hdu 1402（FFT）

数学基础不够QAQ，留着坑以后有水平了再来做。
*以下代码的原形来自kuangbin的博客，本蒟蒻做了一些小小的优化。
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#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;const double PI = acos(-1.0);struct complex{//复数结构体    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);    }};/* * 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){    int i,j,k;    for(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        k = len/2;        while( j >= k) j -= k, k >>= 1;        if(j < k) j += k;    }}/* * 做FFT * len必须为2^k形式， * on==1时是DFT，on==-1时是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 = 200010;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),len = 1;        while(len < len1*2 || len < len2*2)len<<=1;        for(int i = 0;i < len1;i++)            x1[i] = complex(str1[len1-1-i]-'0',0);        for(int i = len1;i < len;i++)            x1[i] = complex(0,0);        for(int i = 0;i < len2;i++)            x2[i] = complex(str2[len2-1-i]-'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("%c",sum[i]+'0');        puts("");    }    return 0;}