FFT水题记录

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hdu1402

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
using namespace std;
const int maxn = 100010;

const double PI = acos(-1.0);
//复数结构体
struct complex
{
    double r,i;
    complex(double _r = 0.0,double _i = 0.0)
    {
        r = _r; i = _i;
    }
    complex operator +(const complex &b)
    {
        return complex(r+b.r,i+b.i);
    }
    complex operator -(const complex &b)
    {
        return complex(r-b.r,i-b.i);
    }
    complex operator *(const complex &b)
    {
        return complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）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]);
        //交换互为小标反转的元素，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;
    }
}
/*
 * 做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].r /= len;
}

complex a[maxn*2],b[maxn*2];
int ans[maxn*2];
int main()
{
    char s1[maxn],s2[maxn];
    while(~scanf("%s%s",s1,s2))
    {
        int len1 = strlen(s1);
        int len2 = strlen(s2);
        int len = 1;
        while(len<len1*2||len<len2*2)len<<=1;
        for(int i = 0;i<len1;i++)
            a[i] = complex(s1[len1-i-1]-'0',0);
        for(int i = len1;i<len;i++)a[i] = complex(0,0);
        for(int i = 0;i<len2;i++)
            b[i] = complex(s2[len2-i-1]-'0',0);
        for(int i = len2;i<len;i++)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].r+0.5);
        }
        for(int i = 0;i<len;i++)
        {
            ans[i+1] += ans[i]/10;
            ans[i] %= 10;
        }
        int p;
        for(int i = len-1;i>0;i--)
        {
            if(ans[i])
            {
                p = i;
                break;
            }
        }
        for(int i = p;i>=0;i--)printf("%d",ans[i]);
        printf("\n");
    }
}

hdu4609

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn = 400010;
const double PI = acos(-1.0);
//复数结构体
struct complex
{
    double r,i;
    complex(double _r = 0.0,double _i = 0.0)
    {
        r = _r; i = _i;
    }
    complex operator +(const complex &b)
    {
        return complex(r+b.r,i+b.i);
    }
    complex operator -(const complex &b)
    {
        return complex(r-b.r,i-b.i);
    }
    complex operator *(const complex &b)
    {
        return complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）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]);
        //交换互为小标反转的元素，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;
    }
}
/*
 * 做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].r /= len;
}
complex a[maxn];
int x[maxn/2];
ll sum[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        for(int i = 0;i<n;i++)
        {
            scanf("%d",&x[i]);
            a[x[i]].r++;
        }
        sort(x,x+n);
        int maxl = x[n-1]+1;  
        int len = 1;
        while(len<maxl*2)len<<=1;
        fft(a,len,1);
        for(int i = 0;i<len;i++)a[i] = a[i]*a[i];
        fft(a,len,-1);
        for(int i = 0;i<len;i++)sum[i] = (ll)(a[i].r+0.5);
        for(int i = 0;i<n;i++)sum[x[i]+x[i]]--;
        for(int i = 0;i<len;i++)sum[i] /= 2;
        for(int i = 1;i<len;i++)sum[i] = sum[i-1]+sum[i];
        ll tot = (ll)n*(n-1)*(n-2)/6;
        ll cnt = 0;
        for(int i = 2;i<n;i++)
        {
            cnt += sum[len-1]-sum[x[i]];
            cnt -= (ll)(n-1-i);
            cnt -= (ll)i;
            cnt -= (ll)i*(n-1-i);
            cnt -= (ll)(n-i-1)*(n-i-2)/2;
        }
        printf("%.7lf\n",(double)cnt/tot);
    }
}

bzoj3527

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn = 400100;
const double PI = acos(-1.0);
//复数结构体
struct complex
{
    double r,i;
    complex(double _r = 0.0,double _i = 0.0)
    {
        r = _r; i = _i;
    }
    complex operator +(const complex &b)
    {
        return complex(r+b.r,i+b.i);
    }
    complex operator -(const complex &b)
    {
        return complex(r-b.r,i-b.i);
    }
    complex operator *(const complex &b)
    {
        return complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）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]);
        //交换互为小标反转的元素，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;
    }
}
/*
 * 做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].r /= len;
}
complex g[maxn],f[maxn],h[maxn];
double q[maxn];
int main()
{   
    memset(f,0,sizeof(f));
    memset(h,0,sizeof(h));
    memset(g,0,sizeof(g));
    int n;
    scanf("%d",&n);
    for(int i = 0;i<n;i++)
    {
        scanf("%lf",&f[i].r);
        h[n-1-i].r = f[i].r;
        q[i] = f[i].r;
    }
    g[0] = 0;
    for(int i = 1;i<n;i++)
    {
        g[i].r = 1.0/(double)i/(double)i;
    }
    int len = 1;
    while(len<n*2)len<<=1;
    fft(g,len,1);
    fft(f,len,1);
    fft(h,len,1);
    for(int i = 0;i<len;i++)
    {
        f[i] = f[i]*g[i];
        h[i] = h[i]*g[i];
    }
    fft(f,len,-1);
    fft(h,len,-1);
    for(int i = 0;i<n;i++)
    {
        printf("%.8lf\n",(f[i].r - h[n-1-i].r));
    }
}