FFT 变化

 

hdu 4609 3-idiots FFT

分类： 模版 数学 stl 2013-09-30 19:01  81人阅读  评论(0)  收藏  举报

 /*
 hdu 4609 3-idiots FFT
 http://www.cnblogs.com/kuangbin/archive/2013/07/24/3210565.html
 */
 #pragma warning(disable : 4786)
 #pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include <string>
 #include <cstring>
 #include <cstdlib>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <functional>
 #include <map>
 #include <set>
 #include <vector>
 #include <stack>
 #include <queue>//priority_queue
 #include <bitset>
 #include <complex>
 #include <utility>
 /*
 **make_pair()
 **/
 using namespace std;
 const double eps=1e-8;
 const double PI=acos(-1.0);
 const int inf=0x7fffffff;
 template<typename T> inline T MIN(T a,T b){return a<b?a:b;}
 template<typename T> inline T MAX(T a,T b){return a>b?a:b;}
 template<typename T> inline T SQR(T a){return a*a;}
 inline int dbcmp(double a){return a>eps?(1):(a<(-eps)?(-1):0);}

 typedef __int64 LL;
 int n;
 const int size=400040;
 int data[size/4];
 LL num[size],sum[size];
 complex<double> x1[size];



 void change(complex<double> 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]);
         k = len/2;
         while( j >= k)
         {
             j -= k;
             k /= 2;
         }
         if(j < k)j += k;
     }
 }
 void fft(complex<double> y[],int len,int on)
 {
     change(y,len);
     for(int h = 2;h <= len;h <<= 1)
     {
         complex<double> wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
         for(int j = 0;j < len;j += h)
         {
             complex<double> w(1,0);
             for(int k = j;k < j+h/2;k++)
             {
                 complex<double> u = y[k];
                 complex<double> 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].real(y[i].real()/len);
 }
 int main() {
     // your code goes here
     int t,i;
     cin>>t;
     while(t--)
     {
         cin>>n;
         memset(num,0,sizeof(num));
         for(i=0;i<n;++i)
         {
             cin>>data[i];
             num[data[i]]++;
         }
         sort(data,data+n);
         int len1=data[n-1]+1;
         int len=1;
         while(len<2*len1) len<<=1;
         for(i=0;i<len1;++i)
         {
             x1[i]=complex<double>(num[i],0);
         }
         for(;i<len;++i)
         {
             x1[i]=complex<double>(0,0);
         }
         fft(x1,len,1);
         for(i=0;i<len;++i)
         {
             x1[i]=x1[i]*x1[i];
         }
         fft(x1,len,-1);

         for(i=0;i<len;++i)
         {
             num[i]=(LL)(x1[i].real()+0.5);
         }

         len=2*data[n-1];
         for(i=0;i<n;++i)
         {
             num[data[i]+data[i]]--;
         }
         for(i=1;i<=len;++i)
         {
             num[i]/=2;
         }
         sum[0]=0;
         for(i=1;i<=len;++i)
         {
             sum[i]=sum[i-1]+num[i];
         }
         LL cnt=0;
         for(i=0;i<n;++i)
         {
             cnt+=sum[len]-sum[data[i]];
             //减掉一个取大一个取小的情况，违背了data[i]是最长边的规则
             cnt-=(LL)(n-1-i)*i;
             //减掉一个取本身，一个取其他数的情况，否则data[i]就被取了两次
             cnt-=(LL)(n-1);
             //减掉取两个比他大的数的情况，违背了data[i]是最长边的规则
             cnt-=(LL)(n-1-i)*(n-2-i)/2;
         }
         LL tot = (LL)n*(n-1)*(n-2)/6;
         printf("%.7lf\n",(double)cnt/tot);
     }
     return 0;
 }

 

hdu 1402 A * B Problem Plus FFT

分类： 模版 stl 数学 2013-09-30 19:02  82人阅读  评论(0)  收藏  举报

 /*
 hdu 1402 A * B Problem Plus FFT

 这是我的第二道FFT的题

 第一题是完全照着别人的代码敲出来的，也不明白是什么意思

 这个代码是在前一题的基础上改的

 做完这个题，我才有点儿感觉，原来FFT在这里就是加速大整数乘法而已

 像前一题，也是一个大整数乘法，然后去掉一些非法的情况

 */
 #pragma warning(disable : 4786)
 #pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include <string>
 #include <cstring>
 #include <cstdlib>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <functional>
 #include <map>
 #include <set>
 #include <vector>
 #include <stack>
 #include <queue>//priority_queue
 #include <bitset>
 #include <complex>
 #include <utility>
 /*
 **make_pair()
 **/
 using namespace std;
 const double eps=1e-8;
 const double PI=acos(-1.0);
 const int inf=0x7fffffff;
 template<typename T> inline T MIN(T a,T b){return a<b?a:b;}
 template<typename T> inline T MAX(T a,T b){return a>b?a:b;}
 template<typename T> inline T SQR(T a){return a*a;}
 inline int dbcmp(double a){return a>eps?(1):(a<(-eps)?(-1):0);}

 typedef __int64 LL;
 int n;
 const int size=400040;
 int data[size/4];
 int sum[size];
 complex<double> x1[size],num1[size],num2[size];



 void change(complex<double> 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]);
         k = len/2;
         while( j >= k)
         {
             j -= k;
             k /= 2;
         }
         if(j < k)j += k;
     }
 }
 void fft(complex<double> y[],int len,int on)
 {
     change(y,len);
     for(int h = 2;h <= len;h <<= 1)
     {
         complex<double> wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
         for(int j = 0;j < len;j += h)
         {
             complex<double> w(1,0);
             for(int k = j;k < j+h/2;k++)
             {
                 complex<double> u = y[k];
                 complex<double> 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].real(y[i].real()/len);
 }
 char s1[size];
 char s2[size];
 int main() {
     // your code goes here
     int t,i,j;
     int len,len1,len2;
     //cin>>t;
     while(gets(s1))
     {
         gets(s2);
         memset(num1,0,sizeof(num1));
         memset(num2,0,sizeof(num2));
         len1=strlen(s1);
         len2=strlen(s2);
         //printf("%d %d",len1,len2);
         len=1;
         while(len<2*len1||len<len2*2) len<<=1;
         for(i=len1-1,j=0;i>=0;--i,++j)
         {
             num1[j]=complex<double>(s1[i]-'0',0);
         }
         while(j<=len)
         {
             num1[j]=complex<double>(0,0);
             ++j;
         }
         for(i=len2-1,j=0;i>=0;--i,++j)
         {
             num2[j]=complex<double>(s2[i]-'0',0);
         }
         while(j<=len)
         {
             num2[j]=complex<double>(0,0);
             ++j;
         }
         fft(num1,len,1);
         fft(num2,len,1);
         for(i=0;i<len;++i)
         {
             num1[i]=num1[i]*num2[i];
         }
         fft(num1,len,-1);

         for(i=0;i<len;++i)
         {
             sum[i]=(int)(num1[i].real()+0.5);
         }
         for(i=0;i<len;++i)
         {
             sum[i+1]+=sum[i]/10;
             sum[i]=sum[i]%10;
         }
         len=len1+len2-1;
         while(sum[len]<=0&&len>0) --len;
         for(;len>=0;--len)
         {
             printf("%c",sum[len]+'0');
         }
         printf("\n");
     }
     return 0;
 }