# BZOJ5300 [CQOI2018]九连环 题解 | 高精度 FFT

今天做了传说中的CQOI六道板子题……有了一种自己很巨的错觉(雾

题面

求n连环的最少步数,n <= 1e5。

题解

首先……我不会玩九连环……

通过找规律(其实是百度搜索)可知,\(n\)连环的最少步数是\(\lfloor\frac{2^{n + 1}}{3}\rfloor\)

(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\(2^{n+1}-1\)!)

然后我们要输出这个\(\lfloor\frac{2^{n + 1}}{3}\rfloor\)就好了。

然后我们发现——毒瘤出题人让我们写高精度。

……那就只能写咯……

用FFT做高精度乘法,封装起来,然后正常进行快速幂即可。

#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>
void read(T &x){
    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(){

    read(T);
    while(T--){
	read(x), x++;
	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