Skip to main content

SPOJ 4300 - Rectangles

Problem Statement

Hint: If the input is 15 squares
rectangles are-
                1x1 1x2 1x3 ...................................1x15
                       2x2 2x3 ............................2x7    
                              3x3   ....................3x5
                        
  • can't make 3x2 cause 3x2 and 2x3 are same.
  • can't make 2x8 cause that requires 16 squares.
  • can't make 4x4 cause that also requires 16 square. 
Solution:

#include<stdio.h>
int main()
{
    int n,c=0,i,j;
    scanf("%d",&n);
    for(i=1;i*i<= n;i++)
    {
        c+= (n/i)-i+1;
    }
    printf("%d\n",c);
}


* If there's any query or mistake, let me know.

Comments

Popular posts from this blog

SPOJ - GSS1

Considering input series: { 4 , -10 , 3 , 100 , -20 , 1 } Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i ≤ j ≤ y } A node contains-  [ START & END is a node's segment limit ] Prefix is the maximum sum starting at START, end can be anywhere. There are two possibilities of the maximum. One, node's leftChild's prefix or two, adding leftChild's sum + rightChild's prefix. (which will make the prefix contiguous) Suffix is the maximum sum ending at END, start can be anywhere. There's two possibility of the maximum. One, node's rightChild's already calculated suffix or two, add rightChild's sum + leftChild's suffix   (which will make the suffix contiguous). Sum : leftChild's sum + rightChild's sum. MAX Maximum of  -  prefix (result is in the node, starts from START but doesn't end in END ) suffix  (result is in the node, doesn't start from START but surely ends in END ) leftChild's max ( result is in left ...

SPOJ 95 - STPAR - Street Parade

Problem Statement Heuristics:   Let, N be the number which is next in fixed order. Take an input X , if it is equal to N then next number will be n++. If not, then search for N in the side lane and after finishing the search, put X in the side lane. After all X is processed but N is not found in either X or side lane, then the order is not possible. Solution:

SPOJ-PIGBANK | Unbounded Knapsack ( Minimization)

Unbounded Knapsack : Given a knapsack weight W and a set of N items with certain value Vi and weight W i , we need to calculate minimum amount that could make up this quantity exactly. To have a more clear understanding of this problem, see SPOJ-PIGBANK Sample Problem : We're given a box which can hold maximum W weight. There are N various coins each of value Vi and Weight Wi. We need to fill the box with total W weight with any number of coins of any value. Note that coins are unlimited.  100 ----- W 2 ------- N 1 1 ----- vi,wi 30 50 --- vi,wi Answer of the problem above is, 60. If we take 2 coins of value 30, weight 50, we'll get exactly weight 100 with a minimum value of 60. Algorithm : This is a DP solution. To get the solution for W=100, we'll solve subproblems first. Let dp[W+1]  be an array which will store the subproblems solutions. In the end, dp[W] will give us the minimum amount. Subproblems soultion will be the minimum of taking the coin Vk ...