OJ Board

mf's output is not correct according to my accepted output.

Brute-force approach can run well within time limit. You need to consider all permutation of columns only. Because Bob will choose columns. We do not need to consider permutation of rows. Ultimately Alice have to cut all rows and how Bob will cut columns is the main question. Alice's order of row cu...

Brute-force approach can run well within time limit. You need to consider all permutation of columns only. Because Bob will choose columns. We do not need to consider permutation of rows. Ultimately Alice have to cut all rows and how Bob will cut columns is the main question. Alice's order of row cu...

Sample input: aasgfndfdsndsbfngferdsfnshhgbshehbnbniavnnbroaegbwaibhubdskeienvbffnhrr ufiuewgbvsusiuvnjakaslcvnirghbnvlodsjdvsbbfsoohbvrrbvdsdfsvdsundvssaamsmdn My accepted program returns: aaaabbbbbbbbdddddeeeffffffggghhhhiiiknnnnnnnnnorrrsssssssuvvw Your program returns: aaaabbbbbbdddddefffgghhiii...

I think there is no test case where all grade value are less than 5 for a particular case.

So i think this input is invalid
I got accepted by considering the fact that at least one grade value is greater than or equal to 5 for a particular course.

Shadek
BUET,CSE-07

what is the max value of n(number of parts of all sticks)?

What are the outputs for the following inputs? First number is number of test cases.

I have used the following dp, dp[n][0]->number of all numbers that can be formed by less than n matches dp[n][1] -> number of all numbers that can be formed by using exactly n matches dp[n][0] = dp[n-1][0] + dp[n-1][1] dp[n][1] = dp[n-2][1] + dp[n-3][1] + dp[n-4][1] + 3*dp[n-5][1] + 2*dp[n-6] + dp[n...

for each (i,j) with i<=j ,I have saved value +value[j] in a set, not multiset. In multiset,same data can be presented more than one. If there is no duplicate of value +value[j] for different pairs(i,j) with (i<=j). then no. of elements in the set would be (n*(n-1))/2 + n = (n*n-n)/2 Otherwise it is ...

The original graph may have many components. Let there are k components. Let the total number of faces is F. Initially F = 0 In input edges we can have- x x for each occurrence of such edges, just F++ Again we can have (x y) or (y x) If we get total p occurrences of (x y) or (y x) F = F+(p-1) Withou...

#include<iostream> #include<string> #include<cstring> #include<sstream> #include<algorithm> #include<cstdio> using namespace std; #define MAX 10005 char com1[10005]; char com2[10005]; int ind1,ind2; int m,n; int coun; int i,j,c[MAX][MAX],b[MAX][MAX]; int len; int LCSlength() { m=len; n=len; for (i=...

what's the problem in this code? #include<iostream> #include<vector> #include<cstring> #include<queue> #include<cstdio> using namespace std; #define un 0 #define dis 1 #define comp 2; vector<int>adj[25]; queue<int> q; int step; int state[25]; bool flag[25]; int in_queue[25]; int p[25]; int src,dest;...

what is the bug in the code. #include<iostream> #include<vector> using namespace std; #define M 32009 vector<int> primeList,v; bool prime[100009]; void sieve() { int i,j,k,l,m; prime[0] = false; prime[1] = false; for(i = 4; i<= M; i += 2) { prime[i] = false; } for(i = 3; i <= M; i += 2) { if(prime[i...

I have found my fault. Input should be like square. So by calculating length of the first string , we can know how many rows will be followed. Then these two line have to use.
if(t > 0)
getchar();

what's the reason of wrong answer? #include<iostream> #include<vector> #include<cstring> using namespace std; #define SIZE 101 vector <int> adj[SIZE][SIZE]; vector <int> adj1[SIZE][SIZE]; bool visited[SIZE][SIZE]; int v; char in[100][101]; int cnt; void visit(int u,int v) { int i; visited[u][v]=true...