Easy Problems List (Updated)

UVa Online Judge

12577 - Hajj-e-Akbar

12289 - One-Two-Three

11942 - Lumberjack Sequencing

12015 - Google is Feeling Lucky

1124 - Celebrity jeopardy

1225 - Digit Counting

12578 - 10:6:2

12342 - Tax Calculator

11936 - The Lazy Lumberjacks

12502 - Three Families

12646 - Zero or One

12478 - Hardest Problem Ever

TIMUS Online Judge (acm.timus.ru)

1000 - A+B Problem

1001 - Reverse Root

1068 - Sum

LightOJ

Lift

How Cow

Setu

February 29

Minimum Arc Distance

Hidden Secret!

IP Checking

Agent J

Unlucky Bird

Double Ended Queue

List of Numbers of Easy Problem’s from UVA [uva.onlinejudge.org]:

100 102 113 119 136 147 160 190
272 299
324 343 344 350 355 357 369 378 382 386 389
401 412 414 438 440 441 442 443 444 458 476 477 478 483 488 489 490 492 494 495 499
537 541 543 575 579 583 591
612 623 673 674 686
706 713
847 895
10008 10013 10035 10038 10050 10055 10062 10070 10071 10079 10082
10101 10110 10127 10168 10170 10194 10195
10220 10222 10235 10260 10286 10293
10300 10302 10323 10324 10340 10346 10370
10409 10424 10450 10469 10473 10499
10509 10573 10579 10591
10611 10696 10699
10714 10773 10783 10789
10812 10894
10905 10921 10922 10924 10929 10931 10935 10940 10945 10970
11000 11044 11074
11172
11233

Comments

Game Theory with examples

Game Theory with examples Introduction In this article I will be covering problems related to two players game in which it is assumed that both players play optimally and we have to find out the winner of the game. First we will look at the basic division of positions to winning and losing . Then we will see the Game of Nim and then see how it will be used to solve the Composite games . Basic Division of positions to winning and losing Problem Statement: Consider a simple game played by two players A and B . There are n stones on the table. Each player can pick 1 , 2 or 5 stones in each turn. Both players pick the stones alternately until the total number of stones left on the table is 0. The player unable to make the move will lose. Assuming that both the players play optimally, output the winner of the game. Solution: As you can see positions 1 , 2 and 5 are winning positions since the player can pick up all the stones and other player will n...

Breaking The Summation Formula (Part 1)

Q. f ( n ) = - 1 + 2 - 3 + .. + ( - 1) n n . Given n, find out f(n) Approach(1)- Bruteforce: 1. Calculation sum=n*(n+1)/2 2. loop[i=1,i=n : i+=2] odd+=i 3.ans=sum-2*odd Code: #include < bits / stdc ++. h > using namespace std ; int main (){ long long x ; cin >> x ; long long p =( x *( x + 1 ))/ 2 ; long long bad = 0 ; for ( long long i = 1 ; i <= x ; i += 2 ) bad += i ; cout << p - 2 * bad << endl ; } Approach(2)-Greedy: Basic: s=1+2+3+4+....+n Formula: sum=n*(n+1)/2= (n/2) + (n+1).2 ...

NT Part 2: Generating Primes, Prime Test, Prime Factorization

Generating primes fast is very important in some problems. Let's cut to the chase and introduce Eratosthenes's Sieve. The main idea is the following. Suppose we want to find all primes between 2 and 50. Iterate from 2 to 50. We start with 2. Since it is not checked, it is a prime number. Now check all numbers that are multiple of except 2. Now we move on, to number 3. It's not checked, so it is a prime number. Now check all numbers that are multiple of , except 3. Now move on to 4. We see that this is checked - this is a multiple of 2! So 4 is not a prime. We continue doing this. Here's the implementation. #include <stdio.h> int primechk [ 21000 ] ; void preprocess ( void ) { int i, j ; for ( i = 2 ; i <= 20000 ; i ++ ) { primechk [ i ] = 1 ; } for ( i = 2 ; i <= 20000 ; i ++ ) { if ( primechk [ i ] == 1 ) { for ( j = 2 ; i * j <= 20000...

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