HDU 1402 A * B (FFT)

题意：

思路：

代码：

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#define PB push_back
#define FT first
#define SD second
#define MP make_pair
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int>  P;
const int N = 2e5 + 10,MOD = 7+1e9;

const double PI = acos(-1.0);

struct Complex
{
double real, image;
Complex(double _real, double _image)
{
real = _real;
image = _image;
}
Complex(){}
};

Complex operator + (const Complex &c1, const Complex &c2)
{
return Complex(c1.real + c2.real, c1.image + c2.image);
}

Complex operator - (const Complex &c1, const Complex &c2)
{
return Complex(c1.real - c2.real, c1.image - c2.image);
}

Complex operator * (const Complex &c1, const Complex &c2)
{
return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}

int rev(int id, int len)
{
int ret = 0;
for(int i = 0; (1 << i) < len; i++)
{
ret <<= 1;
if(id & (1 << i)) ret |= 1;
}
return ret;
}
Complex ret
;
// 非递归形式，从递归的最后一层向上计算
void FFT(Complex* a, int n, int f)
{
for(int i = 0;i < n;++ i) ret[rev(i, n)] = a[i];
for(int s = 1; (1 << s) <= n;++ s)
{
int m = (1 << s);
Complex wm = Complex(cos(f*2*PI/m), sin(f*2*PI/m));
for(int k = 0;k < n;k += m)
{
Complex w = Complex(1, 0);
for(int j = 0;j < (m >> 1); ++j)
{
Complex t = w * ret[k + j + (m >> 1)];
Complex u = ret[k + j];
ret[k + j] = u + t;
ret[k + j + (m >> 1)] = u - t;
w = w * wm;
}
}
}
if(f == -1) for(int i = 0;i < n;++ i) ret[i].real /= n, ret[i].image /= n;
for(int i = 0;i < n; ++i) a[i] = ret[i];
}
char s1
, s2
;
Complex A
, B
;
int ans
;
int main()
{
// freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(~scanf("%s %s", s1, s2))
{
int sa = 0, sb = 0, l1 = strlen(s1), l2 = strlen(s2);
while((1<<sa) < l1) sa ++;
while((1<<sb) < l2) sb ++;
int len = 1 << (max(sa, sb) + 1);
for(int i = 0;i < len;i ++)
{
if(i < l1) A[i] = Complex(s1[l1 - i - 1] - '0', 0);
else A[i] = Complex(0, 0);
if(i < l2) B[i] = Complex(s2[l2 - i - 1] - '0', 0);
else B[i] = Complex(0, 0);
}
FFT(A, len, 1);
FFT(B, len, 1);
for(int i = 0;i < len;i ++) A[i] = A[i] * B[i];
FFT(A, len, -1);
for(int i = 0;i < len;i ++) ans[i] = int(A[i].real + 0.5);
for(int i = 0;i < len - 1;i ++) ans[i+1] += ans[i] / 10, ans[i] %= 10;
bool fg = 0;
for(int i = len - 1;i >= 0;i --)
{
if(ans[i]) printf("%d", ans[i]), fg = 1;
else if(fg || i == 0) printf("0");
}
printf('\n');
}
return 0;
}