版权声明:Andy https://blog.csdn.net/Alibaba_lhl/article/details/82022579
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#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int MAXN = 4e5 + 10;
// 复数结构体
struct Complex
{
double x, y; // 实部和虚部 x + yi
Complex(double _x = 0.0, double _y = 0.0)
{
x = _x;
y = _y;
}
Complex operator - (const Complex &b) const
{
return Complex(x - b.x, y - b.y);
}
Complex operator + (const Complex &b) const
{
return Complex(x + b.x, y + b.y);
}
Complex operator * (const Complex &b) const
{
return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
}
};
// 进行FFT和IFFT前的反转变换
// 位置i和(i二进制反转后的位置)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]);
}
// 交换护卫小标反转的元素,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 /= 2;
}
if (j < k)
{
j += k;
}
}
return ;
}
// 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].x /= len;
}
}
}
// 求卷积
// 用于大数乘法
void conv(Complex a[], Complex b[], int ans[], int len)
{
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);
// 精度复原
for (int i = 0; i < len; i++)
{
ans[i] = a[i].x + 0.5;
}
}
// 进制恢复
// 用语大数乘法
void turn(int ans[], int len, int unit)
{
for (int i = 0; i < len; i++)
{
ans[i + 1] += ans[i] / unit;
ans[i] %= unit;
}
}
char str_1[MAXN], str_2[MAXN];
Complex za[MAXN], zb[MAXN];
int ans[MAXN];
int len;
void init(char str_1[], char str_2[])
{
int len_1 = (int)strlen(str_1);
int len_2 = (int)strlen(str_2);
len = 1;
while (len < 2 * len_1 || len < 2 * len_2)
{
len <<= 1;
}
int i = 0;
for (; i < len_1; i++)
{
za[i].x = str_1[len_1 - i - 1] - '0';
za[i].y = 0.0;
}
while (i < len)
{
za[i].x = za[i].y = 0.0;
i++;
}
for (i = 0; i < len_2; i++)
{
zb[i].x = str_2[len_2 - i - 1] - '0';
zb[i].y = 0.0;
}
while (i < len)
{
zb[i].x = zb[i].y = 0.0;
i++;
}
return ;
}
void solve()
{
conv(za, zb, ans, len);
turn(ans, len, 10);
while (ans[len - 1] == 0)
{
len--;
}
for (int i = len - 1; i >= 0; i--)
{
printf("%d", ans[i]);
}
printf("\n");
return ;
}
int main()
{
while (~scanf("%s%s", str_1, str_2))
{
init(str_1, str_2);
solve();
}
return 0;
}