# bzoj2179 FFT快速傅立叶

## 2179: FFT快速傅立叶

Time Limit: 10 Sec  Memory Limit: 259 MB
Submit: 4155  Solved: 2222
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## Sample Output

12

n<=60000

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
const int maxn = 2e5+5;
const double pai = acos(-1.0);

int len1,len2,len,res[maxn];
char s1[maxn],s2[maxn];
struct node
{
double real, imag;
node(double real = 0.0, double imag = 0.0)
{
this->real = real, this->imag = imag;
}
node operator - (const node&elem) const
{
return node(this->real - elem.real, this->imag - elem.imag);
}
node operator + (const node&elem) const
{
return node(this->real + elem.real, this->imag + elem.imag);
}
node operator * (const node&elem) const
{
return node(this->real * elem.real - this->imag * elem.imag, this->real * elem.imag + this->imag * elem.real);
}
void set(double real = 0.0, double imag = 0.0)
{
this->real = real, this->imag = imag;
}
} A[maxn],B[maxn];

void pre()
{
len1 = strlen(s1),len2 = strlen(s2);
int maxx = max(len1,len2);
len = 1;
while (len < (maxx << 1))
len <<= 1;
for (int i = 0; i < len1; i++)
A[i].set(s1[len1 - i - 1] - '0',0);
for (int i = 0; i < len2; i++)
B[i].set(s2[len2 - i - 1] - '0',0);
for (int i = len1; i < len; i++)
A[i].set();
for (int i = len2; i < len; i++)
B[i].set();
}

void Swap(node &a,node &b)
{
node temp = a;
a = b;
b = temp;
}

void zhuan(node y[])
{
for (int i = 1, j = len >> 1,k; i < len - 1; i++)
{
if (i < j)
Swap(y[i],y[j]);
k = len >> 1;
while (j >= k)
{
j -= k;
k >>= 1;
}
if (j < k)
j += k;
}
}

void FFT(node y[],int op)
{
zhuan(y);
for (int h = 2; h <= len; h <<= 1)
{
node temp(cos(op * 2 * pai / h),sin(op * 2 * pai / h));
for (int i = 0; i < len; i += h)
{
node W(1,0);
for (int j = i; j < i + h / 2; j++)
{
node u = y[j];
node t = W * y[j + h / 2];
y[j] = u + t;
y[j + h / 2] = u - t;
W = W * temp;
}
}
}
if (op == -1)
for (int i = 0; i < len; i++)
y[i].real /= len;
}

void solve(node *A,node *B)
{
FFT(A,1);
FFT(B,1);
for (int i = 0; i < len; i++)
A[i] = A[i] * B[i];
FFT(A,-1);
for (int i = 0; i < len; i++)
res[i] = (int)(A[i].real + 0.5);
}

void calc(int *a)
{
for (int i = 0; i < len; i++)
{
res[i + 1] += res[i] / 10;
res[i] %= 10;
}
while (--len && res[len] == 0);
}

void print(int *a)
{
for (int i = len; i >= 0; i--)
putchar(a[i] + '0');
printf("\n");
}

int main()
{
int nn;
scanf("%d",&nn);
getchar();
gets(s1);
gets(s2);
pre();
solve(A,B);
calc(res);
print(res);

return 0;
}

posted @ 2018-03-24 14:50  zbtrs  阅读(177)  评论(0编辑  收藏  举报