evanellis/Codeforces-Python-Submissions_correct

479

A

Expression

PROGRAMMING

1,000

[ "brute force", "math" ]

null

Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets: - 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9 Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2. It's easy to see that the maximum value that you can obtain is 9. Your task is: given *a*, *b* and *c* print the maximum value that you can get.

The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).

Print the maximum value of the expression that you can obtain.

[ "1\n2\n3\n", "2\n10\n3\n" ]

[ "9\n", "60\n" ]

none

500

[ { "input": "1\n2\n3", "output": "9" }, { "input": "2\n10\n3", "output": "60" }, { "input": "1\n1\n1", "output": "3" }, { "input": "1\n2\n1", "output": "4" }, { "input": "10\n10\n10", "output": "1000" }, { "input": "5\n1\n3", "output": "20" }, { "input": "3\n1\n5", "output": "20" }, { "input": "6\n7\n1", "output": "48" }, { "input": "1\n8\n3", "output": "27" }, { "input": "9\n7\n2", "output": "126" }, { "input": "1\n1\n10", "output": "20" }, { "input": "9\n1\n1", "output": "18" }, { "input": "10\n5\n6", "output": "300" }, { "input": "8\n9\n7", "output": "504" }, { "input": "4\n2\n10", "output": "80" }, { "input": "2\n8\n3", "output": "48" }, { "input": "3\n5\n7", "output": "105" }, { "input": "1\n10\n1", "output": "12" }, { "input": "2\n2\n2", "output": "8" }, { "input": "5\n6\n1", "output": "35" }, { "input": "10\n1\n1", "output": "20" }, { "input": "1\n6\n1", "output": "8" }, { "input": "1\n9\n1", "output": "11" }, { "input": "2\n1\n2", "output": "6" }, { "input": "2\n6\n1", "output": "14" }, { "input": "9\n2\n1", "output": "27" }, { "input": "1\n9\n2", "output": "20" }, { "input": "1\n3\n1", "output": "5" }, { "input": "2\n1\n1", "output": "4" } ]

1,696,000,342

2,147,483,647

Python 3

OK

TESTS

29

46

0

a, b, c = int(input()), int(input()), int(input()) a, c = min([a, c]), max([a, c]) if a == 1 or b == 1: result = a + b if c == 1: result += 1 else: result *= c else: result = a * b * c print(result)

Title: Expression Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets: - 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9 Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2. It's easy to see that the maximum value that you can obtain is 9. Your task is: given *a*, *b* and *c* print the maximum value that you can get. Input Specification: The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). Output Specification: Print the maximum value of the expression that you can obtain. Demo Input: ['1\n2\n3\n', '2\n10\n3\n'] Demo Output: ['9\n', '60\n'] Note: none

```python a, b, c = int(input()), int(input()), int(input()) a, c = min([a, c]), max([a, c]) if a == 1 or b == 1: result = a + b if c == 1: result += 1 else: result *= c else: result = a * b * c print(result) ```

3

281

A

Word Capitalization

PROGRAMMING

800

[ "implementation", "strings" ]

null

Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word. Note, that during capitalization all the letters except the first one remains unchanged.

A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.

Output the given word after capitalization.

[ "ApPLe\n", "konjac\n" ]

[ "ApPLe\n", "Konjac\n" ]

none

500

[ { "input": "ApPLe", "output": "ApPLe" }, { "input": "konjac", "output": "Konjac" }, { "input": "a", "output": "A" }, { "input": "A", "output": "A" }, { "input": "z", "output": "Z" }, { "input": "ABACABA", "output": "ABACABA" }, { "input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX", "output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX" }, { "input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO", "output": 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"CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP" }, { "input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG", "output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG" }, { "input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV", "output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV" }, { "input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI", "output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI" }, { "input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX", "output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX" }, { "input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ", "output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ" }, { "input": "P", "output": "P" }, { "input": "Xyzzy", "output": "Xyzzy" }, { "input": "Zzz", "output": "Zzz" }, { "input": "Zp", "output": "Zp" } ]

1,695,958,154

2,147,483,647

Python 3

OK

TESTS

25

92

0

# -*- coding: utf-8 -*- """ Created on Fri Sep 29 11:21:27 2023 @author: ZHAO XUDI """ w = input() l = list(w) l[0]=l[0].upper() print("".join(l))

Title: Word Capitalization Time Limit: None seconds Memory Limit: None megabytes Problem Description: Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word. Note, that during capitalization all the letters except the first one remains unchanged. Input Specification: A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103. Output Specification: Output the given word after capitalization. Demo Input: ['ApPLe\n', 'konjac\n'] Demo Output: ['ApPLe\n', 'Konjac\n'] Note: none

```python # -*- coding: utf-8 -*- """ Created on Fri Sep 29 11:21:27 2023 @author: ZHAO XUDI """ w = input() l = list(w) l[0]=l[0].upper() print("".join(l)) ```

3

337

A

Puzzles

PROGRAMMING

900

[ "greedy" ]

null

The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.

The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.

Print a single integer — the least possible difference the teacher can obtain.

[ "4 6\n10 12 10 7 5 22\n" ]

[ "5\n" ]

Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

500

[ { "input": "4 6\n10 12 10 7 5 22", "output": "5" }, { "input": "2 2\n4 4", "output": "0" }, { "input": "2 10\n4 5 6 7 8 9 10 11 12 12", "output": "0" }, { "input": "4 5\n818 136 713 59 946", "output": "759" }, { "input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903", "output": "13" }, { "input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94", "output": "3" }, { "input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190", "output": "31" }, { "input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997", "output": "0" }, { "input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176", "output": "412" }, { "input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886", "output": "938" }, { "input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993", "output": "937" }, { "input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7", "output": "0" }, { "input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19", "output": "12" }, { "input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4", "output": "0" }, { "input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17", "output": "31" }, { "input": "2 2\n1000 4", "output": "996" }, { "input": "2 3\n4 502 1000", "output": "498" }, { "input": "3 3\n4 1000 4", "output": "996" } ]

1,677,946,141

2,147,483,647

PyPy 3

OK

TESTS

18

154

0

n, m = [int(i) for i in input().split(' ')] A = [int(i) for i in input().split(' ')] A.sort() res = [] for item in range(m-n+1): res.append(abs(A[item] - A[item+n-1])) print(min(res))

Title: Puzzles Time Limit: None seconds Memory Limit: None megabytes Problem Description: The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. Input Specification: The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. Output Specification: Print a single integer — the least possible difference the teacher can obtain. Demo Input: ['4 6\n10 12 10 7 5 22\n'] Demo Output: ['5\n'] Note: Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

```python n, m = [int(i) for i in input().split(' ')] A = [int(i) for i in input().split(' ')] A.sort() res = [] for item in range(m-n+1): res.append(abs(A[item] - A[item+n-1])) print(min(res)) ```

3

950

A

Left-handers, Right-handers and Ambidexters

PROGRAMMING

800

[ "implementation", "math" ]

null

You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand. The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands. Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand. Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.

The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.

Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.

[ "1 4 2\n", "5 5 5\n", "0 2 0\n" ]

[ "6\n", "14\n", "0\n" ]

In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team. In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.

500

[ { "input": "1 4 2", "output": "6" }, { "input": "5 5 5", "output": "14" }, { "input": "0 2 0", "output": "0" }, { "input": "30 70 34", "output": "128" }, { "input": "89 32 24", "output": "112" }, { "input": "89 44 77", "output": "210" }, { "input": "0 0 0", "output": "0" }, { "input": "100 100 100", "output": "300" }, { "input": "1 1 1", "output": "2" }, { "input": "30 70 35", "output": "130" }, { "input": "89 44 76", "output": "208" }, { "input": "0 100 100", "output": "200" }, { "input": "100 0 100", "output": "200" }, { "input": "100 1 100", "output": "200" }, { "input": "1 100 100", "output": "200" }, { "input": "100 100 0", "output": "200" }, { "input": "100 100 1", "output": "200" }, { "input": "1 2 1", "output": "4" }, { "input": "0 0 100", "output": "100" }, { "input": "0 100 0", "output": "0" }, { "input": "100 0 0", "output": "0" }, { "input": "10 8 7", "output": "24" }, { "input": "45 47 16", "output": "108" }, { "input": "59 43 100", "output": "202" }, { "input": "34 1 30", "output": "62" }, { "input": "14 81 1", "output": "30" }, { "input": "53 96 94", "output": "242" }, { "input": "62 81 75", "output": "218" }, { "input": "21 71 97", "output": "188" }, { "input": "49 82 73", "output": "204" }, { "input": "88 19 29", "output": "96" }, { "input": "89 4 62", "output": "132" }, { "input": "58 3 65", "output": "126" }, { "input": "27 86 11", "output": "76" }, { "input": "35 19 80", "output": "134" }, { "input": "4 86 74", "output": "156" }, { "input": "32 61 89", "output": "182" }, { "input": "68 60 98", "output": "226" }, { "input": "37 89 34", "output": "142" }, { "input": "92 9 28", "output": "74" }, { "input": "79 58 98", "output": "234" }, { "input": "35 44 88", "output": "166" }, { "input": "16 24 19", "output": "58" }, { "input": "74 71 75", "output": "220" }, { "input": "83 86 99", "output": "268" }, { "input": "97 73 15", "output": "176" }, { "input": "77 76 73", "output": "226" }, { "input": "48 85 55", "output": "188" }, { "input": "1 2 2", "output": "4" }, { "input": "2 2 2", "output": "6" }, { "input": "2 1 2", "output": "4" }, { "input": "2 2 1", "output": "4" }, { "input": "3 2 1", "output": "6" }, { "input": "1 2 3", "output": "6" }, { "input": "1 3 2", "output": "6" }, { "input": "2 1 3", "output": "6" }, { "input": "2 3 1", "output": "6" }, { "input": "3 1 2", "output": "6" }, { "input": "99 99 99", "output": "296" }, { "input": "99 99 100", "output": "298" }, { "input": "99 100 99", "output": "298" }, { "input": "99 100 100", "output": "298" }, { "input": "100 99 99", "output": "298" }, { "input": "100 99 100", "output": "298" }, { "input": "100 100 99", "output": "298" }, { "input": "89 32 23", "output": "110" }, { "input": "4 5 0", "output": "8" }, { "input": "3 0 3", "output": "6" }, { "input": "0 0 2", "output": "2" }, { "input": "97 97 0", "output": "194" }, { "input": "1 4 0", "output": "2" }, { "input": "5 2 0", "output": "4" }, { "input": "0 5 10", "output": "14" }, { "input": "0 1 2", "output": "2" }, { "input": "5 2 3", "output": "10" }, { "input": "5 5 0", "output": "10" }, { "input": "0 0 10", "output": "10" }, { "input": "0 1 1", "output": "2" }, { "input": "0 0 1", "output": "0" } ]

1,620,206,432

2,147,483,647

Python 3

OK

TESTS

79

62

0

l,r,a=map(int,input().split()) x=min(l,r) d=abs(l-r) if d>a: ans=(x+a)*2 elif d<=a: a=a-d ans=(x+d+a//2)*2 print(ans)

Title: Left-handers, Right-handers and Ambidexters Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand. The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands. Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand. Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively. Input Specification: The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training. Output Specification: Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players. Demo Input: ['1 4 2\n', '5 5 5\n', '0 2 0\n'] Demo Output: ['6\n', '14\n', '0\n'] Note: In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team. In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.

