# # BZOJ5300 [CQOI2018]九连环 题解 | 高精度 FFT

## 题解

（实际上，九连环的步骤恰好是一个叫【格雷码】的编码方式中的$1$一直到$2^{n+1}-1$！）

……那就只能写咯……

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#define enter putchar('\n')
#define space putchar(' ')
using namespace std;
typedef long long ll;
template <class T>
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op == 1) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 150000;
const double PI = acos(-1);
int T, x;

struct cp {
double a, b;
cp(){}
cp(double x, double y): a(x), b(y){}
cp operator + (const cp &obj) const {
return cp(a + obj.a, b + obj.b);
}
cp operator - (const cp &obj) const {
return cp(a - obj.a, b - obj.b);
}
cp operator * (const cp &obj) const {
return cp(a * obj.a - b * obj.b, a * obj.b + b * obj.a);
}
} inv[N], omg[N];

void init(int n){
for(int i = 0; i < n; i++){
omg[i] = cp(cos(2 * PI / n * i), sin(2 * PI / n * i));
inv[i] = cp(omg[i].a, -omg[i].b);
}
}
void fft(cp *a, int n, cp *omg){
int lim = 0;
while((1 << lim) < n) lim++;
for(int i = 0; i < n; i++){
int t = 0;
for(int j = 0; j < lim; j++)
if(i >> j & 1) t |= 1 << (lim - j - 1);
if(i < t) swap(a[i], a[t]);
}
for(int l = 2; l <= n; l <<= 1){
int m = l / 2;
for(cp *p = a; p != a + n; p += l)
for(int i = 0; i < m; i++){
cp t = omg[n / l * i] * p[i + m];
p[i + m] = p[i] - t;
p[i] = p[i] + t;
}
}
}

struct big {
int g[N], len;
big(){
memset(g, 0, sizeof(g));
len = 1;
}
big(int x){
memset(g, 0, sizeof(g));
len = 0;
if(!x){
len = 1;
return;
}
while(x) g[len++] = x % 10, x /= 10;
}
void out(){
for(int i = len - 1; i >= 0; i--)
printf("%d", g[i]);
enter;
}
void operator /= (int x){
int sum = 0, newlen = 0;
for(int i = len - 1; i >= 0; i--){
sum = sum * 10 + g[i];
if(sum < x) g[i] = 0;
else{
if(!newlen) newlen = i + 1;
g[i] = sum / x;
sum %= x;
}
}
len = max(newlen, 1);
}
void operator *= (const big &b){
static cp A[N], B[N];
int newlen = len + b.len - 1, n = 1;
while(n < newlen) n <<= 1;
for(int i = 0; i < n; i++){
A[i] = cp(i < len ? g[i] : 0, 0);
B[i] = cp(i < b.len ? b.g[i] : 0, 0);
}
init(n);
fft(A, n, omg);
fft(B, n, omg);
for(int i = 0; i < n; i++)
A[i] = A[i] * B[i];
fft(A, n, inv);
for(int i = 0; i < newlen; i++)
g[i] = (int)floor(A[i].a / n + 0.5);
g[len = newlen] = 0;
for(int i = 0; i < len; i++)
g[i + 1] += g[i] / 10, g[i] %= 10;
if(g[len]) len++;
}
} ret, a;

int main(){

while(T--){
ret = big(1), a = big(2);
while(x){
if(x & 1) ret *= a;
a *= a;
x >>= 1;
}
ret /= 3;
ret.out();
}

return 0;
}

posted @ 2018-04-20 19:05  胡小兔  阅读(552)  评论(0编辑  收藏