wenbao与FFT

FFT入门模板

http://acm.hdu.edu.cn/showproblem.php?pid=1402

#include <iostream>
#include <string.h>
#include <cmath>
using namespace std;

const double PI = acos(-1.0);

//复数结构体
struct complex{
    double r, i;
    complex(double _r = 0.0, double _i = 0.0) {r = _r, i = _i;}
    complex operator + (const complex &a) { return complex(r+a.r, i+a.i); }
    complex operator - (const complex &a) { return complex(r-a.r, i-a.i); }
    complex operator * (const complex &a) { return complex(r*a.r-i*a.i, r*a.i+i*a.r); }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）互换
 * len必须去2的幂
 */
void change(complex y[], int len){
    for(int i = 1, j = len/2; i < len-1; ++i){
        if(i < j) swap(y[i], y[j]);
        //交换互为小标反转的元素，i<j保证交换一次
        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的
        int k = len/2;
        while(j >= k) j -= k, k /= 2;
        if(j < k) j += k;
    }
}
//on == 1 DFT on == -1 IDFT
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;
}

const int maxn = 2e5+10;
complex x1[maxn], x2[maxn];
char str1[maxn/2], str2[maxn/2];
int sum[maxn];

int main(){
    while(~scanf("%s%s", str1, str2)){
        int len1 = strlen(str1), len2 = strlen(str2);
        int len = 1;
        while(len < len1*2 || len < len2*2) len <<= 1;
        for(int i = 0; str1[i]; ++i) x1[i] = complex(str1[len1-i-1]-'0', 0);
        for(int i = len1; i < len; ++i) x1[i] = complex(0, 0);
        for(int i = 0; str2[i]; ++i) x2[i] = complex(str2[len2-i-1]-'0', 0);
        for(int i = len2; i < len; ++i) x2[i] = complex(0, 0);
        //求DFT
        fft(x1, len, 1);
        fft(x2, len, 1);
        for(int i = 0; i < len; ++i) x1[i] = x1[i] * x2[i];
        fft(x1, len, -1);
        for(int i = 0; i < len; ++i) sum[i] = (int)(x1[i].r+0.5);
        for(int i = 0; i < len; ++i){
            sum[i+1] += sum[i]/10;
            sum[i] %= 10;
        }
        len = len1 + len2 - 1;
        while(sum[len] <= 0 && len > 0) --len;
        for(int i = len; i >= 0; --i){
            printf("%d", sum[i]);
        }
        printf("
");
    }
    return 0;
}

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