac{(k^{2^{r-2}}+1)\times (k^{2^{r-1}}+1)\times (k^{2^r}+1)}{G(k)\times G(k)})\\ & =...\end{array}$$

$$\begin{array}{rl}{\mathrm{l}\mathrm{c}\mathrm{m}}_{i=l}^r(k^{2^i}+1)\mathrm{m}\mathrm{o}\mathrm{d}p& =\frac{{\prod }_{i=l}^r(k^{2^i}+1)}{G(K{)}^{r-l}}\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =\frac{{\prod }_{i=l}^r(k^{2^i}+1)}{G^{r-l}(K)}\mathrm{m}\mathrm{o}\mathrm{d}p\end{array}$$

发现这个小东西 ${\prod }_{i=l}^r((k^{2^l}{)}^{i-l}+1)$ 就是 $\sum _{i=0}^{r=2^{r-l+1}-1}(k^{2^l}{)}^i$ (二进制拆分的唯一性)
$$\begin{array}{r}\frac{(k^{2^l}{)}^{2^{r-l+1}}-1}{k^{2^l}-1}\\ =\frac{k^{2^{r+1}}-1}{k^{2^l}-1}\end{array}$$

$$\begin{array}{rl}{\mathrm{l}\mathrm{c}\mathrm{m}}_{i=l}^r(k^{2^i}+1)\mathrm{m}\mathrm{o}\mathrm{d}p& =\frac{k^{2^{r+1}}-1}{(k^{2^l}-1)\times G^{r-l}(K)}\end{array}$$

然后有一个很 ？？ 的 ？？ 特判， $k\mathrm{m}\mathrm{o}\mathrm{d}p==0$ 时要搞很多奇怪的东西，去 ？ 好了。

因为当 $k\mathrm{m}\mathrm{o}\mathrm{d}p==0$ 时， $2^{r+1}\mathrm{m}\mathrm{o}\mathrm{d}(p-1)$ 如果为零，则它会算出来一个 $k^0\mathrm{m}\mathrm{o}\mathrm{d}p=1$，所以不对，这时候改为返回值为 $0$ 就好了。

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
    x = 0; T f = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + ch - '0';
        ch = getchar ();
    }
    x *= f;
}
template <typename T, typename... Args>
void read (T &x, Args&... args) {
    read (x); read (args...);
}
char For_Print[25];
template <typename T>
void write (T x) {
    if (x == 0) { putchar ('0'); return; }
    if (x < 0) { putchar ('-'); x = -x; }
    int poi = 0;
    while (x) {
        For_Print[++poi] = x % 10 + '0';
        x /= 10;
    }
    while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
    write (x); putchar (ch);
}

const LL Mod = 1e9 + 7;

LL sqre (LL x) { return x * x % Mod; }
void del (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void add (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

LL t, k, l, r, p;

LL quick_pow (LL x, LL y, LL M) {
    LL res = 1;
    while (y) {
        if (y & 1) res = (res * x) % M;
        x = (x * x) % M; y >>= 1;
    }
    return res;
}
LL inv (LL x, LL M) {
    return quick_pow (x, M - 2, M);
}
LL G (LL x) {
    return (x % 2 == 0 ? 1 : 2);
}
LL f (LL x) {
	return k % p == 0 ? 0 : quick_pow (k, quick_pow (2, x, p - 1), p);
}

int main () {
    // freopen ("D:\\lihan\\1.in", "r", stdin);
    // freopen ("D:\\lihan\\1.out", "w", stdout);
    read (t);
    while (t--) {
        read (k, l, r, p);
        
		if (p == 2) {
			printf ("%d\n", (k % 2 == 1 ? 0 : 1));
			continue;
		}
		
        LL res = 0;
		if (f (l) == 1) {
			res = quick_pow (2, r - l + 1, p);
		}
		else {
			
//			printf ("f (r + 1) = %lld, f (l) = %lld\n", f (r + 1), f (l));
			
			res = f (r + 1) - 1 + p;
			res = (res * inv (f (l) - 1 + p, p)) % p;
		}
		
//		cout << res << endl;
		
        res *= inv (quick_pow (G (k), r - l, p) % p, p);
        res = (res % p + p) % p;
        print (res, '\n');
    }
    return 0;
}

3.一些编辑很久的错误式子，pi用没有😓，下次我就要用这个出题 /wx

$$\begin{array}{rl}{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l}^r(k^{2^i}+1)\mathrm{m}\mathrm{o}\mathrm{d}p& =gcd((k^{2^l}+1),{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l+1}^r((k^{2^i}+1)-(k^{2^{i-1}}+1)\left)\right)\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =gcd((k^{2^l}+1),{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l+1}^r(k^{2^i}-k^{2^{i-1}}\left)\right)\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =gcd((k^{2^l}+1),{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l+1}^r(k^{2^{i-1}\cdot 2}-k^{2^{i-1}}\left)\right)\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =gcd((k^{2^l}+1),{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l+1}^r(k^{2^{i-1}}\times k^{2^{i-1}}-k^{2^{i-1}}\left)\right)\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =gcd((k^{2^l}+1),{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l+1}^r(k^{2^{i-1}}\times (k^{2^{i-1}}-1)\left)\right)\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =gcd((k^{2^l}+1),k^{2^l}\cdot {\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l+1}^r(k^{2^{i-1}}-1\left)\right)\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =gcd((k^{2^l}+1),{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l}^{r-1}(k^{2^i}-1\left)\right)\mathrm{m}\mathrm{o}\mathrm{d}p\end{array}$$

$$\begin{array}{rl}{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l}^r(k^{2^i}-1)& =gcd(k^{2^l}-1,{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l+1}^r(k^{2^i}-k^{2^{i-1}}))\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =gcd(k^{2^l}-1,k^{2^l}\cdot {\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l}^{r-1}(k^{2^i}-1))\mathrm{m}\mathrm{o}\mathrm{d}p\\ & =(k^{2^l}-1)\mathrm{m}\mathrm{o}\mathrm{d}p\end{array}$$

所以原式可以化为

$$\begin{array}{rl}{\mathrm{g}\mathrm{c}\mathrm{d}}_{i=l}^r(k^{2^i}+1)\mathrm{m}\mathrm{o}\mathrm{d}p& =gcd((k^{2^l}+1),(k^{2^l}-1))\mathrm{m}\mathrm{o}\mathrm{d}p\end{array}$$