HDU 1402 A * B Problem Plus（高精度乘法FFT模板题）

题目链接：点这里

题目大意：A*B=？？？

解题思路：FFT模板，别问为什么，问就是不会（脑部滑稽）

#include  
using namespace std;  
typedef long long LL;  
const double PI=acos(-1.0),eps=1e-8;  
#define L(x) (1 << (x))  
const int maxn = 200000;  
double ax[maxn], ay[maxn], bx[maxn], by[maxn];  
int revv(int x, int bits)  
{  
    int ret = 0;  
    for (int i = 0; i < bits; i++)  
    {  
        ret <<= 1;  
        ret |= x & 1;  
        x >>= 1;  
    }  
    return ret;  
}  
void fft(double * a, double * b, int n, bool rev)  
{  
    int bits = 0;  
    while (1 << bits < n) ++bits;  
    for (int i = 0; i < n; i++)  
    {  
        int j = revv(i, bits);  
        if (i < j)  
            swap(a[i], a[j]), swap(b[i], b[j]);  
    }  
    for (int len = 2; len <= n; len <<= 1)  
    {  
        int half = len >> 1;  
        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);  
        if (rev) wmy = -wmy;  
        for (int i = 0; i < n; i += len)  
        {  
            double wx = 1, wy = 0;  
            for (int j = 0; j < half; j++)  
            {  
                double cx = a[i + j], cy = b[i + j];  
                double dx = a[i + j + half], dy = b[i + j + half];  
                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;  
                a[i + j] = cx + ex, b[i + j] = cy + ey;  
                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;  
                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;
                wx = wnx, wy = wny;  
            }  
        }  
    }  
    if (rev)  
    {  
        for (int i = 0; i < n; i++)  
            a[i] /= n, b[i] /= n;  
    }  
}  
  
int solve(int a[],int na,int b[],int nb,int ans[])  
{  
    int len = max(na, nb), ln;  
    for(ln=0; L(ln)= na) ax[i] = 0, ay[i] =0;  
        else ax[i] = a[i], ay[i] = 0;  
    }  
    fft(ax, ay, len, 0);  
    for (int i = 0; i < len; ++i)  
    {  
        if (i >= nb) bx[i] = 0, by[i] = 0;  
        else bx[i] = b[i], by[i] = 0;  
    }  
    fft(bx, by, len, 0);  
    for (int i = 0; i < len; ++i)  
    {  
        double cx = ax[i] * bx[i] - ay[i] * by[i];  
        double cy = ax[i] * by[i] + ay[i] * bx[i];  
        ax[i] = cx, ay[i] = cy;  
    }  
    fft(ax, ay, len, 1);  
    for (int i = 0; i < len; ++i)  
        ans[i] = (int)(ax[i] + 0.5);  
    return len;  
}  
  
char s[maxn],t[maxn];  
int a[maxn],b[maxn],c[maxn],lena,lenb;  
  
int main()  
{  
    while(~scanf("%s%s",s,t))  
    {  
        memset(c,0,sizeof c);  
        lena=strlen(s); lenb=strlen(t);  
        if(lena==1&&s[0]=='0') {printf("0\n"); continue;}//其中一个为0；  
        if(lenb==1&&t[0]=='0') {printf("0\n"); continue;}  
        for(int i=lena-1;i>=0;i--) a[lena-1-i]=s[i]-'0';  
        for(int i=lenb-1;i>=0;i--) b[lenb-1-i]=t[i]-'0';  
        solve(a,lena,b,lenb,c);//核心a*b存于c[]中；  
        int k=0;
        for(int i=0;i<=100000;i++){int p=c[i]+k;c[i]=p%10,k=p/10; }//进位  
        int len; for(int i=0;i<=100000;i++) if(c[i]!=0) len=i;//去前导0；  
        for(int i=len;i>=0;i--) printf("%d",c[i]); printf("\n");  
    }  
    return 0;  
}