HDU 1402 A * B Problem Plus FFT入门题

最新推荐文章于 2020-01-13 20:32:17 发布

77458

最新推荐文章于 2020-01-13 20:32:17 发布

阅读量783

点赞数 1

CC 4.0 BY-SA版权

分类专栏： ACM-FFT

本文链接：https://blog.youkuaiyun.com/qq_18661257/article/details/53048043

ACM-FFT 专栏收录该内容

1 篇文章

订阅专栏

本文深入浅出地介绍了快速傅立叶变换（FFT）的基本原理及其在多项式乘法中的应用，并提供了一段完整的C++代码实现，展示了如何通过FFT高效地计算两个多项式的系数。

FFT的理解

大牛的博客 $1$ ：Ichimei
大牛的博客 $2$ ：sdj222555

简单的理解

就是对两个多项式乘法求系数

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;


const int MAXN = 2e5 + 5;
const double PI = acos(-1.0);

struct complex {
    double r, i;
    complex(double _r = 0.0, double _i = 0.0) {
        r = _r;
        i = _i;
    }
    complex operator +(const complex &b) {
        return complex(r + b.r, i + b.i);
    }
    complex operator -(const complex &b) {
        return complex(r - b.r, i - b.i);
    }
    complex operator *(const complex &b) {
        return complex(r * b.r - i * b.i, r * b.i + i * b.r);
    }
};

void change(complex y[], int len) {
    int i, j, k;
    for(i = 1, j = len / 2; i < len - 1; i++) {
        if(i < j)swap(y[i], y[j]);
        k = len / 2;
        while( j >= k) {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}

void fft(complex y[], int len, int on) {
    change(y, len);
    for(int h = 2; h <= len; h <<= 1) {
        complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        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(on == -1)
        for(int i = 0; i < len; i++)
            y[i].r /= len;
}

char A[MAXN], B[MAXN];
int ans[MAXN];
complex x[MAXN], y[MAXN];

int main() {
    while(~scanf("%s%s", A, B)) {
        int lena = 0, len1 = strlen(A);
        int lenb = 0, len2 = strlen(B);
        while(1 << lena < len1) lena ++;
        while(1 << lenb < len2) lenb ++;
        int len = 1 << max(lena, lenb) + 1;
        for(int i = 0; i < len; i ++) {
            if(i < len1) x[i] = complex(A[len1 - i - 1] - '0', 0);
            else x[i] = complex(0, 0);
            if(i < len2) y[i] = complex(B[len2 - i - 1] - '0', 0);
            else y[i] = complex(0, 0);
        }
        fft(x, len, 1);
        fft(y, len, 1);
        for(int i = 0; i < len; i ++) {
            x[i] = x[i] * y[i];
        }
        fft(x, len, -1);
        for(int i = 0; i < len; i ++) {
            ans[i] = (int)(x[i].r + 0.5);
        }
        for(int i = 0; i < len; i ++) {
            ans[i + 1] += ans[i] / 10;
            ans[i] %= 10;
        }
        int flag = 0;
        for(int i = len - 1; i >= 0; i --) {
            if(ans[i] > 0) {
                flag = 1;
                printf("%d", ans[i]);
                continue;
            }
            if(flag || i == 0) printf("0");
        }
        printf("\n");
    }

    return 0;
}