SWUST OJ 2540 少女输出问题

最新推荐文章于 2023-05-25 11:58:21 发布

原创最新推荐文章于 2023-05-25 11:58:21 发布 · 708 阅读

0 ·

CC 4.0 BY-SA版权

ACM/ICPC_FFT 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一道关于快速傅立叶变换（FFT）的算法题目，并提供了完整的代码实现。通过本例，读者可以了解到FFT的基本原理及其在多项式乘法等场景中的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：见这里
题意: 中文题目， FFT裸题。
代码如下：

//
//Created by BLUEBUFF 2016/1/11
//Copyright (c) 2016 BLUEBUFF.All Rights Reserved
//

#pragma comment(linker,"/STACK:102400000,102400000")
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/hash_policy.hpp>
//#include <bits/stdc++.h>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstdlib>
#include <cstring>
#include <complex>
#include <sstream> //isstringstream
#include <iostream>
#include <algorithm>
using namespace std;
//using namespace __gnu_pbds;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, LL> pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b)     memset(a, b, sizeof(a))
#define MP(x, y)      make_pair(x,y)
template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }
const int maxn = 270010;
const int maxm = 1e5+5;
const int maxs = 10;
const int maxp = 1e3 + 10;
const int INF  = 1e9;
const int UNF  = -1e9;
const int mod  = 1e9 + 7;
const int rev = (mod + 1) >> 1; // FWT
const double PI = acos(-1);
//head

struct FastIO
{
    static const int S = 1310720;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar()
    {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len) return -1;
        return buf[pos ++];
    }
    inline int xuint()
    {
        int c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x;
    }
    inline int xint()
    {
        int s = 1, c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        if (c == '-') s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while (c <= 32) c = xchar();
        for (; c > 32; c = xchar()) * s++ = c;
        *s = 0;
    }
    inline void wchar(int x)
    {
        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos ++] = x;
    }
    inline void wint(LL x)
    {
        if (x < 0) wchar('-'), x = -x;
        char s[24];
        int n = 0;
        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;
        while (n--) wchar(s[n]);
    }
    inline void wstring(const char *s)
    {
        while (*s) wchar(*s++);
    }
    ~FastIO()
    {
        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;

typedef complex <double> Complex;

void rader(Complex *y, int len) {
    for(int i = 1, j = len / 2; i < len - 1; i++) {
        if(i < j) swap(y[i], y[j]);
        int k = len / 2;
        while(j >= k) {j -= k; k /= 2;}
        if(j < k) j += k;
    }
}
void fft(Complex *y, int len, int op) {
    rader(y, len);
    for(int h = 2; h <= len; h <<= 1) {
        double ang = op * 2 * PI / h;
        Complex wn(cos(ang), sin(ang));
        for(int j = 0; j < len; j += h) {
            Complex w(1, 0);
            for(int k = j; k < j + h / 2; k++) {
                Complex u = y[k];
                Complex t = w * y[k + h / 2];
                y[k] = u + t;
                y[k + h / 2] = u - t;
                w = w * wn;
            }
        }
    }
    if(op == -1) for(int i = 0; i < len; i++) y[i] /= len;
}

int n, m, x;
Complex x1[maxn], x2[maxn];

int main()
{
    //scanf("%d%d", &n, &m);
    n = io.xint();
    m = io.xint();
//    REP1(i, 0, n) a[i] = io.xint();
//    REP1(i, 0, m) b[i] = io.xint();
    int len = 1;
    while(len < n * 2 || len < m * 2) len <<= 1;
    REP1(i, 0, n){
        x = io.xint();
        x1[i] = Complex(x, 0);
    }
    REP1(i, n, len) x1[i] = Complex(0, 0);
    REP1(i, 0, m){
        x = io.xint();
        x2[i] = Complex(x, 0);
    }
    REP1(i, m, len) x2[i] = Complex(0, 0);
    fft(x1, len, 1);
    fft(x2, len, 1);
    REP1(i, 0, len) x1[i] = x1[i] * x2[i];
    fft(x1, len, -1);
    len = n + m - 1;
    for(int i = 0; i < len - 1; i++) printf("%d ", (int) (x1[i].real() + 0.5));
    printf("%d\n", (int) (x1[len - 1].real() + 0.5));
    return 0;
}