Lego Blocks
Problem Statement :
You have an infinite number of 4 types of lego blocks of sizes given as (depth x height x width): d h w 1 1 1 1 1 2 1 1 3 1 1 4 Using these blocks, you want to make a wall of height n and width m. Features of the wall are: - The wall should not have any holes in it. - The wall you build should be one solid structure, so there should not be a straight vertical break across all rows of bricks. - The bricks must be laid horizontally. How many ways can the wall be built? Example n = 2 m = 3 The height is 2 and the width is 3. Here are some configurations: image These are not all of the valid permutations. There are 9 valid permutations in all. Function Description Complete the legoBlocks function in the editor below. legoBlocks has the following parameter(s): int n: the height of the wall int m: the width of the wall Returns - int: the number of valid wall formations modulo (10^9 + 7) Input Format The first line contains the number of test cases t. Each of the next t lines contains two space-separated integers n and m. Constraints 1 <= t <= 100 1 <= n,m <= 1000
Solution :
Solution in C :
In C++ :
#include <iostream>
using namespace std;
#define MOD 1000000007
typedef unsigned long long ull;
ull dp[1001]; // with rep
ull dpr1[1001]; // with rep
ull dpr2[1001]; // with rep
ull poww ( ull x, int n){
if(n==0) return 1;
ull t = poww(x, n/2);
t = t * t;
if(t>=MOD) t%= MOD;
if(n&1){
t = t * x;
if(t>=MOD) t%= MOD;
}
return t;
}
void pre(){
dp[1] = 1; dp[2] = 2; dp[3] = 4; dp[4] = 8;
for(int i=5;i<=1000;i++){
dp[i] = dp[i-1] + dp[i-2] + dp[i-3] + dp[i-4];
if(dp[i] >= MOD)
dp[i] %= MOD;
}
}
ull calc(int N, int M){
for(int i=1;i<=N;i++){
dpr1[i] = poww(dp[i], M);
}
for(int i=1;i<=N;i++){
dpr2[i] = dpr1[i];
for(int k=1;k<i;k++){
ull x = dpr2[k];
x = x * dpr1[i-k];
if(x >= MOD)
x %= MOD;
dpr2[i] -= x;
if(dpr2[i] >= MOD)
dpr2[i] += MOD;
}
}
return dpr2[N];
}
int main(){
pre();
int T;
int N,M;
cin >> T;
while(T--){
cin >> N >> M;
cout << calc(M,N) << endl;
//cout << dp[N][M] << endl;
}
}
In Java :
import java.util.*;
public class Solution {
int md=1000000007;
int[][] ways = new int[1001][1001];
int[][] waysRestrict = new int[1001][1001];
public Solution(){
for(int[] w : ways) Arrays.fill(w, -1);
for(int[] w : waysRestrict) Arrays.fill(w, -1);
}
public int solve(int n, int m)
{
if (ways[n][m] != -1) return ways[n][m];
long ans;
if (m==1) ans = 1;
else if (n==1){
if (m<=4) {
ans = 2*solve(1,m-1);
}
else {
ans = solve(1,m-1);
ans = (ans + solve(1,m-2))%md;
ans = (ans + solve(1,m-3))%md;
ans = (ans + solve(1,m-4))%md;
}
}
else{
ans = 1; int one = solve(1,m);
for (int i=0; i<n; i++) ans = (ans * one)%md;
}
ways[n][m] = (int)ans;
return ways[n][m];
}
public int solveRestrict(int n, int m)
{
if (waysRestrict[n][m] != -1) return waysRestrict[n][m];
long ans;
if (m==1) ans = 1;
// else if (n==1) ans = solve(n,m);
else {
ans = solve(n,m);
for (int i=1; i<m; i++) {
long a = solve(n,i);
a = (a*solveRestrict(n,m-i))%md;
ans -= a;
if (ans<0) ans+=md;
}
}
waysRestrict[n][m] = (int)ans;
return waysRestrict[n][m];
}
public static void main (String[] args) {
Solution o = new Solution();
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for (int i=0; i<n; i++) {
int a, b;
a = sc.nextInt(); b = sc.nextInt();
System.out.println(o.solveRestrict(a,b));
}
sc.close();
}
}
In C :
#include<stdio.h>
unsigned long long int call(int,int);
unsigned long long int f[1001];
unsigned long long int to[1001][1001];
unsigned long long int res[1001][1001];
int main()
{
int h,w,i,j,t;
unsigned long long int r=0;
f[1]=1;
f[2]=2;
f[3]=4;
f[4]=8;
for(i=1;i<=1000;i++)
{
for(j=1;j<=1000;j++)
{
res[i][j]=0;
}
}
for(i=5;i<=10000;i++)
{
f[i]=((f[i-4]+f[i-3])%1000000007+(f[i-2]+f[i-1])%1000000007)%1000000007;
}
for(j=1;j<=1000;j++)
{
to[1][j]=f[j];
}
for(i=2;i<=1000;i++)
{
for(j=1;j<=1000;j++)
{
to[i][j]=(to[i-1][j]*f[j])%1000000007;
}
}
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&h,&w);
//printf("f[%d] = %llu\n",w,f[w]);
r=call(h,w);
printf("%llu\n",r);
}
return 0;
}
unsigned long long int call(int h,int w)
{
int i=0;
unsigned long long int r=0;
if(h==1)
{
if(w>4)
return 0;
else
return 1;
}
else if(res[h][w]!=0)
{
return res[h][w];
}
else
{
for(i=1;i<=w-1;i++)
{
r=(r+(call(h,i)*to[h][w-i])%1000000007)%1000000007;
}
if(to[h][w]>r)
r=to[h][w]-r;
else
{
r=r-to[h][w];
r=1000000007-r;
}
res[h][w]=r;
return res[h][w];
}
}
In Python3 :
m = 1000000007
dp1 = [1,1,2,4]
for i in range(4,1001):
dp1.append((dp1[-1]+dp1[-2]+dp1[-3]+dp1[-4])%m)
def f(h,w):
dp2 = []
dp3 = [pow(dp1[i],h,m) for i in range(w+1)]
for i in range(w+1):
tmp = dp3[i]
for j in range(1,i):
tmp = (tmp - dp2[j]*dp3[i-j])%m
dp2.append(tmp)
return dp2[-1]
for e in range(int(input())):
print(f(*list(map(int,input().split()))))
View More Similar Problems
Insert a node at a specific position in a linked list
Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e
View Solution →Delete a Node
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo
View Solution →Print in Reverse
Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing
View Solution →Reverse a linked list
Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio
View Solution →Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →
