解题报告：HDU_4093 Xavier is Learning to Count FFT

题目链接

题意：

给定一个长度为n的数列，选其中p个数相加，要求输出可以得到所有结果和得到方案数。

思路：

不写了。。。贴代码留恋。。。写的难受，一直MLE，还卡精度，这题出现在现场赛里。。有毒

#include<bits/stdc++.h>const long double PI = acos(-1.0);using namespace std;struct comple{   long double r , i;   comple(long double rr=0,long double ii=0){      r = rr; i = ii;   }comple operator + (const comple& a){      return comple(r+a.r,i+a.i);   }comple operator - (const comple& a){      return comple(r-a.r,i-a.i);   }comple operator * (const comple& a){      return comple(r*a.r-i*a.i , r*a.i+i*a.r);   }};int getLen(int x){   int res = 1;   while(res<x)res<<=1;   return res;}void brc(comple *a,int l){   for(int i=1,j=l/2;i<l-1;i++){      if(i<j)swap(a[i],a[j]);      int k = l/2;      while(j>=k){         j-=k;         k>>=1;      }if(j<k)j+=k;   }}void fft(comple *y,int l,int on){   brc(y,l);   comple u,t;   for(int h=2;h<=l;h<<=1){      comple wn( cos(on*2*PI/h),sin(on*2*PI/h) );      for(int j=0;j<l;j+=h){         comple w(1,0);         for(int k=j;k<j+h/2;k++){            u = y[k];            t = w*y[k+h/2];            y[k] = u+t;            y[k+h/2] = u-t;            w  = w*wn;         }      }   }if(on<0){      for(int i=0;i<l;i++){         y[i].r/=l;      }   }}const int MAXL = (1<<16);int n;int A[13005];comple F[12][MAXL];long long tmp[12][MAXL];void dfs(int now , int need ,const int& l){   if(now>need)return ;   if(now==1){      for(int i=0;i<13001;i++){         F[10][i<<2].r = F[1][i<<1].r = F[4][i*3].r = F[0][i].r = tmp[10][i<<2] = tmp[1][i<<1] = tmp[4][i*3] = tmp[0][i] = A[i];         tmp[11][i*5]=A[i];         if(now==need){            if(tmp[0][i])printf("%d: %I64d\n",i,tmp[0][i]);         }      }return dfs(now+1,need,l);   }   if(now==2){      fft(F[0],l,1);      for(int i=0;i<l;i++){         F[2][i] = F[0][i] * F[0][i];      }fft(F[2],l,-1);      for(int i=0;i<l;i++){         tmp[2][i] = (long long)(F[2][i].r+0.5);         F[2][i].i = 0;         tmp[2][i] = (tmp[2][i]+tmp[1][i])/2;         F[2][i].r = tmp[2][i];         if(now==need){            long long ans = tmp[2][i] - tmp[1][i];            if(ans)printf("%d: %I64d\n",i,ans);         }      }return dfs(now+1,need,l);   }if(now==3){      fft(F[2],l,1);fft(F[1],l,1);      for(int i=0;i<l;i++){         F[5][i] = F[0][i] * F[2][i];         F[3][i] = F[0][i] * F[1][i];      }fft(F[3],l,-1);fft(F[5],l,-1);      for(int i=0;i<l;i++){         F[3][i].i = F[5][i].i = 0;         tmp[3][i] = (long long)(F[3][i].r+0.5);         F[3][i].r = tmp[3][i];         tmp[5][i] = (long long)(F[5][i].r+0.5);         tmp[5][i] = ( tmp[5][i] + tmp[3][i]  + tmp[4][i] ) / 3;         F[5][i].r = tmp[5][i];         if(now==need){            long long ans = (tmp[5][i] - tmp[3][i]);            if(ans)printf("%d: %I64d\n",i,ans);         }      }return dfs(now+1,need,l);   }if(now==4){      fft(F[4],l,1);fft(F[5],l,1);fft(F[10],l,1);      for(int i=0;i<l;i++){         F[6][i] = F[5][i] * F[0][i];         F[7][i] = F[1][i] * F[2][i];         F[8][i] = F[1][i] * F[1][i];         F[9][i] = F[4][i] * F[0][i];      }fft(F[6],l,-1);fft(F[7],l,-1);fft(F[8],l,-1);fft(F[9],l,-1);      for(int i=0;i<l;i++){         F[6][i].i = F[7][i].i = F[8][i].i = F[9][i].i = 0;         tmp[6][i] = (long long)(F[6][i].r+0.5);         tmp[7][i] = (long long)(F[7][i].r+0.5);         tmp[8][i] = (long long)(F[8][i].r+0.5);         tmp[8][i] = (tmp[8][i]+tmp[10][i])/2;         tmp[9][i] = (long long)(F[9][i].r+0.5);         tmp[6][i] = (tmp[6][i]+tmp[7][i]+tmp[9][i]+tmp[10][i])/4;         if(now==need){            long long ans = (  tmp[6][i] - tmp[7][i]   + tmp[8][i] - tmp[10][i] );            if(ans)printf("%d: %I64d\n",i,ans);         }         F[6][i].i = 0;         F[6][i].r = tmp[6][i];         F[8][i].r = tmp[8][i];      }return dfs(now+1,need,l);   }if(now==5){      fft(F[6],l,1);fft(F[8],l,1);      for(int i=0;i<l;i++){         F[6][i] = F[6][i] * F[0][i];         F[7][i] = F[1][i] * F[5][i];         F[9][i] = F[0][i] * F[8][i];         F[8][i] = F[4][i] * F[2][i];         F[3][i] = F[10][i] * F[0][i];      }fft(F[7],l,-1);fft(F[8],l,-1);      fft(F[9],l,-1);fft(F[6],l,-1);      fft(F[3],l,-1);      for(int i=0;i<l;i++){         tmp[6][i] = (long long)(F[6][i].r+0.5);         tmp[7][i] = (long long)(F[7][i].r+0.5);         tmp[8][i] = (long long)(F[8][i].r+0.5);         tmp[9][i] = (long long)(F[9][i].r+0.5);         tmp[3][i] = (long long)(F[3][i].r+0.5);         long long cnt = (tmp[6][i]+tmp[7][i]+tmp[8][i]+tmp[3][i]+tmp[11][i]);         tmp[6][i] = cnt/5;         if(now==need){            long long ans = tmp[6][i] - tmp[7][i] + tmp[9][i]-tmp[3][i];            if(ans)printf("%d: %I64d\n",i,ans);         }      }return dfs(now+1,need,l);   }}int main(){   int T,k,t=0;   scanf("%d",&T);   while(T--){      printf("Case #%d:\n",++t);      scanf("%d%d",&n,&k);      memset(A,0,sizeof(A));      memset(tmp,0,sizeof(tmp));      memset(F,0,sizeof(F));      int l = 0;      for(int i=0,x;i<n;i++){         scanf("%d",&x);         l = max(l,x);         A[x]++;      }l = l * k ;      l = getLen(l);      dfs(1,k,l);      printf("\n");   }return 0;}