[数学推导]对称轴
- 源自校内模拟赛
Statement
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求
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\[(\sum_{i=0}^{p-1}\binom{2i}im^i)\bmod p \]
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\(1\le m<p\le 10^{14}\) ,\(p\) 为质数
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多组数据,数据组数不超过 \(10^4\)
Solution
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神仙题
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一个转化:
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\[\binom{2n}n=\frac 1{(n!)^2}(\prod_{i=1}^n(2i-1))(\prod_{i=1}^n2i)=\frac{2^n}{n!}\prod_{i=1}^n(2i-1) \]
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这看上去没什么用
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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 \(2\)
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\[ans=\sum_{i=0}^{p-1}\frac{(2m)^i}{i!}\prod_{j=1}^i(2j-1)=\sum_{i=0}^{p-1}\frac{(-4m)^i}{i!}\prod_{j=1}^i(\frac{p+1}2-j) \]
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\[=\sum_{i=0}^{p-1}(-4m)^i\frac{\prod_{j=1}^i(\lfloor\frac p2\rfloor-j+1)}{i!}=\sum_{i=0}^{\lfloor\frac p2\rfloor}\binom{\lfloor\frac p2\rfloor}i(-4m)^i=(1-4m)^{\lfloor\frac p2\rfloor} \]
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于是快速幂套快 (gui) 速乘即可,每组数据 \(O(\log^2p)\)
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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\n", qpow((1ll - (m << 2) + (rqy << 2)) % rqy, rqy - 1 >> 1));
}
int main()
{
int T; read(T);
while (T--) work();
return 0;
}