```python l,r,a=map(int,input().split()) x=min(l,r) d=abs(l-r) if d>a: ans=(x+a)*2 elif d<=a: a=a-d ans=(x+d+a//2)*2 print(ans) ```

3

59

A

Word

PROGRAMMING

800

[ "implementation", "strings" ]

A. Word

2

256

Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.

The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.

Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.

[ "HoUse\n", "ViP\n", "maTRIx\n" ]

[ "house\n", "VIP\n", "matrix\n" ]

none

500

[ { "input": "HoUse", "output": "house" }, { "input": "ViP", "output": "VIP" }, { "input": "maTRIx", "output": "matrix" }, { "input": "BNHWpnpawg", "output": "bnhwpnpawg" }, { "input": "VTYGP", "output": "VTYGP" }, { "input": "CHNenu", "output": "chnenu" }, { "input": "ERPZGrodyu", "output": "erpzgrodyu" }, { "input": "KSXBXWpebh", "output": "KSXBXWPEBH" }, { "input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv", "output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv" }, { "input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd", "output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd" }, { "input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv", "output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv" }, { "input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg", "output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg" }, { "input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak", "output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK" }, { "input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw", "output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW" }, { "input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB", "output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB" }, { "input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge", "output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge" }, { "input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw", "output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw" }, { "input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks", "output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks" }, { "input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz", "output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ" }, { "input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg", "output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG" }, { "input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO", "output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO" }, { "input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd", "output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD" }, { "input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas", "output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas" }, { "input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm", "output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM" }, { "input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn", "output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn" }, { "input": "t", "output": "t" }, { "input": "N", "output": "N" }, { "input": "kv", "output": "kv" }, { "input": "Ur", "output": "ur" }, { "input": "CN", "output": "CN" } ]

1,696,168,349

2,147,483,647

Python 3

OK

TESTS

30

92

0

a=input() M=0 for i in a: if ord(i)<91: M=M+1 if 2*M>len(a): print(a.upper()) else: print(a.lower())

Title: Word Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. Input Specification: The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. Output Specification: Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. Demo Input: ['HoUse\n', 'ViP\n', 'maTRIx\n'] Demo Output: ['house\n', 'VIP\n', 'matrix\n'] Note: none

```python a=input() M=0 for i in a: if ord(i)<91: M=M+1 if 2*M>len(a): print(a.upper()) else: print(a.lower()) ```

3.977

609

A

USB Flash Drives

PROGRAMMING

800

[ "greedy", "implementation", "sortings" ]

null

Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes. Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.

The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives. The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file. Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes. It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.

Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.

[ "3\n5\n2\n1\n3\n", "3\n6\n2\n3\n2\n", "2\n5\n5\n10\n" ]

[ "2\n", "3\n", "1\n" ]

In the first example Sean needs only two USB flash drives — the first and the third. In the second example Sean needs all three USB flash drives. In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.

0

[ { "input": "3\n5\n2\n1\n3", "output": "2" }, { "input": "3\n6\n2\n3\n2", "output": "3" }, { "input": "2\n5\n5\n10", "output": "1" }, { "input": "5\n16\n8\n1\n3\n4\n9", "output": "2" }, { "input": "10\n121\n10\n37\n74\n56\n42\n39\n6\n68\n8\n100", "output": "2" }, { "input": "12\n4773\n325\n377\n192\n780\n881\n816\n839\n223\n215\n125\n952\n8", "output": "7" }, { "input": "15\n7758\n182\n272\n763\n910\n24\n359\n583\n890\n735\n819\n66\n992\n440\n496\n227", "output": "15" }, { "input": "30\n70\n6\n2\n10\n4\n7\n10\n5\n1\n8\n10\n4\n3\n5\n9\n3\n6\n6\n4\n2\n6\n5\n10\n1\n9\n7\n2\n1\n10\n7\n5", "output": "8" }, { "input": "40\n15705\n702\n722\n105\n873\n417\n477\n794\n300\n869\n496\n572\n232\n456\n298\n473\n584\n486\n713\n934\n121\n303\n956\n934\n840\n358\n201\n861\n497\n131\n312\n957\n96\n914\n509\n60\n300\n722\n658\n820\n103", "output": "21" }, { "input": "50\n18239\n300\n151\n770\n9\n200\n52\n247\n753\n523\n263\n744\n463\n540\n244\n608\n569\n771\n32\n425\n777\n624\n761\n628\n124\n405\n396\n726\n626\n679\n237\n229\n49\n512\n18\n671\n290\n768\n632\n739\n18\n136\n413\n117\n83\n413\n452\n767\n664\n203\n404", "output": "31" }, { "input": "70\n149\n5\n3\n3\n4\n6\n1\n2\n9\n8\n3\n1\n8\n4\n4\n3\n6\n10\n7\n1\n10\n8\n4\n9\n3\n8\n3\n2\n5\n1\n8\n6\n9\n10\n4\n8\n6\n9\n9\n9\n3\n4\n2\n2\n5\n8\n9\n1\n10\n3\n4\n3\n1\n9\n3\n5\n1\n3\n7\n6\n9\n8\n9\n1\n7\n4\n4\n2\n3\n5\n7", "output": "17" }, { "input": "70\n2731\n26\n75\n86\n94\n37\n25\n32\n35\n92\n1\n51\n73\n53\n66\n16\n80\n15\n81\n100\n87\n55\n48\n30\n71\n39\n87\n77\n25\n70\n22\n75\n23\n97\n16\n75\n95\n61\n61\n28\n10\n78\n54\n80\n51\n25\n24\n90\n58\n4\n77\n40\n54\n53\n47\n62\n30\n38\n71\n97\n71\n60\n58\n1\n21\n15\n55\n99\n34\n88\n99", "output": "35" }, { "input": "70\n28625\n34\n132\n181\n232\n593\n413\n862\n887\n808\n18\n35\n89\n356\n640\n339\n280\n975\n82\n345\n398\n948\n372\n91\n755\n75\n153\n948\n603\n35\n694\n722\n293\n363\n884\n264\n813\n175\n169\n646\n138\n449\n488\n828\n417\n134\n84\n763\n288\n845\n801\n556\n972\n332\n564\n934\n699\n842\n942\n644\n203\n406\n140\n37\n9\n423\n546\n675\n491\n113\n587", "output": "45" }, { "input": "80\n248\n3\n9\n4\n5\n10\n7\n2\n6\n2\n2\n8\n2\n1\n3\n7\n9\n2\n8\n4\n4\n8\n5\n4\n4\n10\n2\n1\n4\n8\n4\n10\n1\n2\n10\n2\n3\n3\n1\n1\n8\n9\n5\n10\n2\n8\n10\n5\n3\n6\n1\n7\n8\n9\n10\n5\n10\n10\n2\n10\n1\n2\n4\n1\n9\n4\n7\n10\n8\n5\n8\n1\n4\n2\n2\n3\n9\n9\n9\n10\n6", "output": "27" }, { "input": "80\n2993\n18\n14\n73\n38\n14\n73\n77\n18\n81\n6\n96\n65\n77\n86\n76\n8\n16\n81\n83\n83\n34\n69\n58\n15\n19\n1\n16\n57\n95\n35\n5\n49\n8\n15\n47\n84\n99\n94\n93\n55\n43\n47\n51\n61\n57\n13\n7\n92\n14\n4\n83\n100\n60\n75\n41\n95\n74\n40\n1\n4\n95\n68\n59\n65\n15\n15\n75\n85\n46\n77\n26\n30\n51\n64\n75\n40\n22\n88\n68\n24", "output": "38" }, { "input": "80\n37947\n117\n569\n702\n272\n573\n629\n90\n337\n673\n589\n576\n205\n11\n284\n645\n719\n777\n271\n567\n466\n251\n402\n3\n97\n288\n699\n208\n173\n530\n782\n266\n395\n957\n159\n463\n43\n316\n603\n197\n386\n132\n799\n778\n905\n784\n71\n851\n963\n883\n705\n454\n275\n425\n727\n223\n4\n870\n833\n431\n463\n85\n505\n800\n41\n954\n981\n242\n578\n336\n48\n858\n702\n349\n929\n646\n528\n993\n506\n274\n227", "output": "70" }, { "input": "90\n413\n5\n8\n10\n7\n5\n7\n5\n7\n1\n7\n8\n4\n3\n9\n4\n1\n10\n3\n1\n10\n9\n3\n1\n8\n4\n7\n5\n2\n9\n3\n10\n10\n3\n6\n3\n3\n10\n7\n5\n1\n1\n2\n4\n8\n2\n5\n5\n3\n9\n5\n5\n3\n10\n2\n3\n8\n5\n9\n1\n3\n6\n5\n9\n2\n3\n7\n10\n3\n4\n4\n1\n5\n9\n2\n6\n9\n1\n1\n9\n9\n7\n7\n7\n8\n4\n5\n3\n4\n6\n9", "output": "59" }, { "input": "90\n4226\n33\n43\n83\n46\n75\n14\n88\n36\n8\n25\n47\n4\n96\n19\n33\n49\n65\n17\n59\n72\n1\n55\n94\n92\n27\n33\n39\n14\n62\n79\n12\n89\n22\n86\n13\n19\n77\n53\n96\n74\n24\n25\n17\n64\n71\n81\n87\n52\n72\n55\n49\n74\n36\n65\n86\n91\n33\n61\n97\n38\n87\n61\n14\n73\n95\n43\n67\n42\n67\n22\n12\n62\n32\n96\n24\n49\n82\n46\n89\n36\n75\n91\n11\n10\n9\n33\n86\n28\n75\n39", "output": "64" }, { "input": "90\n40579\n448\n977\n607\n745\n268\n826\n479\n59\n330\n609\n43\n301\n970\n726\n172\n632\n600\n181\n712\n195\n491\n312\n849\n722\n679\n682\n780\n131\n404\n293\n387\n567\n660\n54\n339\n111\n833\n612\n911\n869\n356\n884\n635\n126\n639\n712\n473\n663\n773\n435\n32\n973\n484\n662\n464\n699\n274\n919\n95\n904\n253\n589\n543\n454\n250\n349\n237\n829\n511\n536\n36\n45\n152\n626\n384\n199\n877\n941\n84\n781\n115\n20\n52\n726\n751\n920\n291\n571\n6\n199", "output": "64" }, { "input": "100\n66\n7\n9\n10\n5\n2\n8\n6\n5\n4\n10\n10\n6\n5\n2\n2\n1\n1\n5\n8\n7\n8\n10\n5\n6\n6\n5\n9\n9\n6\n3\n8\n7\n10\n5\n9\n6\n7\n3\n5\n8\n6\n8\n9\n1\n1\n1\n2\n4\n5\n5\n1\n1\n2\n6\n7\n1\n5\n8\n7\n2\n1\n7\n10\n9\n10\n2\n4\n10\n4\n10\n10\n5\n3\n9\n1\n2\n1\n10\n5\n1\n7\n4\n4\n5\n7\n6\n10\n4\n7\n3\n4\n3\n6\n2\n5\n2\n4\n9\n5\n3", "output": "7" }, { "input": "100\n4862\n20\n47\n85\n47\n76\n38\n48\n93\n91\n81\n31\n51\n23\n60\n59\n3\n73\n72\n57\n67\n54\n9\n42\n5\n32\n46\n72\n79\n95\n61\n79\n88\n33\n52\n97\n10\n3\n20\n79\n82\n93\n90\n38\n80\n18\n21\n43\n60\n73\n34\n75\n65\n10\n84\n100\n29\n94\n56\n22\n59\n95\n46\n22\n57\n69\n67\n90\n11\n10\n61\n27\n2\n48\n69\n86\n91\n69\n76\n36\n71\n18\n54\n90\n74\n69\n50\n46\n8\n5\n41\n96\n5\n14\n55\n85\n39\n6\n79\n75\n87", "output": "70" }, { "input": "100\n45570\n14\n881\n678\n687\n993\n413\n760\n451\n426\n787\n503\n343\n234\n530\n294\n725\n941\n524\n574\n441\n798\n399\n360\n609\n376\n525\n229\n995\n478\n347\n47\n23\n468\n525\n749\n601\n235\n89\n995\n489\n1\n239\n415\n122\n671\n128\n357\n886\n401\n964\n212\n968\n210\n130\n871\n360\n661\n844\n414\n187\n21\n824\n266\n713\n126\n496\n916\n37\n193\n755\n894\n641\n300\n170\n176\n383\n488\n627\n61\n897\n33\n242\n419\n881\n698\n107\n391\n418\n774\n905\n87\n5\n896\n835\n318\n373\n916\n393\n91\n460", "output": "78" }, { "input": "100\n522\n1\n5\n2\n4\n2\n6\n3\n4\n2\n10\n10\n6\n7\n9\n7\n1\n7\n2\n5\n3\n1\n5\n2\n3\n5\n1\n7\n10\n10\n4\n4\n10\n9\n10\n6\n2\n8\n2\n6\n10\n9\n2\n7\n5\n9\n4\n6\n10\n7\n3\n1\n1\n9\n5\n10\n9\n2\n8\n3\n7\n5\n4\n7\n5\n9\n10\n6\n2\n9\n2\n5\n10\n1\n7\n7\n10\n5\n6\n2\n9\n4\n7\n10\n10\n8\n3\n4\n9\n3\n6\n9\n10\n2\n9\n9\n3\n4\n1\n10\n2", "output": "74" }, { "input": "100\n32294\n414\n116\n131\n649\n130\n476\n630\n605\n213\n117\n757\n42\n109\n85\n127\n635\n629\n994\n410\n764\n204\n161\n231\n577\n116\n936\n537\n565\n571\n317\n722\n819\n229\n284\n487\n649\n304\n628\n727\n816\n854\n91\n111\n549\n87\n374\n417\n3\n868\n882\n168\n743\n77\n534\n781\n75\n956\n910\n734\n507\n568\n802\n946\n891\n659\n116\n678\n375\n380\n430\n627\n873\n350\n930\n285\n6\n183\n96\n517\n81\n794\n235\n360\n551\n6\n28\n799\n226\n996\n894\n981\n551\n60\n40\n460\n479\n161\n318\n952\n433", "output": "42" }, { "input": "100\n178\n71\n23\n84\n98\n8\n14\n4\n42\n56\n83\n87\n28\n22\n32\n50\n5\n96\n90\n1\n59\n74\n56\n96\n77\n88\n71\n38\n62\n36\n85\n1\n97\n98\n98\n32\n99\n42\n6\n81\n20\n49\n57\n71\n66\n9\n45\n41\n29\n28\n32\n68\n38\n29\n35\n29\n19\n27\n76\n85\n68\n68\n41\n32\n78\n72\n38\n19\n55\n83\n83\n25\n46\n62\n48\n26\n53\n14\n39\n31\n94\n84\n22\n39\n34\n96\n63\n37\n42\n6\n78\n76\n64\n16\n26\n6\n79\n53\n24\n29\n63", "output": "2" }, { "input": "100\n885\n226\n266\n321\n72\n719\n29\n121\n533\n85\n672\n225\n830\n783\n822\n30\n791\n618\n166\n487\n922\n434\n814\n473\n5\n741\n947\n910\n305\n998\n49\n945\n588\n868\n809\n803\n168\n280\n614\n434\n634\n538\n591\n437\n540\n445\n313\n177\n171\n799\n778\n55\n617\n554\n583\n611\n12\n94\n599\n182\n765\n556\n965\n542\n35\n460\n177\n313\n485\n744\n384\n21\n52\n879\n792\n411\n614\n811\n565\n695\n428\n587\n631\n794\n461\n258\n193\n696\n936\n646\n756\n267\n55\n690\n730\n742\n734\n988\n235\n762\n440", "output": "1" }, { "input": "100\n29\n9\n2\n10\n8\n6\n7\n7\n3\n3\n10\n4\n5\n2\n5\n1\n6\n3\n2\n5\n10\n10\n9\n1\n4\n5\n2\n2\n3\n1\n2\n2\n9\n6\n9\n7\n8\n8\n1\n5\n5\n3\n1\n5\n6\n1\n9\n2\n3\n8\n10\n8\n3\n2\n7\n1\n2\n1\n2\n8\n10\n5\n2\n3\n1\n10\n7\n1\n7\n4\n9\n6\n6\n4\n7\n1\n2\n7\n7\n9\n9\n7\n10\n4\n10\n8\n2\n1\n5\n5\n10\n5\n8\n1\n5\n6\n5\n1\n5\n6\n8", "output": "3" }, { "input": "100\n644\n94\n69\n43\n36\n54\n93\n30\n74\n56\n95\n70\n49\n11\n36\n57\n30\n59\n3\n52\n59\n90\n82\n39\n67\n32\n8\n80\n64\n8\n65\n51\n48\n89\n90\n35\n4\n54\n66\n96\n68\n90\n30\n4\n13\n97\n41\n90\n85\n17\n45\n94\n31\n58\n4\n39\n76\n95\n92\n59\n67\n46\n96\n55\n82\n64\n20\n20\n83\n46\n37\n15\n60\n37\n79\n45\n47\n63\n73\n76\n31\n52\n36\n32\n49\n26\n61\n91\n31\n25\n62\n90\n65\n65\n5\n94\n7\n15\n97\n88\n68", "output": "7" }, { "input": "100\n1756\n98\n229\n158\n281\n16\n169\n149\n239\n235\n182\n147\n215\n49\n270\n194\n242\n295\n289\n249\n19\n12\n144\n157\n92\n270\n122\n212\n97\n152\n14\n42\n12\n198\n98\n295\n154\n229\n191\n294\n5\n156\n43\n185\n184\n20\n125\n23\n10\n257\n244\n264\n79\n46\n277\n13\n22\n97\n212\n77\n293\n20\n51\n17\n109\n37\n68\n117\n51\n248\n10\n149\n179\n192\n239\n161\n13\n173\n297\n73\n43\n109\n288\n198\n81\n70\n254\n187\n277\n1\n295\n113\n95\n291\n293\n119\n205\n191\n37\n34\n116", "output": "6" }, { "input": "100\n20562\n721\n452\n11\n703\n376\n183\n197\n203\n406\n642\n346\n446\n256\n760\n201\n360\n702\n707\n388\n779\n653\n610\n497\n768\n670\n134\n780\n306\n661\n180\n259\n256\n362\n6\n121\n415\n747\n170\n67\n439\n728\n193\n622\n481\n38\n225\n343\n303\n253\n436\n305\n68\n794\n247\n291\n600\n750\n188\n199\n757\n28\n776\n749\n253\n351\n53\n629\n129\n578\n209\n89\n651\n262\n638\n353\n469\n31\n144\n460\n176\n535\n562\n366\n639\n234\n577\n364\n761\n617\n303\n450\n778\n311\n289\n221\n274\n239\n626\n194\n36", "output": "30" }, { "input": "100\n32630\n548\n21\n756\n138\n56\n719\n97\n86\n599\n531\n464\n137\n737\n239\n626\n179\n170\n271\n335\n178\n298\n597\n616\n104\n190\n216\n333\n606\n621\n310\n70\n286\n593\n655\n472\n556\n77\n625\n744\n395\n487\n762\n316\n469\n258\n475\n596\n651\n180\n737\n101\n511\n708\n163\n713\n475\n557\n649\n705\n583\n97\n447\n119\n91\n372\n194\n394\n667\n384\n635\n611\n746\n758\n613\n54\n508\n405\n188\n570\n254\n28\n599\n767\n115\n737\n194\n148\n686\n705\n236\n762\n613\n770\n633\n401\n773\n448\n274\n141\n718", "output": "52" }, { "input": "1\n1\n1", "output": "1" } ]

1,567,814,306

2,147,483,647

Python 3

OK

TESTS

34

109

0

n = int(input()) list_size = [] for i in range(n+1): if i == 0: file_size = int(input()) else: size = int(input()) list_size.append(size) list_size.sort(reverse=True) for i in range(n): file_size = file_size - list_size[i] if file_size <= 0: print("{}".format(i+1)) break

Title: USB Flash Drives Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes. Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives. Input Specification: The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives. The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file. Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes. It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*. Output Specification: Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives. Demo Input: ['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n'] Demo Output: ['2\n', '3\n', '1\n'] Note: In the first example Sean needs only two USB flash drives — the first and the third. In the second example Sean needs all three USB flash drives. In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.

