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;
}

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