# BZOJ-2179 FFT快速傅立叶

FFT模版题。。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cctype>
#include <complex>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define down(i, l, r) for(int i=l; i>=r; i--)
#define cd complex <double>
#define PI acos(0.0)*2.0
#define ll long long
#define base 10000
#define maxn 50009
using namespace std;
{
int x=0, f=1; char ch=getchar();
while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();
return x*f;
}
cd a[maxn], b[maxn], c[maxn], A[maxn];
ll ans[maxn];
int n, m, len, n1[maxn], n2[maxn], s[maxn], ansn=0;
char num[maxn];
void fft(cd *a, bool flag)
{
rep(i, 0, n-1) s[i]=0;
for(int i=1, j=n; i<n; i*=2, j/=2) rep(h, j/2, j-1) s[h]+=i;
for(int i=1; i<n; i*=2) rep(j, 0, i-1) s[j+i]+=s[j];
rep(i, 0, n-1) A[i]=a[s[i]];
double pi=flag?PI:-PI;
for(int step=1; step<n; step*=2)
{
cd e=exp(cd(0, 2.0*pi/double(step*2))), w=cd(1, 0);
for(int pos=0; pos<step; ++pos, w*=e)
for(int i=pos; i<n; i+=step*2)
{
cd ret=A[i], rec=w*A[i+step];
A[i]=ret+rec, A[i+step]=ret-rec;
}
}
if (!flag) rep(i, 0, n-1) A[i]/=n;
rep(i, 0, n-1) a[i]=A[i];
}
int main()
{
scanf("%s", num);
rep(i, 0, m-1) n1[(m-1-i)/4]=n1[(m-1-i)/4]*10+num[i]-'0';
scanf("%s", num);
rep(i, 0, m-1) n2[(m-1-i)/4]=n2[(m-1-i)/4]*10+num[i]-'0';
n=1; m=len*2; while (n<m) n*=2;
rep(i, 0, n-1) a[i]=cd(n1[i], 0); fft(a, true);
rep(i, 0, n-1) b[i]=cd(n2[i], 0); fft(b, true);
rep(i, 0, n-1) c[i]=a[i]*b[i]; fft(c, false);
rep(i, 0, n-1) ans[i]=(ll)(c[i].real()+0.5);
rep(i, 0, n-1) ans[i+1]+=ans[i]/base, ans[i]%=base;
ansn=n; while (ansn>0 && !ans[ansn]) ansn--;
printf("%lld", ans[ansn]);
down(i, ansn-1, 0)
if (ans[i]>=1000) printf("%lld", ans[i]);
else if (ans[i]>=100) printf("0%lld", ans[i]);
else if (ans[i]>=10) printf("00%lld", ans[i]);
else printf("000%lld", ans[i]);
printf("\n");
return 0;
}
posted @ 2015-05-05 16:00  NanoApe  阅读(139)  评论(0编辑  收藏  举报