Won't we be able to choose from (n+m-2*k)C(n-k)? As the number of choices are n-k down and m-k right, and we have to choose n-k options. Also, min(m,n)C(k) are the number of choices for diagonals for each k.
Please correct me if I'm wrong.
Problem in short: In a grid of size R x C, you need to find the number of paths from the top-left to the bottom-right. You may only move 1 block to the right or one block down in each step. Yo...
Need help! My solution fails for the some of the testcases in the second subtask(d=2).
#include <iostream>#include <algorithm>#include <fstream>#include <vector>#include <deque>#include <assert.h>#include <bitset>#include <queue>#include <stack>#include <set>#include <map>#include <tuple>#include <stdio.h>#include <string.h>#include <utility>#include <math.h>using namespace std;// This is 4D DP with bottom up approachlong long dp[305][305][305][2], g[305][305];int n, m, D;
Since the last few days, the IARCS website has been down and giving fatal error for C++ submissions. For the time being, I request people to skip the IARCS problems. I've contacted the maintai...
Getting WA in some test cases:
#include <stdio.h>int main(){int n, k;scanf("%d %d", &n, &k);int ar[n];while(k--){for (int i = 0; i < n; ++i) {scanf("%d ", &ar[i]) ;}if (n == 1)printf("%d ",ar[0]);for (int i = n - 2; i >= 0; --i) {if (ar[i] < ar[i + 1]){for (int j = n - 1; j >i; --j) {if (ar[j] > ar[i]){int t = ar[i];ar[i] = ar[j];ar[j] = t;// printf("%d %d \n", ar[i], ar[j]);}}
No. The insertion first checks if the element is lesser than or greater than the root at each level. In short, if its in left sub-tree, it must have been smaller than the root. Irrespective of whether it ends up in left or right sub-tree.
ICO Online Judge is the Online Judge provided by Indian Computing Olympiad
This is frustrating, it only shows fatal. I don't know if its an error in the code. Or something else. Here, is the link, will anybody help? IARCS Problem Archive
