标签：com class 应该 一个 scan i++ n+1 mat oid
给你一个 \(n\) 连环，游戏规则是：
现在 \(m\) 组询问，每组询问给你 \(n\) 连环，问你至少多少步取下所有的环。
\(1\leq n\leq 10^5,1\leq m\leq 10\)
数学书上推导很清楚了：
值得注意的是第二张图片中的 \(n\) 为奇数的推导式中 \(\frac{2(1-2^{n+1})}{1-2^2}\) 应该是 \(\frac{1-2^{n+1}}{1-2^2}\)
有幸能指出数学书的错误。
然后 \(\text{FFT}\) 快速幂乱搞就好了。不开 \(-O2\) 玩 \(\text{FFT}\) 不就是在玩火吗？？？
这个瓜皮代码常数过大在 b 站上过不了。
#include <bits/stdc++.h>
#define dob complex<double>
using namespace std;
const int N = (100000<<2)+5;
const double pi = acos(-1.);
int n, nn, m, len, L, R[N], A[N];
dob a[N], b[N];
void FFT(dob *A, int o) {
for (int i = 0; i < len; i++) if (i < R[i]) swap(A[i], A[R[i]]);
for (int i = 1; i < len; i <<= 1) {
dob wn(cos(pi/i), sin(pi*o/i)), x, y;
for (int j = 0; j < len; j += (i<<1)) {
dob w(1, 0);
for (int k = 0; k < i; k++, w = w*wn) {
x = A[j+k], y = w*A[i+j+k];
A[j+k] = x+y, A[i+j+k] = x-y;
}
}
}
}
void work() {
scanf("%d", &n); nn = n; ++n; m = log(2)*n+5;
for (L = 0, len = 1; len <= m; len <<= 1) ++L;
for (int i = 0; i < len; i++) a[i] = b[i] = 0;
for (int i = 0; i < len; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
a[0] = 1, b[0] = 2;
while (n) {
FFT(a, 1), FFT(b, 1);
if (n&1) for (int i = 0; i <= len; i++) a[i] = a[i]*b[i];
for (int i = 0; i <= len; i++) b[i] = b[i]*b[i]; n >>= 1;
FFT(a, -1); FFT(b, -1);
for (int i = 0; i < len; i++) A[i] = a[i].real()/len+0.5;
int loc = 0; while (A[loc] && loc < len) A[loc+1] += A[loc]/10, A[loc] %= 10, ++loc;
for (int i = 0; i < len; i++) a[i] = A[i];
for (int i = 0; i < len; i++) A[i] = b[i].real()/len+0.5;
loc = 0; while (A[loc] && loc < len) A[loc+1] += A[loc]/10, A[loc] %= 10, ++loc;
for (int i = 0; i < len; i++) b[i] = A[i];
}
for (int i = 0; i < len; i++) A[i] = a[i].real();
A[0] -= 1+(!(nn&1));
for (int i = len-1, flag = 0, sum = 0; i >= 0; i--) {
sum = sum*10+A[i]; if (sum/3) flag = 1;
if (flag) printf("%d", sum/3), sum %= 3;
}
puts("");
}
int main() {int t; cin >> t; while (t--) work(); return 0; }标签：com class 应该 一个 scan i++ n+1 mat oid
原文地址：https://www.cnblogs.com/NaVi-Awson/p/8989286.html
