# Two Two

### Problem Statement :

```Prof. Twotwo as the name suggests is very fond powers of 2. Moreover he also has special affinity to number 800. He is known for carrying quirky experiments on powers of 2.

One day he played a game in his class. He brought some number plates on each of which a digit from 0 to 9 is written. He made students stand in a row and gave a number plate to each of the student. Now turn by turn, he called for some students who are standing continuously in the row say from index i to index j (i<=j) and asked them to find their strength.

The strength of the group of students from i to j is defined as:

strength(i , j)
{
if a[i] = 0
return 0; //If first child has value 0 in the group, strength of group is zero
value = 0;
for k from i to j
value = value*10 + a[k]
return value;
}
Prof called for all possible combinations of i and j and noted down the strength of each group. Now being interested in powers of 2, he wants to find out how many strengths are powers of two. Now its your responsibility to get the answer for prof.

Input Format

First line contains number of test cases T
Next T line contains the numbers of number plates the students were having when standing in the row in the form of a string A.

Constraints

1 ≤ T ≤ 100
1 ≤ len(A) ≤ 105
0 ≤ A[i] ≤ 9

Output Format

Output the total number of strengths of the form 2x such that 0 ≤ x ≤ 800.```

### Solution :

```                            ```Solution in C :

In   C++ :

#include <iostream>
#include <cstdio>
#include <ctime>

using namespace std;

struct node
{
int e;
struct node *path;
};

int prodf,prodlen;
char bstr;
struct node *posStack;
int stackSize=0;

struct node* addEdge(struct node *start, int edge)
{
if (start->path[edge]==NULL)
{
struct node *tempNode = new node;
tempNode->e=0;
for(int i=0;i<10;++i)
tempNode->path[i]=NULL;
start->path[edge]=tempNode;
}
return (start->path[edge]);
}

void markEnd(struct node *vnode)
{
vnode->e=1;
}

int main()
{
clock_t t_start =clock();
prodlen=1;
prodf=1;
struct node *automaton = new struct node;
automaton->e=0;
for(int i=0;i<10;++i)
automaton->path[i]=NULL;
struct node *ppos=automaton;
markEnd(ppos);
for(int n=1;n<=800;++n)
{
int rem=0;
for(int i=0;i<prodlen;++i)
{
rem=(2*prodf[i]+rem);
prodf[i]=rem%10;
rem/=10;
}
if(rem!=0)
prodf[prodlen++]=rem;
ppos=automaton;
for(int i=prodlen-1;i>=0;--i)
{
}
markEnd(ppos);
}
clock_t t_end=clock();
//   cout<<(t_end+0.0-t_start)/CLOCKS_PER_SEC<<endl;

int T,k;
cin>>T;
while(T--)
{
long long ans=0;
stackSize=0;
scanf("%s",bstr);
for(int i=0,j;bstr[i]!='\0';++i)
{
k=bstr[i]-'0';
for(j=0;j<stackSize;)
{
if(posStack[j]->path[k]!=NULL)
{
posStack[j]=posStack[j]->path[k];
if(posStack[j]->e==1)
ans++;
++j;
}
else
posStack[j]=posStack[--stackSize];
}
if(automaton->path[k]!=NULL)
{
posStack[stackSize]=automaton->path[k];
if(posStack[stackSize]->e==1)
ans++;
++stackSize;
}
}
cout<<ans<<endl;
}

return 0;
}

In   Java  :

import java.io.*;
import java.math.BigInteger;
import java.util.ArrayList;

public class Solution {

static class Node {
Node[] edges = new Node;

int weight;
}

public static void solve(Input in, PrintWriter out) throws IOException {
Node root = new Node();
for (int pow = 0; pow <= 800; ++pow) {
String s = BigInteger.ONE.shiftLeft(pow).toString();
Node cur = root;
for (char c : s.toCharArray()) {
if (cur.edges[c - '0'] == null) {
cur.edges[c - '0'] = new Node();
}
cur = cur.edges[c - '0'];
}
cur.weight++;
}
ArrayList<Node> q = new ArrayList<Node>();
for (int it = 0; it < q.size(); ++it) {
Node n = q.get(it);
for (int i = 0; i < 10; ++i) {
Node n1 = n.edges[i];
if (n1 == null) {
continue;
}
}
} else {
}
}
}
for (Node n : q) {
}
}
for (Node n : q) {
for (int i = 0; i < 10; ++i) {
if (n.edges[i] == null) {
n.edges[i] = root;
} else {
}
}
}
}
int tests = in.nextInt();
for (int test = 0; test < tests; ++test) {
String s = in.next();
int ans = 0;
Node cur = root;
for (char c : s.toCharArray()) {
cur = cur.edges[c - '0'];
ans += cur.weight;
}
out.println(ans);
}
}