```python n = int(input()) list_size = [] for i in range(n+1): if i == 0: file_size = int(input()) else: size = int(input()) list_size.append(size) list_size.sort(reverse=True) for i in range(n): file_size = file_size - list_size[i] if file_size <= 0: print("{}".format(i+1)) break ```

3

641

A

Little Artem and Grasshopper

PROGRAMMING

1,000

[ "implementation" ]

null

Little Artem found a grasshopper. He brought it to his house and constructed a jumping area for him. The area looks like a strip of cells 1<=×<=*n*. Each cell contains the direction for the next jump and the length of that jump. Grasshopper starts in the first cell and follows the instructions written on the cells. Grasshopper stops immediately if it jumps out of the strip. Now Artem wants to find out if this will ever happen.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — length of the strip. Next line contains a string of length *n* which consists of characters "<" and ">" only, that provide the direction of the jump from the corresponding cell. Next line contains *n* integers *d**i* (1<=≤<=*d**i*<=≤<=109) — the length of the jump from the *i*-th cell.

Print "INFINITE" (without quotes) if grasshopper will continue his jumps forever. Otherwise print "FINITE" (without quotes).

[ "2\n><\n1 2\n", "3\n>><\n2 1 1\n" ]

[ "FINITE\n", "INFINITE" ]

In the first sample grasshopper starts from the first cell and jumps to the right on the next cell. When he is in the second cell he needs to jump two cells left so he will jump out of the strip. Second sample grasshopper path is 1 - 3 - 2 - 3 - 2 - 3 and so on. The path is infinite.

500

[ { "input": "2\n><\n1 2", "output": "FINITE" }, { "input": "3\n>><\n2 1 1", "output": "INFINITE" }, { "input": "1\n>\n1000000000", "output": "FINITE" }, { "input": "1\n<\n1000000000", "output": "FINITE" }, { "input": "2\n>>\n1 1", "output": "FINITE" }, { "input": "5\n>><><\n1 2 3 1 2", "output": "FINITE" }, { "input": "5\n>><><\n1 2 2 1 2", "output": "INFINITE" }, { "input": "10\n>>>>>>>>><\n1 1 1 1 1 1 1 1 1 10", "output": "FINITE" }, { "input": "10\n>>>>>>>>><\n1 1 1 1 1 1 1 1 1 5", "output": "INFINITE" }, { "input": "10\n>>>>>>>>><\n1 1 1 1 1 1 1 1 1 1", "output": "INFINITE" }, { "input": "3\n><<\n2 1 1", "output": "INFINITE" }, { "input": "10\n>>>>>>>>>>\n1 1 1 1 1 1 1 1 1 100", "output": "FINITE" }, { "input": "3\n><<\n2 100 2", "output": "INFINITE" }, { "input": "3\n><<\n2 3 2", "output": "INFINITE" }, { "input": "3\n>><\n2 3 2", "output": "INFINITE" }, { "input": "3\n><>\n1 1 1", "output": "INFINITE" }, { "input": "3\n>><\n2 2 2", "output": "INFINITE" }, { "input": "4\n>><<\n2 100 2 100", "output": "INFINITE" }, { "input": "3\n><>\n1 1 3", "output": "INFINITE" }, { "input": "5\n><>>>\n1 1 10 10 10", "output": "INFINITE" }, { "input": "5\n><<<<\n500 1 2 3 4", "output": "FINITE" }, { "input": "4\n><><\n2 1000 1 1", "output": "INFINITE" }, { "input": "3\n><<\n2 10 2", "output": "INFINITE" }, { "input": "3\n><<\n2 5 2", "output": "INFINITE" }, { "input": "2\n><\n5 5", "output": "FINITE" }, { "input": "3\n>><\n1 10 10", "output": "FINITE" }, { "input": "4\n><<>\n2 1 1 5", "output": "INFINITE" }, { "input": "5\n<<>>>\n1 1 1 1 1", "output": "FINITE" }, { "input": "3\n><>\n1 1 12345", "output": "INFINITE" }, { "input": "4\n>>><\n3 2 2 3", "output": "INFINITE" }, { "input": "3\n><>\n1 1 100000", "output": "INFINITE" } ]

1,462,115,713

2,147,483,647

Python 3

OK

TESTS

60

124

12,288,000

n, s = int(input()), input() a = [int(i) for i in input().split()] c = i = 0 while True: if s[i] == '>': i += a[i] else: i -= a[i] if i < 0 or i >= n: print("FINITE") exit() c += 1 if c > n: print("INFINITE") exit()

Title: Little Artem and Grasshopper Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Artem found a grasshopper. He brought it to his house and constructed a jumping area for him. The area looks like a strip of cells 1<=×<=*n*. Each cell contains the direction for the next jump and the length of that jump. Grasshopper starts in the first cell and follows the instructions written on the cells. Grasshopper stops immediately if it jumps out of the strip. Now Artem wants to find out if this will ever happen. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — length of the strip. Next line contains a string of length *n* which consists of characters "<" and ">" only, that provide the direction of the jump from the corresponding cell. Next line contains *n* integers *d**i* (1<=≤<=*d**i*<=≤<=109) — the length of the jump from the *i*-th cell. Output Specification: Print "INFINITE" (without quotes) if grasshopper will continue his jumps forever. Otherwise print "FINITE" (without quotes). Demo Input: ['2\n><\n1 2\n', '3\n>><\n2 1 1\n'] Demo Output: ['FINITE\n', 'INFINITE'] Note: In the first sample grasshopper starts from the first cell and jumps to the right on the next cell. When he is in the second cell he needs to jump two cells left so he will jump out of the strip. Second sample grasshopper path is 1 - 3 - 2 - 3 - 2 - 3 and so on. The path is infinite.

```python n, s = int(input()), input() a = [int(i) for i in input().split()] c = i = 0 while True: if s[i] == '>': i += a[i] else: i -= a[i] if i < 0 or i >= n: print("FINITE") exit() c += 1 if c > n: print("INFINITE") exit() ```

3

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate.

There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.

Write one line — the corresponding answer. Do not omit the leading 0s.

[ "1010100\n0100101\n", "000\n111\n", "1110\n1010\n", "01110\n01100\n" ]

[ "1110001\n", "111\n", "0100\n", "00010\n" ]

none

500

[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": "101010101111010001\n001001111101111101", "output": "100011010010101100" }, { "input": "0110010101111100000\n0011000101000000110", "output": "0101010000111100110" }, { "input": "11110100011101010111\n00001000011011000000", "output": "11111100000110010111" }, { "input": "101010101111101101001\n111010010010000011111", "output": "010000111101101110110" }, { "input": "0000111111100011000010\n1110110110110000001010", "output": "1110001001010011001000" }, { "input": "10010010101000110111000\n00101110100110111000111", "output": "10111100001110001111111" }, { "input": "010010010010111100000111\n100100111111100011001110", "output": "110110101101011111001001" }, { "input": "0101110100100111011010010\n0101100011010111001010001", "output": "0000010111110000010000011" }, { "input": "10010010100011110111111011\n10000110101100000001000100", "output": "00010100001111110110111111" }, { "input": "000001111000000100001000000\n011100111101111001110110001", "output": "011101000101111101111110001" }, { "input": "0011110010001001011001011100\n0000101101000011101011001010", "output": "0011011111001010110010010110" }, { "input": "11111000000000010011001101111\n11101110011001010100010000000", "output": "00010110011001000111011101111" }, { "input": "011001110000110100001100101100\n001010000011110000001000101001", "output": "010011110011000100000100000101" }, { "input": "1011111010001100011010110101111\n1011001110010000000101100010101", "output": "0000110100011100011111010111010" }, { "input": "10111000100001000001010110000001\n10111000001100101011011001011000", "output": "00000000101101101010001111011001" }, { "input": "000001010000100001000000011011100\n111111111001010100100001100000111", "output": "111110101001110101100001111011011" }, { "input": "1101000000000010011011101100000110\n1110000001100010011010000011011110", "output": "0011000001100000000001101111011000" }, { "input": "01011011000010100001100100011110001\n01011010111000001010010100001110000", "output": "00000001111010101011110000010000001" }, { "input": "000011111000011001000110111100000100\n011011000110000111101011100111000111", "output": "011000111110011110101101011011000011" }, { "input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000", "output": "1011001001111001001011101010101000010" }, { "input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011", "output": "10001110000010101110000111000011111110" }, { "input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100", "output": "000100001011110000011101110111010001110" }, { "input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001", "output": "1101110101010110000011000000101011110011" }, { "input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100", "output": "11001011110010010000010111001100001001110" }, { "input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110", "output": "001100101000011111111101111011101010111001" }, { "input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001", "output": "0111010010100110110101100010000100010100000" }, { "input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100", "output": "11111110000000100101000100110111001100011001" }, { "input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011", "output": "101011011100100010100011011001101010100100010" }, { "input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001", "output": "1101001100111011010111110110101111001011110111" }, { "input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001", "output": "10010101000101000000011010011110011110011110001" }, { "input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100", "output": "011011011100000000010101110010000000101000111101" }, { "input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100", "output": "0101010111101001011011110110011101010101010100011" }, { "input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011", "output": "11001011010010111000010110011101100100001110111111" }, { "input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011", "output": "111011101010011100001111101001101011110010010110001" }, { "input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001", "output": "0100111110110011111110010010010000110111100101101101" }, { "input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100", "output": "01011001110111010111001100010011010100010000111011000" }, { "input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111", "output": "100011101001001000011011011001111000100000010100100100" }, { "input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110", "output": "1100110010000101101010111111101001001001110101110010110" }, { "input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110", "output": "01000111100111001011110010100011111111110010101100001101" }, { "input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010", "output": "110001010001000011000101110101000100001011111001011001001" }, { "input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111", "output": "1110100010111000101001001011101110011111100111000011011011" }, { "input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110", "output": "01110110101110100100110011010000001000101100101111000111011" }, { "input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011", "output": "111100101000000011101011011001110010101111000110010010000000" }, { "input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111", "output": "0100100010111110010011101010000011111110001110010110010111001" }, { "input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111", "output": "00110100000011001101101100100010110010001100000001100110011101" }, { "input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011", "output": "000000011000111011110011101000010000010100101000000011010110010" }, { "input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010", "output": "0010100110110100111100100100101101010100100111011010001001010101" }, { "input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111", "output": "11010110111100101111101001100001110100010110010110110111100110100" }, { "input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111", "output": "111111010011011100101110100110111111111001111110011010111111110000" }, { "input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110", "output": "1010101010100010001001001001100000111000010010010100010011000100000" }, { "input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000", "output": "00011111011111001000011100010011100011010100101011011000001001111110" }, { "input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111", "output": "001111000011001110100111010101111111011100110011001010010010000111011" }, { "input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101", "output": "0110001100110100010000110111000010011010011000011001010011010100010100" }, { "input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010", "output": "00010000000110110101000011001000000100100110111010011111101010001010000" }, { "input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001", "output": "000100100000000110011100100001010110101001100101110010010011111001110111" }, { "input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000", "output": "1000111100010011010110011101000000101010101100011111100001101111001010010" }, { "input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011", "output": "01000010011110111001011011110000001011000111101101101010010110001010100100" }, { "input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101", "output": "101101110110110010011100001011111100100001110000101100110000100011011100110" }, { "input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001", "output": "1110111111110010111000001100101010101011010100101010100101100011001001111111" }, { "input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111", "output": "10111011000111000101110100101000100111011011100011110110000101010001111010111" }, { "input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110", "output": "110100100110000101010010011010011001100110000111010000010100001011110111111101" }, { "input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111", "output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111" }, { "input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001", "output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001" }, { "input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110", "output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011" }, { "input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111", "output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101" }, { "input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010", "output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101" }, { "input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000", "output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001" }, { "input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011", "output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110" }, { "input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011", "output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011" }, { "input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011", "output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011" }, { "input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010", "output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011" }, { "input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001", "output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110" }, { "input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111", "output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010" }, { "input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011", "output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011" }, { "input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100", "output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000" }, { "input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001", "output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101" }, { "input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110", "output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101" }, { "input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011", "output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110" }, { "input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111", "output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110" }, { "input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000", "output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001" }, { "input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010", "output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101" }, { "input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101", "output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011" }, { "input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011", "output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010", "output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011" }, { "input": "0\n0", "output": "0" }, { "input": "0\n1", "output": "1" } ]

1,592,416,339

2,147,483,647

PyPy 3

OK

TESTS

102

156

0

def func(x,y): s = '' for i in range(len(x)): s += str(int(x[i])^int(y[i])) print(s) x = input() y = input() func(x,y)

Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none

```python def func(x,y): s = '' for i in range(len(x)): s += str(int(x[i])^int(y[i])) print(s) x = input() y = input() func(x,y) ```

3.961

13

A

Numbers

PROGRAMMING

1,000

[ "implementation", "math" ]

A. Numbers

1

64

Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18. Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1. Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.

Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).

Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.

[ "5\n", "3\n" ]

[ "7/3\n", "2/1\n" ]

In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.

0

[ { "input": "5", "output": "7/3" }, { "input": "3", "output": "2/1" }, { "input": "1000", "output": "90132/499" }, { "input": "927", "output": "155449/925" }, { "input": "260", "output": "6265/129" }, { "input": "131", "output": "3370/129" }, { "input": "386", "output": "857/12" }, { "input": "277", "output": "2864/55" }, { "input": "766", "output": "53217/382" }, { "input": "28", "output": "85/13" }, { "input": "406", "output": "7560/101" }, { "input": "757", "output": "103847/755" }, { "input": "6", "output": "9/4" }, { "input": "239", "output": "10885/237" }, { "input": "322", "output": "2399/40" }, { "input": "98", "output": "317/16" }, { "input": "208", "output": "4063/103" }, { "input": "786", "output": "55777/392" }, { "input": "879", "output": "140290/877" }, { "input": "702", "output": "89217/700" }, { "input": "948", "output": "7369/43" }, { "input": "537", "output": "52753/535" }, { "input": "984", "output": "174589/982" }, { "input": "934", "output": "157951/932" }, { "input": "726", "output": "95491/724" }, { "input": "127", "output": "3154/125" }, { "input": "504", "output": "23086/251" }, { "input": "125", "output": "3080/123" }, { "input": "604", "output": "33178/301" }, { "input": "115", "output": "2600/113" }, { "input": "27", "output": "167/25" }, { "input": "687", "output": "85854/685" }, { "input": "880", "output": "69915/439" }, { "input": "173", "output": "640/19" }, { "input": "264", "output": "6438/131" }, { "input": "785", "output": "111560/783" }, { "input": "399", "output": "29399/397" }, { "input": "514", "output": "6031/64" }, { "input": "381", "output": "26717/379" }, { "input": "592", "output": "63769/590" }, { "input": "417", "output": "32002/415" }, { "input": "588", "output": "62723/586" }, { "input": "852", "output": "131069/850" }, { "input": "959", "output": "5059/29" }, { "input": "841", "output": "127737/839" }, { "input": "733", "output": "97598/731" }, { "input": "692", "output": "87017/690" }, { "input": "69", "output": "983/67" }, { "input": "223", "output": "556/13" }, { "input": "93", "output": "246/13" }, { "input": "643", "output": "75503/641" }, { "input": "119", "output": "2833/117" }, { "input": "498", "output": "1459/16" }, { "input": "155", "output": "4637/153" }, { "input": "305", "output": "17350/303" }, { "input": "454", "output": "37893/452" }, { "input": "88", "output": "1529/86" }, { "input": "850", "output": "32645/212" }, { "input": "474", "output": "20581/236" }, { "input": "309", "output": "17731/307" }, { "input": "762", "output": "105083/760" }, { "input": "591", "output": "63761/589" }, { "input": "457", "output": "38317/455" }, { "input": "141", "output": "3832/139" }, { "input": "385", "output": "27232/383" }, { "input": "387", "output": "27628/385" }, { "input": "469", "output": "40306/467" }, { "input": "624", "output": "35285/311" }, { "input": "330", "output": "487/8" }, { "input": "31", "output": "222/29" }, { "input": "975", "output": "171679/973" }, { "input": "584", "output": "62183/582" }, { "input": "668", "output": "81127/666" }, { "input": "331", "output": "20297/329" }, { "input": "189", "output": "6789/187" }, { "input": "251", "output": "11939/249" }, { "input": "876", "output": "69196/437" }, { "input": "615", "output": "68987/613" }, { "input": "451", "output": "37258/449" }, { "input": "499", "output": "45727/497" }, { "input": "699", "output": "89117/697" }, { "input": "619", "output": "70019/617" }, { "input": "413", "output": "10515/137" }, { "input": "197", "output": "7399/195" }, { "input": "794", "output": "14281/99" }, { "input": "659", "output": "79403/657" }, { "input": "653", "output": "77695/651" }, { "input": "23", "output": "45/7" }, { "input": "430", "output": "16985/214" }, { "input": "249", "output": "11659/247" }, { "input": "837", "output": "126869/835" }, { "input": "258", "output": "12373/256" }, { "input": "995", "output": "59665/331" }, { "input": "102", "output": "504/25" }, { "input": "989", "output": "177124/987" }, { "input": "376", "output": "13008/187" }, { "input": "657", "output": "15715/131" }, { "input": "746", "output": "50509/372" }, { "input": "602", "output": "13177/120" } ]

