HDU 3924 extend-tree [a good sub-governance question]

A multi-school joint question.

Given is the one-to-one correspondence between the tree shape and specific numbers.

Given a number N (1 <= n <= 10 ^ 14), the structure of the corresponding tree must be output.

There was something I didn't want to understand during the training, but I didn't do it.

Through preprocessing, a maximum of 20 nodes can be obtained, which can reach 10 ^ 14 bits.

Preprocessing and finding

Num [I] indicates the number of tree types containing I nodes.

Total [I] indicates the total number of different types of I nodes.

Rig [I] indicates the number of data types that the neutron tree and the right subtree can represent for each node.

Then, the output process function DFS (int l, LLn) indicates the N class Tree Containing L nodes.

In the middle, you need to find the knots in the left subtree, the neutron tree, and the right subtree, and the number of course trees to be output respectively.

This step should be carefully calculated based on the preprocessing above.

// joy   hdu3924#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;typedef __int64 LL;const int MAX=30;const int LIM=21;////////////////////////////////////LL num[MAX],rig[MAX],total[MAX];int b[4];LL gs(int L){LL res=0;for(int i=0;i<=L;i++){for(int j=i;j<=L;j++){int k=L-i-j;if(j>k) continue;b[0]=i;b[1]=j;b[2]=k;do{res+=num[b[0]]*num[b[1]]*num[b[2]];}while(next_permutation(b,b+3));}}return res;}LL getR(int L){LL res=0;for(int i=0;i<=L;i++){res+=num[i]*num[L-i];}return res;}void init(){num[0]=num[1]=1;for(int i=2;i<=LIM;i++) num[i]=gs(i-1);total[0]=0;for(int i=1;i<=LIM;i++) total[i]=total[i-1]+num[i];rig[0]=1;for(int i=1;i<=LIM;i++) rig[i]=getR(i);}//////////////////////////int getNN(LL N){for(int i=1;i<=LIM;i++) if(total[i]>=N) return i;}void dfs(int L,LL N){if(L==1) {printf("X");return ;}int x=L-1;int ll,mm,rr;for(ll=0;N>num[ll]*rig[x-ll];ll++) N-=num[ll]*rig[x-ll];if(ll){printf("(");dfs(ll,(N-1)/rig[x-ll]+1);printf(")");}N%=rig[x-ll];if(N==0) N=rig[x-ll];x-=ll;for(mm=0;N>num[mm]*num[x-mm];mm++) N-=num[mm]*num[x-mm];if(mm){printf("(");dfs(mm,(N-1)/num[x-mm]+1);printf(")");}N%=num[x-mm];if(N==0) N=num[x-mm];x-=mm;if(x){printf("(");dfs(x,N);printf(")");}printf("X");}int main(){init();int T;scanf("%d",&T);int CN=0;while(T--){LL N;scanf("%I64d",&N);int nn=getNN(N);printf("Case #%d: ",++CN);dfs(nn,N-total[nn-1]);puts("");}return 0;}