As far as I understand, you want to place n rectangles with fixed C=W/H ratio on the wall with given Width and Height Let rectangle height is h (unknown yet), width is w = C * h Every row of grid contains nr = Floor(Width / (C * h)) // rounding down Every column contains … Read more
Here’s a hint: 23 : 11 + 11+ 1 ( 3 magic numbers) 120: 110+ 10 (2 magic numbers) The highest digit in the target number is the answer, since you need exactly k magic numbers (all having 1 in the relevant position) in order for the sum to contain the digit k. So the … Read more
With (val>>1)-(val & 1) you get your first result: 0 -> 0 1 -> -1 2 -> 1 The alternative result can be got with (val & 1)-(val>>1): 0 -> 0 1 -> 1 2 -> -1 0 solved Formula for Mapping three values [closed]
As Mark Baker pointed out, you’ll need to use something like Dijkstra’s algorithm to traverse a weighted graph (which is what you’ll have when you include distances). Here’s a simple distance function I found here: <?php /*::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/ /*:: :*/ /*:: This routine calculates the distance between two points (given the :*/ /*:: latitude/longitude of those … Read more
As discussed in the comments to the question, it’s not very clear exactly what epitaph‘s setup is, and for some reason s/he seems unwilling to clarify. But let’s suppose that the “graph” is actually simply represented by a distance matrix (which may possibly be what the mess of numbers and Xs in the question is … Read more
#include<stdio.h> #include<conio.h> int main() { int n=3,i,j,k,l,m; clrscr(); for(i=0;i<n;i++) { for(j=0;j<n-i-1;j++) printf(” “); for(k=0;k<=i-1;k++) printf(“*”,k+1); for(m=0;m<1;m++) { if(i<=0) printf(“*”); else printf(” “); } for(l=i-1;l>=0;l–) printf(“*”,l+1); printf(“\n”); } for(i=0;i<n-1;i++) { for(j=0;j<=i;j++) printf(” “); for(k=0;k<n-i-2;k++) printf(“*”,k+1); for(m=0;m<1;m++) { if(i==1) {} else printf(” “); } for(l=1;l>0;l–) printf(“*”,l+1); printf(“\n”); } getch(); return 0; } Finally I got my solution … Read more
Introduction Solved pattern programming is a type of programming that involves solving a problem by breaking it down into smaller, more manageable pieces. It is a powerful tool for problem solving and can be used to solve a wide variety of problems. Solved pattern programming is often used in computer science, engineering, and mathematics. It … Read more
NOTE: I am assuming that we do not have use every color. If we do, additional checks can easily be applied. Here is some code that can solve your problem (I think)… It uses a brute force / backtracking algo. Here is the output for 3 colors, and 2 by 3 grid. It prints every … Read more
#include <algorithm> void myfunction (int &i) { // function: i=1000001; } int arrayJmp ( const vector<int> &A ) { int N=A.size(); vector<int> indexes; for_each (indexes.begin(), indexes.end(), myfunction); int index=0; while(std::find(indexes.begin(), indexes.end(), index) == indexes.end()){ indexes.push_back(index); index+=A[index]; if (index==(N-1)&&A[index]>0) return indexes.size()+1; } return -1; } 9 solved simulating pawn jump in the array/vector [closed]
Lets assume that the inner loop actually does something. FredrikRedin and niculare are, of course, correct, that the number of inner loop executions will be the same. Given that this is the case, the second example is preferred (slightly) because the second example has to set up the inner loop ten times, while the first … Read more
Taking a look at the string constructor: … (7) template <class InputIterator> string (InputIterator first, InputIterator last); You can see it is possible to create a string via iterators. The rbegin/rend iterators are InputIterators that points to the reverse positions that they refer: rend() –>”My string”<– rbegin() That said, when you pass the rbegin() and … Read more
Let’s consider two ways to solve it that are two slow, and then merge them into one solution, that will be guaranteed to finish in milliseconds. Approach 1 (slow) Allocate an array v of size 2^16. Every time you add an element, do the following: void add(int s) { for (int i = 0; i … Read more
I could not post it to comment section. Please see below: List<Integer> integers = Arrays.asList(1, 2, 3, 4); int i = 0; StringBuilder sb = new StringBuilder(“[“); for (int value : integers) { sb.append((i == 0) ? “” : (i % 2 == 0 ? “,” : “:”)).append(value); i++; } sb.append(“]”); System.out.println(sb.toString()); Output is: [1:2,3:4] … Read more
We have: 1 : if statement 3 : * operator 3 : function statement Totally: 7 So we have: T(n)=t(n-1) + t(n-2) + t(n-3) + 7 T(99)=1 T(98)=1 … T(1)=1 Then to reduce T(n), we have T(n-3)<T(n-2)<T(n-1) therefore: T(n) <= T(n-1)+T(n-1)+T(n-1) + 7 T(n) <= 3T(n-1) + 7 So we can solve T(n) = 3T(n-1) … Read more
There is a problem in this line: double v[k] = log(log(sqrt(e[k]+1)+1)+1); in the first iteration k == 0 so you are trying to declare a zero sized array. I dont understand the code enough to tell you how to fix it, but a zero sized array is definitely not what you want in that place. … Read more