1,678,432,528

2,147,483,647

Python 3

OK

TESTS

99

92

0

import math def convert_base(n, base): digits = [] while n > 0: digits.append(n % base) n //= base return digits def sum_digits(n): return sum(n) A = int(input()) total_sum = 0 count = 0 for base in range(2, A): digits = convert_base(A, base) if digits: total_sum += sum_digits(digits) count += 1 if count > 0: gcd = math.gcd(total_sum, count) numerator = total_sum // gcd denominator = count // gcd print(f"{numerator}/{denominator}") else: print("0")

Title: Numbers Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18. Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1. Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10. Input Specification: Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000). Output Specification: Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator. Demo Input: ['5\n', '3\n'] Demo Output: ['7/3\n', '2/1\n'] Note: In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.

```python import math def convert_base(n, base): digits = [] while n > 0: digits.append(n % base) n //= base return digits def sum_digits(n): return sum(n) A = int(input()) total_sum = 0 count = 0 for base in range(2, A): digits = convert_base(A, base) if digits: total_sum += sum_digits(digits) count += 1 if count > 0: gcd = math.gcd(total_sum, count) numerator = total_sum // gcd denominator = count // gcd print(f"{numerator}/{denominator}") else: print("0") ```

3.954

63

A

Sinking Ship

PROGRAMMING

900

[ "implementation", "sortings", "strings" ]

A. Sinking Ship

2

256

The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically: The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last. If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less). For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew.

The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain.

Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship.

[ "6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n" ]

[ "Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n" ]

none

500

[ { "input": "6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman", "output": "Teddy\nAlice\nBob\nJulia\nCharlie\nJack" }, { "input": "1\nA captain", "output": "A" }, { "input": "1\nAbcdefjhij captain", "output": "Abcdefjhij" }, { "input": "5\nA captain\nB man\nD woman\nC child\nE rat", "output": "E\nD\nC\nB\nA" }, { "input": "10\nCap captain\nD child\nC woman\nA woman\nE child\nMan man\nB child\nF woman\nRat rat\nRatt rat", "output": "Rat\nRatt\nD\nC\nA\nE\nB\nF\nMan\nCap" }, { "input": "5\nJoyxnkypf captain\nDxssgr woman\nKeojmnpd rat\nGdv man\nHnw man", "output": "Keojmnpd\nDxssgr\nGdv\nHnw\nJoyxnkypf" }, { "input": "11\nJue rat\nWyglbyphk rat\nGjlgu child\nGi man\nAttx rat\nTheorpkgx man\nYm rat\nX child\nB captain\nEnualf rat\nKktsgyuyv woman", "output": "Jue\nWyglbyphk\nAttx\nYm\nEnualf\nGjlgu\nX\nKktsgyuyv\nGi\nTheorpkgx\nB" }, { "input": "22\nWswwcvvm woman\nBtmfats rat\nI rat\nOcmtsnwx man\nUrcqv rat\nYghnogt woman\nWtyfc man\nWqle child\nUjfrelpu rat\nDstixj man\nAhksnio woman\nKhkvaap woman\nSjppvwm rat\nEgdmsv rat\nDank rat\nNquicjnw rat\nLh captain\nTdyaqaqln rat\nQtj rat\nTfgwijvq rat\nNbiso child\nNqthvbf woman", "output": "Btmfats\nI\nUrcqv\nUjfrelpu\nSjppvwm\nEgdmsv\nDank\nNquicjnw\nTdyaqaqln\nQtj\nTfgwijvq\nWswwcvvm\nYghnogt\nWqle\nAhksnio\nKhkvaap\nNbiso\nNqthvbf\nOcmtsnwx\nWtyfc\nDstixj\nLh" }, { "input": "36\nKqxmtwmsf child\nIze woman\nDlpr child\nK woman\nF captain\nRjwfeuhba rat\nBbv rat\nS rat\nMnmg woman\nSmzyx woman\nSr man\nQmhroracn rat\nSoqpuqock rat\nPibdq man\nIlrkrptx rat\nZaecfyqka man\nMmersfs child\nVvvocqi man\nHjeqxvq rat\nMpmb woman\nWmgu woman\nCerelmhoxi child\nA man\nDylv man\nXrdgmmtcpq woman\nXj woman\nCeh child\nOfccguo man\nNevr man\nAknkwdkq child\nBmo man\nLqo woman\nVej child\nT child\nRlijc man\nNsdtpdg woman", "output": "Rjwfeuhba\nBbv\nS\nQmhroracn\nSoqpuqock\nIlrkrptx\nHjeqxvq\nKqxmtwmsf\nIze\nDlpr\nK\nMnmg\nSmzyx\nMmersfs\nMpmb\nWmgu\nCerelmhoxi\nXrdgmmtcpq\nXj\nCeh\nAknkwdkq\nLqo\nVej\nT\nNsdtpdg\nSr\nPibdq\nZaecfyqka\nVvvocqi\nA\nDylv\nOfccguo\nNevr\nBmo\nRlijc\nF" }, { "input": "2\nA captain\nB man", "output": "B\nA" }, { "input": "2\nB woman\nA captain", "output": "B\nA" }, { "input": "2\nX child\nY captain", "output": "X\nY" }, { "input": "2\nX captain\nY rat", "output": "Y\nX" }, { "input": "6\nAbdelrhman captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman", "output": "Teddy\nAlice\nBob\nJulia\nCharlie\nAbdelrhman" }, { "input": "4\nAratos captain\nMind child\nAman woman\nCarrot man", "output": "Mind\nAman\nCarrot\nAratos" } ]

1,619,277,401

2,147,483,647

PyPy 3

OK

TESTS

26

216

0

n=int(input()) rat=[] wc=[] man=[] cap=[] for i in range(n): s=input().split() if(s[1]=="rat"): rat.append(s[0]) elif(s[1]=="woman" or s[1]=="child"): wc.append(s[0]) elif(s[1]=="man"): man.append(s[0]) else: cap.append(s[0]) for i in rat: print(i) for i in wc: print(i) for i in man: print(i) for i in cap: print(i)

Title: Sinking Ship Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically: The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last. If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less). For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew. Input Specification: The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain. Output Specification: Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship. Demo Input: ['6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n'] Demo Output: ['Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n'] Note: none

```python n=int(input()) rat=[] wc=[] man=[] cap=[] for i in range(n): s=input().split() if(s[1]=="rat"): rat.append(s[0]) elif(s[1]=="woman" or s[1]=="child"): wc.append(s[0]) elif(s[1]=="man"): man.append(s[0]) else: cap.append(s[0]) for i in rat: print(i) for i in wc: print(i) for i in man: print(i) for i in cap: print(i) ```

3.946

25

B

Phone numbers

PROGRAMMING

1,100

[ "implementation" ]

B. Phone numbers

2

256

Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.

The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups.

Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.

[ "6\n549871\n", "7\n1198733\n" ]

[ "54-98-71", "11-987-33\n" ]