public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
out.close();
}

static class Input {
StringBuilder sb = new StringBuilder();

this.in = in;
}

public Input(String s) {
}

public String next() throws IOException {
sb.setLength(0);
while (true) {
if (c == -1) {
return null;
}
if (" \n\r\t".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
if (c == -1 || " \n\r\t".indexOf(c) != -1) {
break;
}
sb.append((char)c);
}
return sb.toString();
}

public int nextInt() throws IOException {
return Integer.parseInt(next());
}

public long nextLong() throws IOException {
return Long.parseLong(next());
}

public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}

In   C  :

#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int ret;
scanf("%d", &ret);
return ret;
}
int *ret = malloc(n * sizeof(int));
for (int i = 0; i < n; ++i) {
scanf("%d", ret + i);
}
return ret;
}
static int intcomp(const void *v1, const void *v2) {
return *(const int *)v1 - *(const int *)v2;
}
#define BASE 801
static char powers[BASE];
static int power_nums;
static void build_powers() {
powers = '1';
powers = '\0';
power_nums = 1;
for (int i = 1; i <= 800; ++i) {
int previous = power_nums[i - 1];
char *ppower  = powers[i - 1];
int start_pos = previous;
if (ppower >= '5') {
start_pos ++;
}
power_nums[i] = start_pos;
char *current = powers[i];
current[start_pos] = '\0';
int rem = 0;
for (int j = previous - 1; j >= 0; --j) {
int val = (ppower[j] - '0') * 2 + rem;
rem = val / 10;
current[start_pos - 1] = '0' + (val % 10);
start_pos --;
}
if (rem != 0) {
current = '0' + rem;
}
}
}
struct node {
struct node *childs;
char child_count;
char c;
char end;
};
static void init_node(struct node *n, char c, char end) {
n->c = c;
n->end = end;
}
static struct node *build_node() {
struct node * root = malloc(sizeof(struct node));
init_node(root, 0, -1);
for (int i = 0; i < BASE; ++i) {
struct node *n = root;
for (const char *iter = powers[i]; *iter; iter++) {
if (!n->childs) {
n->childs = malloc(10 * sizeof(struct node));
n->child_count = 10;
for (int i = 0; i < 10; ++i) {
n->childs[i].c = '0' + i;
}
}
n = n->childs + (*iter - '0');
}
n->end = 1;
}
return root;
}
static long long int count_appearance(const char *haystack, const char *needle) {
long long int result = 0;
while (1) {
if ((haystack = strstr(haystack, needle))) {
result ++;
haystack++;
} else {
return result;
}
}
}
static long long int solve(const char *buffer, struct node *root) {
long long result = 0;
for (const char *iter = buffer; *iter; ++iter) {
struct node *n = root;
for (const char *iter2 = iter; *iter2; ++iter2) {
if (!n->childs) {
break;
} else {
n = &n->childs[*iter2 - '0'];
if (n->end) {
result++;
}
}
}
}
return result;
}
int main(int argc, char *argv[]) {
build_powers();
struct node *root = build_node();
char buffer;
for (int i = 0; i < t; ++i) {
scanf(" %s", buffer);
printf("%lld\n", solve(buffer, root));
}
return 0;
}

In   Python3  :

tree = [False, {}]

current = tree
for c in word :
try :
current = current[c]
except KeyError :
current[c] = [False, {}]
current = current[c]
current = True

def count(word) :
count = 0
for start in range(len(word)) :
current, index = tree, start
while True :
if current :
count += 1
try :
current = current[word[index]]
index += 1
except (KeyError, IndexError) :
break
return count

v = 1
for x in range(801) :
v <<= 1

Done = {}
T = int(input())
for t in range(T) :
A = input()
if not A in Done :
Done[A] = count(A[::-1])
print(Done[A])```
```

## Square-Ten Tree

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## Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

## Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

## Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