Trees
Memory limit: 32 MB
Trees occur very often in computer science. As opposed to trees in nature, computer science trees "grow upside down"; the root is up and the leaves are down.
A tree consists of elements named nodes. A root is one of the nodes. Each node 
(except for the root) has its (exactly one) father 
. If the node 
is the father of 
, then 
is a son of 
. Nodes that have no sons are called leaves. Sons of the node 
, their sons, sons of their sons, and so on, are called descendants of the node 
. Every node - except for the root - is a descendant of the root.
Each node has a level number assigned to it. The level number of the root is 
, and the level number of sons is greater by 
then that of their father.
A tree is a complete binary tree if and only if each node has exactly two or zero sons. In binary trees sons are named left and right.
The following picture shows an example of a complete binary tree. The nodes of that tree are numbered in a special order called preorder. In this order the root has the number 
, a father precedes its sons, and the left son and any its descendant have smaller numbers than the right son and every its descendant.
There are many written representations of complete binary trees having such numbering of nodes. Three ones follow.
Genealogical representation.
It is a sequence of numbers. The first element of the sequence equals
(zero), and for 
, the 
-th element of
the sequence is the number of the father of the node 
.
Bracket representation.
Each node corresponds to a string composed of brackets. Leaves correspond
to
. Each other node 
corresponds to a string 
, where 
and 
denote the strings that left and
right sons of 
respectively correspond to. The string the root
corresponds to is the bracket representation of the tree.
Level representation.
It is a sequence of level numbers of successive tree leaves (according to
the
assumed numbering).
The tree in the picture may be described as follows:
| Genealogical representation | 0 1 2 2 4 4 1 7 7 |
| Bracket representation | ((()(()()))(()())) |
| Level representation | 2 3 3 2 2 |
Task
Write a program that reads from the standard input a sequence of numbers and
examines whether it is the level representation of a complete binary tree. If
not, the program writes one word NIE ("
") in the standard
output. If so, the program finds two other representations of this tree
(genealogical and bracket ones), and writes them in the standard output.
Input
In the first line of the standard input there is a positive number of the
sequence elements (not greater than 2500). In the second line there are
successive elements of the sequence separated by single spaces.
The numbers in the standard input are written correctly. Your program need not
verify that.
Output
The standard output should contain:
- either only one word NIE,
- or in the first line - the consecutive elements of the genealogical
representation, separated by single spaces;
in the second line - the bracket representation, i.e. a sequence of left and
right brackets with no spaces between them.
Examples
For the input data:
5
2 2 3 3 2
the correct result is:
0 1 2 2 1 5 6 6 5
((()())((()())()))
For the input data:
4
1 2 2 3
the correct result is:
NIE
Task author: Wojciech Rytter.
这个题目不难，用贪心的方法递归就可以了，注意细节。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N=;
int st[N],fa[N],np,n,rt;
int ch[N][],tot,flag=;
void DFS(int &x,int d){
x=++tot;
if(np<=n&&d==st[np]){
np+=;return;
}
if(d<st[np])DFS(ch[x][],d+);
if(np<=n&&d<st[np])DFS(ch[x][],d+);
else flag=;
} void Print1(int x,int f){
if(!x)return;
cout<<f<<" ";
Print1(ch[x][],x);
Print1(ch[x][],x);
} void Print2(int x){
if(!x)return;
cout<<"(";
Print2(ch[x][]);
Print2(ch[x][]);
cout<<")";
} int main(){
ios::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
cin>>n;np=;
for(int i=;i<=n;i++)cin>>st[i];
DFS(rt,);
if(np<=n||!flag){
cout<<"NIE\n";
return ;
}
Print1(rt,);
cout<<"\n";
Print2(rt);
cout<<"\n";
return ;
}
Ones And Zeros
Memory limit: 32 MB
Certain positive integers have their decimal representation consisting only of ones and zeros, and having at least one digit one, e.g. 101. If a positive integer has not such a property, one can try to multiply it by some positive integer to find out whether the product has this property.
Task
Write a program which:
- reads from the standard input positive integers
not greater than 
, - for each integer read computes a correct answer,
- writes the answer to the standard output.
The answer is either a positive multiple of 
whose decimal representation consists of at most 100 (a hundred) digits, only zeros or ones, or the word BRAK ("absence"), if there is no such multiple.
Input
The standard input contains in the first line a positive integer 
. In consecutive lines there is a sequence of 
numbers in the range of [
], one number per line. The numbers in the standard input are written correctly, and your program need not verify that.
Output
Each line of the standard output, starting with the first, should contain:
- either only one word BRAK,
- or exactly one positive integer being a multiple of a successive number given in the input; each multiple must be a number composed only of digits
and 
, and has to be written with no spaces between the digits.
The answers are to be written in standard output in the same order as the corresponding numbers in standard input.
Example
For the input data:
6
17
11011
17
999
125
173
the correct result is:
11101
11011
11101
111111111111111111111111111
1000
1011001101
Task author: Andrzej Walat.
这道题目不知是不会出现BARK还是数据没有BARK，自己目前不会证明。
枚举位DP就好了。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N=;
int n,k,tmp,pre[N],val[N],num[N],st[N],top,rnd;
bool vis[N];
void Print(){
int p=,Mx=;
do{
num[val[p]]=;
Mx=max(Mx,val[p]);
p=pre[p];
}while(p);
for(int i=Mx;i>=;i--)
cout<<num[i];
cout<<"\n";
}
int main(){
ios::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
cin>>n;
while(n--){
cin>>k;
memset(vis,,sizeof(vis));
memset(num,,sizeof(num));
if(k==){cout<<<<"\n";continue;}
tmp=%k;vis[]=true;
val[]=;pre[]=;rnd=;
while(true){
int flag=;rnd+=;
for(int i=k-;i>=;i--)
if(vis[i]==true||!i){
if(!vis[(i+tmp)%k]){
pre[(i+tmp)%k]=i;
val[(i+tmp)%k]=rnd;
}
st[++top]=(i+tmp)%k;
if((i+tmp)%k==){flag=;break;}
}
while(top)vis[st[top--]]=true;
tmp=tmp*%k;
if(flag){Print();break;}
}
}
return ;
}