none

0

[ { "input": "6\n549871", "output": "54-98-71" }, { "input": "7\n1198733", "output": "119-87-33" }, { "input": "2\n74", "output": "74" }, { "input": "2\n33", "output": "33" }, { "input": "3\n074", "output": "074" }, { "input": "3\n081", "output": "081" }, { "input": "4\n3811", "output": "38-11" }, { "input": "5\n21583", "output": "215-83" }, { "input": "8\n33408349", "output": "33-40-83-49" }, { "input": "9\n988808426", "output": "988-80-84-26" }, { "input": "10\n0180990956", "output": "01-80-99-09-56" }, { "input": "15\n433488906230138", "output": "433-48-89-06-23-01-38" }, { "input": "22\n7135498415686025907059", "output": "71-35-49-84-15-68-60-25-90-70-59" }, { "input": "49\n2429965524999668169991253653390090510755018570235", "output": "242-99-65-52-49-99-66-81-69-99-12-53-65-33-90-09-05-10-75-50-18-57-02-35" }, { "input": "72\n491925337784111770500147619881727525570039735507439360627744863794794290", "output": "49-19-25-33-77-84-11-17-70-50-01-47-61-98-81-72-75-25-57-00-39-73-55-07-43-93-60-62-77-44-86-37-94-79-42-90" }, { "input": "95\n32543414456047900690980198395035321172843693417425457554204776648220562494524275489599199209210", "output": "325-43-41-44-56-04-79-00-69-09-80-19-83-95-03-53-21-17-28-43-69-34-17-42-54-57-55-42-04-77-66-48-22-05-62-49-45-24-27-54-89-59-91-99-20-92-10" }, { "input": "97\n9362344595153688016434451101547661156123505108492010669557671355055642365998461003851354321478898", "output": "936-23-44-59-51-53-68-80-16-43-44-51-10-15-47-66-11-56-12-35-05-10-84-92-01-06-69-55-76-71-35-50-55-64-23-65-99-84-61-00-38-51-35-43-21-47-88-98" }, { "input": "98\n65521815795893886057122984634320900545031770769333931308009346017867969790810907868670369236928568", "output": "65-52-18-15-79-58-93-88-60-57-12-29-84-63-43-20-90-05-45-03-17-70-76-93-33-93-13-08-00-93-46-01-78-67-96-97-90-81-09-07-86-86-70-36-92-36-92-85-68" }, { "input": "99\n455213856470326729480192345541970106407563996625458559297407682539801838244443866898560852503660390", "output": "455-21-38-56-47-03-26-72-94-80-19-23-45-54-19-70-10-64-07-56-39-96-62-54-58-55-92-97-40-76-82-53-98-01-83-82-44-44-38-66-89-85-60-85-25-03-66-03-90" }, { "input": "100\n4004223124942730640235383244438257614581534320356060987241659784249551110165034719443327659510644224", "output": "40-04-22-31-24-94-27-30-64-02-35-38-32-44-43-82-57-61-45-81-53-43-20-35-60-60-98-72-41-65-97-84-24-95-51-11-01-65-03-47-19-44-33-27-65-95-10-64-42-24" } ]

1,681,686,646

2,147,483,647

PyPy 3-64

OK

TESTS

20

154

0

n = int(input()) phone_number = input() if n % 2 == 0: groups = [phone_number[i:i+2] for i in range(0, n, 2)] else: groups = [phone_number[0:3]] + [phone_number[i:i+2] for i in range(3, n, 2)] print("-".join(groups))

Title: Phone numbers Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits. Input Specification: The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups. Output Specification: Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any. Demo Input: ['6\n549871\n', '7\n1198733\n'] Demo Output: ['54-98-71', '11-987-33\n'] Note: none

```python n = int(input()) phone_number = input() if n % 2 == 0: groups = [phone_number[i:i+2] for i in range(0, n, 2)] else: groups = [phone_number[0:3]] + [phone_number[i:i+2] for i in range(3, n, 2)] print("-".join(groups)) ```

3.9615

368

B

Sereja and Suffixes

PROGRAMMING

1,100

[ "data structures", "dp" ]

null

Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.? Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.

The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements. Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).

Print *m* lines — on the *i*-th line print the answer to the number *l**i*.

[ "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n" ]

[ "6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n" ]

none

1,000

[ { "input": "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10", "output": "6\n6\n6\n6\n6\n5\n4\n3\n2\n1" }, { "input": "8 3\n8 6 4 3 4 2 4 8\n6\n4\n2", "output": "3\n4\n5" }, { "input": "7 10\n1 3 8 6 2 2 7\n4\n2\n6\n3\n4\n4\n6\n2\n7\n4", "output": "3\n5\n2\n4\n3\n3\n2\n5\n1\n3" }, { "input": "10 2\n2 6 5 7 2 2 3 2 4 8\n1\n2", "output": "7\n7" }, { "input": "7 1\n68346 10956 76708 23018 84063 34833 80407\n1", "output": "7" }, { "input": "2 2\n8 4\n1\n1", "output": "2\n2" }, { "input": "1 5\n5\n1\n1\n1\n1\n1", "output": "1\n1\n1\n1\n1" }, { "input": "4 7\n3 1 4 2\n4\n1\n2\n3\n2\n4\n4", "output": "1\n4\n3\n2\n3\n1\n1" }, { "input": "4 3\n9 1 7 1\n1\n4\n2", "output": "3\n1\n2" }, { "input": "8 3\n9280 6676 2720 6172 8329 10413 3975 1394\n5\n6\n7", "output": "4\n3\n2" }, { "input": "1 1\n1\n1", "output": "1" } ]

1,667,951,516

2,147,483,647

PyPy 3-64

OK

TESTS

30

998

16,896,000

n,m=map(int,input().split()) l=list(map(int,input().split())) l.reverse() s,l1={l[0]},[1] for i in range(1,len(l)): if l[i] not in s: s.add(l[i]) l1.append(l1[i-1]+1) else: l1.append(l1[i-1]) for j in range(m): x=int(input()) print(l1[n-x])

Title: Sereja and Suffixes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.? Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements. Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*). Output Specification: Print *m* lines — on the *i*-th line print the answer to the number *l**i*. Demo Input: ['10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n'] Demo Output: ['6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n'] Note: none

```python n,m=map(int,input().split()) l=list(map(int,input().split())) l.reverse() s,l1={l[0]},[1] for i in range(1,len(l)): if l[i] not in s: s.add(l[i]) l1.append(l1[i-1]+1) else: l1.append(l1[i-1]) for j in range(m): x=int(input()) print(l1[n-x]) ```

3

168

A

Wizards and Demonstration

PROGRAMMING

900

[ "implementation", "math" ]

null

Some country is populated by wizards. They want to organize a demonstration. There are *n* people living in the city, *x* of them are the wizards who will surely go to the demonstration. Other city people (*n*<=-<=*x* people) do not support the wizards and aren't going to go to the demonstration. We know that the city administration will react only to the demonstration involving at least *y* percent of the city people. Having considered the matter, the wizards decided to create clone puppets which can substitute the city people on the demonstration. So all in all, the demonstration will involve only the wizards and their puppets. The city administration cannot tell the difference between a puppet and a person, so, as they calculate the percentage, the administration will consider the city to be consisting of only *n* people and not containing any clone puppets. Help the wizards and find the minimum number of clones to create to that the demonstration had no less than *y* percent of the city people.

The first line contains three space-separated integers, *n*, *x*, *y* (1<=≤<=*n*,<=*x*,<=*y*<=≤<=104,<=*x*<=≤<=*n*) — the number of citizens in the city, the number of wizards and the percentage the administration needs, correspondingly. Please note that *y* can exceed 100 percent, that is, the administration wants to see on a demonstration more people that actually live in the city (<=><=*n*).

Print a single integer — the answer to the problem, the minimum number of clones to create, so that the demonstration involved no less than *y* percent of *n* (the real total city population).

[ "10 1 14\n", "20 10 50\n", "1000 352 146\n" ]

[ "1\n", "0\n", "1108\n" ]

In the first sample it is necessary that at least 14% of 10 people came to the demonstration. As the number of people should be integer, then at least two people should come. There is only one wizard living in the city and he is going to come. That isn't enough, so he needs to create one clone. In the second sample 10 people should come to the demonstration. The city has 10 wizards. They will all come to the demonstration, so nobody has to create any clones.

500

[ { "input": "10 1 14", "output": "1" }, { "input": "20 10 50", "output": "0" }, { "input": "1000 352 146", "output": "1108" }, { "input": "68 65 20", "output": "0" }, { "input": "78 28 27", "output": "0" }, { "input": "78 73 58", "output": "0" }, { "input": "70 38 66", "output": "9" }, { "input": "54 4 38", "output": "17" }, { "input": "3 1 69", "output": "2" }, { "input": "11 9 60", "output": "0" }, { "input": "71 49 65", "output": "0" }, { "input": "78 55 96", "output": "20" }, { "input": "2765 768 9020", "output": "248635" }, { "input": "3478 1728 9727", "output": "336578" }, { "input": "9678 6173 5658", "output": "541409" }, { "input": "1138 570 6666", "output": "75290" }, { "input": "7754 204 9038", "output": "700603" }, { "input": "8403 7401 4769", "output": "393339" }, { "input": "4909 2111 8860", "output": "432827" }, { "input": "6571 6449 8965", "output": "582642" }, { "input": "9178 2255 7996", "output": "731618" }, { "input": "6151 6148 3746", "output": "224269" }, { "input": "8890 5449 8734", "output": "771004" }, { "input": "7879 2590 2818", "output": "219441" }, { "input": "8484 6400 547", "output": "40008" }, { "input": "7835 6710 1639", "output": "121706" }, { "input": "2379 1436 9663", "output": "228447" }, { "input": "6487 5670 8", "output": "0" }, { "input": "4470 2543 6", "output": "0" }, { "input": "7261 5328 10", "output": "0" }, { "input": "3871 3795 7", "output": "0" }, { "input": "9620 6557 6", "output": "0" }, { "input": "4890 1112 5", "output": "0" }, { "input": "1 1 1", "output": "0" }, { "input": "1 1 10000", "output": "99" }, { "input": "10000 1 1", "output": "99" }, { "input": "10000 1 10000", "output": "999999" }, { "input": "10000 10000 1", "output": "0" }, { "input": "10000 10000 10000", "output": "990000" }, { "input": "67 1 3", "output": "2" }, { "input": "7878 4534 9159", "output": "717013" }, { "input": "1 1 1", "output": "0" }, { "input": "3271 5 50", "output": "1631" } ]

1,612,370,586

2,147,483,647

Python 3

OK

TESTS

43

154

0

a,b,c=map(int,input().split()) x=a*c y=0 if x%100!=0: y=1 y+=(int(x/100)) if b>=y:print(0) else:print(y-b)

