3517 -- And Then There Was One Let’s play a stone removing game. Initially, n stones are arranged on...

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3517 -- And Then There Was One



Let’s play a stone removing game.

Initially, n stones are arranged on a circle and numbered 1, …, n clockwise (Figure 1). You are also given two numbers k and m. From this state, remove stones one by one following the rules explained below, until only one remains. In step 1, remove stone m. In step 2, locate the k-th next stone clockwise from m and remove it. In subsequent steps, start from the slot of the stone removed in the last step, make k hops clockwise on the remaining stones and remove the one you reach. In other words, skip (k − 1) remaining stones clockwise and remove the next one. Repeat this until only one stone is left and answer its number.

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3517 -- And Then There Was One

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1012 -- Joseph Suppose that there are k good guys and k bad guys. In the circle the first k are good...

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1012 -- Joseph

Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.

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1012 -- Joseph

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Josephus problem | Set 1 (A O(n) Solution) - GeeksforGeeks There are n people standing in a circle waiting...

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Josephus problem | Set 1 (A O(n) Solution) - GeeksforGeeks

There are n people standing in a circle waiting to be executed. The counting out begins at some point in the circle and proceeds around the circle in a fixed direction. In each step, a certain number of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle. The task is to choose the place in the initial circle so that you are the last one remaining and so survive.

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Josephus problem | Set 1 (A O(n) Solution) - GeeksforGeeks

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Dynamic Programming | Set 18 (Partition problem) - GeeksforGeeks Partition problem is to determine whether...

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Dynamic Programming | Set 18 (Partition problem) - GeeksforGeeks

Partition problem is to determine whether a given set can be partitioned into two subsets such that the sum of elements in both subsets is same.

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Dynamic Programming | Set 18 (Partition problem) - GeeksforGeeks

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Tree List Recursion Problem http://ift.tt/1lZDJgt

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Tree List Recursion Problem

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Tree List Recursion Problem

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Developer's Reading List | Dr Dobb's http://ift.tt/1sx5QUG?...

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Developer's Reading List | Dr Dobb's

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Developer's Reading List | Dr Dobb's

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. . . About Mathematical Induction http://ift.tt/1ueTvGM

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. . . About Mathematical Induction

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. . . About Mathematical Induction

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Fors: Search for multiple labels on Blogger http://ift.tt/1mfp7NJ...

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Fors: Search for multiple labels on Blogger

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Fors: Search for multiple labels on Blogger

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3211 -- Washing Clothes Dearboy was so busy recently that now he has piles of clothes to wash. Luckily...

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3211 -- Washing Clothes



Dearboy was so busy recently that now he has piles of clothes to wash. Luckily, he has a beautiful and hard-working girlfriend to help him. The clothes are in varieties of colors but each piece of them can be seen as of only one color. In order to prevent the clothes from getting dyed in mixed colors, Dearboy and his girlfriend have to finish washing all clothes of one color before going on to those of another color.

From experience Dearboy knows how long each piece of clothes takes one person to wash. Each piece will be washed by either Dearboy or his girlfriend but not both of them. The couple can wash two pieces simultaneously. What is the shortest possible time they need to finish the job?

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3211 -- Washing Clothes

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2184 -- Cow Exhibition The cows want to prove to the public that they are both smart and fun. In order...

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2184 -- Cow Exhibition

The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.

Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si's and, likewise, the total funness TF of the group is the sum of the Fi's. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.

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2184 -- Cow Exhibition

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3624 -- Charm Bracelet Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course...

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3624 -- Charm Bracelet



Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

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3624 -- Charm Bracelet

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Navigate and fix errors and warnings in a class with Eclipse keyboard shortcuts | Eclipse On E http:...

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Navigate and fix errors and warnings in a class with Eclipse keyboard shortcuts | Eclipse On E

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Navigate and fix errors and warnings in a class with Eclipse keyboard shortcuts | Eclipse On E

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Convert string concatenations into StringBuilder or MessageFormat calls with Eclipse’s Quick Fix | Eclipse...

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Convert string concatenations into StringBuilder or MessageFormat calls with Eclipse’s Quick Fix | Eclipse On E

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Convert string concatenations into StringBuilder or MessageFormat calls with Eclipse’s Quick Fix | Eclipse On E

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Eclipse: Save Time When Debugging With Run-To-Line http://ift.tt/1u01uLw

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Eclipse: Save Time When Debugging With Run-To-Line

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Eclipse: Save Time When Debugging With Run-To-Line

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Java Colloection: Using subList subList(fromIndex, toIndex): [fromIndex, toIndex) List.subList returns...

Java Collection: Using Collections API Common Collections Utils Collections.sort(), max(), min(), .reverse...

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Java Collection: Using Collections API

Common Collections Utils Collections.sort(), max(), min(), .reverse(), shuffle(),rotate() Collections.swap(),fill(), copy(),replaceAll(), return Collections.emptyList(), unmodifiabkeList(), synchronizedList(),checkedList() Collections.disjoint(c1,c2) -- ret...

Common Collections Utils Collections.sort(), max(), min(), .reverse(), shuffle(),rotate() Collections.swap(),fill(), copy(),replaceAll(), return Collections.emptyList(), unmodifiabkeList(), synchronizedList(),checkedList() ...

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Reshared post from Alex Shar:

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Original Post from Alex Shar:

Fenghuang, Hunan, China

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Reshared post from Google:

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Original Post from Google:

Today we're introducing the first family of Android One phones in India—our effort to bring high-quality smartphones to as many people as possible: http://goo.gl/rj3Dsb

Introducing Android One

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