[数学推导]对称轴

• 源自校内模拟赛

Statement

• 求

• [(sum_{i=0}^{p-1}inom{2i}im^i)mod p ]

• (1le m<ple 10^{14}) ，(p) 为质数

• 多组数据，数据组数不超过 (10^4)

Solution

• 神仙题

• 一个转化：

• [inom{2n}n=frac 1{(n!)^2}(prod_{i=1}^n(2i-1))(prod_{i=1}^n2i)=frac{2^n}{n!}prod_{i=1}^n(2i-1) ]

• 这看上去没什么用

• 但我们可 (bu) 以 (neng) 想到把 (prod) 里面的每个数都取反后加上 (p) 再除以 (2)

• [ans=sum_{i=0}^{p-1}frac{(2m)^i}{i!}prod_{j=1}^i(2j-1)=sum_{i=0}^{p-1}frac{(-4m)^i}{i!}prod_{j=1}^i(frac{p+1}2-j) ]

• [=sum_{i=0}^{p-1}(-4m)^ifrac{prod_{j=1}^i(lfloorfrac p2 floor-j+1)}{i!}=sum_{i=0}^{lfloorfrac p2 floor}inom{lfloorfrac p2 floor}i(-4m)^i=(1-4m)^{lfloorfrac p2 floor} ]

• 于是快速幂套快 (gui) 速乘即可，每组数据 (O(log^2p))

• 当然如果你有高超的打表和猜想技巧，这题是可以被打表水过去的

Code

#include <bits/stdc++.h>

template <class T>
inline void read(T &res)
{
	res = 0; bool bo = 0; char c;
	while (((c = getchar()) < '0' || c > '9') && c != '-');
	if (c == '-') bo = 1; else res = c - 48;
	while ((c = getchar()) >= '0' && c <= '9')
		res = (res << 3) + (res << 1) + (c - 48);
	if (bo) res = ~res + 1;
}

typedef long long ll;

ll rqy, m;

ll prod(ll a, ll b)
{
	ll res = 0;
	while (b)
	{
		if (b & 1) res = (res + a) % rqy;
		a = (a + a) % rqy;
		b >>= 1;
	}
	return res;
}

ll qpow(ll a, ll b)
{
	ll res = 1;
	while (b)
	{
		if (b & 1) res = prod(res, a);
		a = prod(a, a);
		b >>= 1;
	}
	return res;
}

void work()
{
	read(rqy); read(m);
	printf("%lld
", qpow((1ll - (m << 2) + (rqy << 2)) % rqy, rqy - 1 >> 1));
}

int main()
{
	int T; read(T);
	while (T--) work();
	return 0;
}