HDU 1402 A * B Problem Plus FFT入门题

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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;}
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