【FFT求卷积】Problem D. Duel

【AC】

 #include <stdio.h>
 #include <iostream>
 #include <string.h>
 #include <algorithm>
 #include <math.h>
 using namespace std;
 typedef long long ll;
 const double PI = acos(-1.0);
 struct complex
 {
     double r,i;
     complex(double _r = ,double _i = )
     {
         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);
     }
 };
 void change(complex y[],int len)
 {
     int i,j,k;
     for(i = , j = len/;i < len-;i++)
     {
         if(i < j)swap(y[i],y[j]);
         k = len/;
         while( j >= k)
         {
             j -= k;
             k /= ;
         }
         if(j < k)j += k;
     }
 }
 void fft(complex y[],int len,int on)
 {
     change(y,len);
     for(int h = ;h <= len;h <<= )
     {
         complex wn(cos(-on**PI/h),sin(-on**PI/h));
         for(int j = ;j < len;j += h)
         {
             complex w(,);
             for(int k = j;k < j+h/;k++)
             {
                 complex u = y[k];
                 complex t = w*y[k+h/];
                 y[k] = u+t;
                 y[k+h/] = u-t;
                 w = w*wn;
             }
         }
     }
     if(on == -)
         for(int i = ;i < len;i++)
             y[i].r /= len;
 }

 const int MAXN = ;
 complex x1[MAXN];
 int a[MAXN/];
 long long num[MAXN];//100000*100000会超int
 char str[MAXN/];
 int main()
 {
     freopen("duel.in","r",stdin);
     freopen("duel.out","w",stdout);
     int n=;
     scanf("%s",str+);
     int ls=strlen(str+);
     for(int i=;i<=ls;i++)
     {
         if(str[i]=='')
         {
             a[n++]=i;
         }
     }
     memset(num,,sizeof(num));
     for(int i = ;i < n;i++)
     {
         num[a[i]]++;
     }
     sort(a,a+n);
     int len1 = a[n-]+;
     int len = ;
     while( len < *len1 )len <<= ;
     for(int i = ;i < len1;i++)
         x1[i] = complex(num[i],);
     for(int i = len1;i < len;i++)
         x1[i] = complex(,);
     fft(x1,len,);
     for(int i = ;i < len;i++)
         x1[i] = x1[i]*x1[i];
     fft(x1,len,-);
     for(int i = ;i < len;i++)
         num[i] = (long long)(x1[i].r+0.5);
     len = *a[n-];
     //减掉取两个相同的组合
     for(int i = ;i < n;i++)
         num[a[i]+a[i]]--;
     //选择的无序，除以2
     for(int i = ;i <= len;i++)
     {
         num[i]/=;
     }
     ll ans=;
     for(int i=;i<=len;i++)
     {
         if(i%!=) continue;
         if(str[i/]=='') continue;
         ans+=num[i];
     }
     cout<<ans<<endl;
     return ;
 }

http://www.cnblogs.com/kuangbin/archive/2013/07/24/3210565.html