Title: Wizards and Demonstration Time Limit: None seconds Memory Limit: None megabytes Problem Description: Some country is populated by wizards. They want to organize a demonstration. There are *n* people living in the city, *x* of them are the wizards who will surely go to the demonstration. Other city people (*n*<=-<=*x* people) do not support the wizards and aren't going to go to the demonstration. We know that the city administration will react only to the demonstration involving at least *y* percent of the city people. Having considered the matter, the wizards decided to create clone puppets which can substitute the city people on the demonstration. So all in all, the demonstration will involve only the wizards and their puppets. The city administration cannot tell the difference between a puppet and a person, so, as they calculate the percentage, the administration will consider the city to be consisting of only *n* people and not containing any clone puppets. Help the wizards and find the minimum number of clones to create to that the demonstration had no less than *y* percent of the city people. Input Specification: The first line contains three space-separated integers, *n*, *x*, *y* (1<=≤<=*n*,<=*x*,<=*y*<=≤<=104,<=*x*<=≤<=*n*) — the number of citizens in the city, the number of wizards and the percentage the administration needs, correspondingly. Please note that *y* can exceed 100 percent, that is, the administration wants to see on a demonstration more people that actually live in the city (<=><=*n*). Output Specification: Print a single integer — the answer to the problem, the minimum number of clones to create, so that the demonstration involved no less than *y* percent of *n* (the real total city population). Demo Input: ['10 1 14\n', '20 10 50\n', '1000 352 146\n'] Demo Output: ['1\n', '0\n', '1108\n'] Note: In the first sample it is necessary that at least 14% of 10 people came to the demonstration. As the number of people should be integer, then at least two people should come. There is only one wizard living in the city and he is going to come. That isn't enough, so he needs to create one clone. In the second sample 10 people should come to the demonstration. The city has 10 wizards. They will all come to the demonstration, so nobody has to create any clones.

```python a,b,c=map(int,input().split()) x=a*c y=0 if x%100!=0: y=1 y+=(int(x/100)) if b>=y:print(0) else:print(y-b) ```

3

764

A

Taymyr is calling you

PROGRAMMING

800

[ "brute force", "implementation", "math" ]

null

Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist. Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.

The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).

Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.

[ "1 1 10\n", "1 2 5\n", "2 3 9\n" ]

[ "10\n", "2\n", "1\n" ]

Taymyr is a place in the north of Russia. In the first test the artists come each minute, as well as the calls, so we need to kill all of them. In the second test we need to kill artists which come on the second and the fourth minutes. In the third test — only the artist which comes on the sixth minute.

500

[ { "input": "1 1 10", "output": "10" }, { "input": "1 2 5", "output": "2" }, { "input": "2 3 9", "output": "1" }, { "input": "4 8 9", "output": "1" }, { "input": "7 9 2", "output": "0" }, { "input": "10000 10000 10000", "output": "1" }, { "input": "24 22 9235", "output": "34" }, { "input": "74 8 417", "output": "1" }, { "input": "972 1 203", "output": "0" }, { "input": "550 1 754", "output": "1" }, { "input": "860 1 884", "output": "1" }, { "input": "358 2 809", "output": "2" }, { "input": "33 27 216", "output": "0" }, { "input": "2940 1 9311", "output": "3" }, { "input": "4624 1 1953", "output": "0" }, { "input": "2696 2 7345", "output": "2" }, { "input": "3443 2 6701", "output": "0" }, { "input": "3 613 2275", "output": "1" }, { "input": "1 10000 10000", "output": "1" }, { "input": "10000 1 10000", "output": "1" }, { "input": "1 1 1", "output": "1" }, { "input": "1 1 10000", "output": "10000" }, { "input": "34 27 10000", "output": "10" }, { "input": "2 2 9999", "output": "4999" }, { "input": "2 2 1", "output": "0" }, { "input": "6 4 36", "output": "3" }, { "input": "33 6 3005", "output": "45" }, { "input": "5 1 20", "output": "4" }, { "input": "1 2 10", "output": "5" }, { "input": "2 1 100", "output": "50" }, { "input": "10 20 10000", "output": "500" }, { "input": "8 12 12", "output": "0" } ]

1,656,607,025

2,147,483,647

Python 3

OK

TESTS

32

46

0

a = list(map(int,input().split())) count = 0 for i in range(1,a[2]+1): if(i%a[0]==0 and i%a[1]==0): count = count + 1 print(count)

Title: Taymyr is calling you Time Limit: None seconds Memory Limit: None megabytes Problem Description: Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist. Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. Input Specification: The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). Output Specification: Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. Demo Input: ['1 1 10\n', '1 2 5\n', '2 3 9\n'] Demo Output: ['10\n', '2\n', '1\n'] Note: Taymyr is a place in the north of Russia. In the first test the artists come each minute, as well as the calls, so we need to kill all of them. In the second test we need to kill artists which come on the second and the fourth minutes. In the third test — only the artist which comes on the sixth minute.

```python a = list(map(int,input().split())) count = 0 for i in range(1,a[2]+1): if(i%a[0]==0 and i%a[1]==0): count = count + 1 print(count) ```

3

50

A

Domino piling

PROGRAMMING

800

[ "greedy", "math" ]

A. Domino piling

2

256

You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions.

In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).

Output one number — the maximal number of dominoes, which can be placed.

[ "2 4\n", "3 3\n" ]

[ "4\n", "4\n" ]

none

500

[ { "input": "2 4", "output": "4" }, { "input": "3 3", "output": "4" }, { "input": "1 5", "output": "2" }, { "input": "1 6", "output": "3" }, { "input": "1 15", "output": "7" }, { "input": "1 16", "output": "8" }, { "input": "2 5", "output": "5" }, { "input": "2 6", "output": "6" }, { "input": "2 7", "output": "7" }, { "input": "2 14", "output": "14" }, { "input": "2 15", "output": "15" }, { "input": "1 4", "output": "2" }, { "input": "2 16", "output": "16" }, { "input": "3 5", "output": "7" }, { "input": "3 6", "output": "9" }, { "input": "3 10", "output": "15" }, { "input": "3 14", "output": "21" }, { "input": "3 15", "output": "22" }, { "input": "3 16", "output": "24" }, { "input": "5 7", "output": "17" }, { "input": "16 16", "output": "128" }, { "input": "15 16", "output": "120" }, { "input": "2 3", "output": "3" }, { "input": "15 15", "output": "112" }, { "input": "14 16", "output": "112" }, { "input": "11 13", "output": "71" }, { "input": "5 16", "output": "40" }, { "input": "8 15", "output": "60" }, { "input": "2 2", "output": "2" }, { "input": "3 4", "output": "6" }, { "input": "4 4", "output": "8" }, { "input": "1 1", "output": "0" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "14 15", "output": "105" } ]

1,613,657,754

2,147,483,647

Python 3

OK

TESTS

35

124

0

dim=input() m,n = dim.split(" ") m=int(m) n=int(n) result = (m*n)//2 print(result)

Title: Domino piling Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions: 1. Each domino completely covers two squares. 2. No two dominoes overlap. 3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board. Find the maximum number of dominoes, which can be placed under these restrictions. Input Specification: In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). Output Specification: Output one number — the maximal number of dominoes, which can be placed. Demo Input: ['2 4\n', '3 3\n'] Demo Output: ['4\n', '4\n'] Note: none

```python dim=input() m,n = dim.split(" ") m=int(m) n=int(n) result = (m*n)//2 print(result) ```

3.969

263

A

Beautiful Matrix

PROGRAMMING

800

[ "implementation" ]

null

You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix: 1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5). You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.

The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.

Print a single integer — the minimum number of moves needed to make the matrix beautiful.

[ "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n", "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n" ]

[ "3\n", "1\n" ]

none

500

[ { "input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "3" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "1" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "0" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1", "output": "4" }, { "input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "3" }, { "input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "4" }, { "input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "2" }, { "input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "3" }, { "input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "4" }, { "input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "3" }, { "input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "2" }, { "input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "1" }, { "input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "2" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0", "output": "2" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0", "output": "1" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0", "output": "2" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0", "output": "3" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0", "output": "2" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0", "output": "1" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0", "output": "2" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0", "output": "3" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0", "output": "4" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0", "output": "3" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0", "output": "2" }, { "input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0", "output": "3" } ]

1,695,725,568

2,147,483,647

Python 3

OK

TESTS

25

92

0

def search(): for i in range(5): line=input().split() for j in range(5): if line[j]=='1': return abs(i-2)+abs(j-2) print(search())

Title: Beautiful Matrix Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix: 1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5). You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful. Input Specification: The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one. Output Specification: Print a single integer — the minimum number of moves needed to make the matrix beautiful. Demo Input: ['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n'] Demo Output: ['3\n', '1\n'] Note: none

```python def search(): for i in range(5): line=input().split() for j in range(5): if line[j]=='1': return abs(i-2)+abs(j-2) print(search()) ```

3

43

A

Football

PROGRAMMING

1,000

[ "strings" ]

A. Football

2

256

One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.

Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.

[ "1\nABC\n", "5\nA\nABA\nABA\nA\nA\n" ]

[ "ABC\n", "A\n" ]

none

500

[ { "input": "1\nABC", "output": "ABC" }, { "input": "5\nA\nABA\nABA\nA\nA", "output": "A" }, { "input": "2\nXTSJEP\nXTSJEP", "output": "XTSJEP" }, { "input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ", "output": "XZYDJAEDZ" }, { "input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD", "output": "QCCYXL" }, { "input": "3\nAZID\nEERWBC\nEERWBC", "output": "EERWBC" }, { "input": "3\nHNCGYL\nHNCGYL\nHNCGYL", "output": "HNCGYL" }, { "input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG", "output": "ZZWZTG" }, { "input": "4\nA\nA\nKUDLJMXCSE\nA", "output": "A" }, { "input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW", "output": "PHBTW" }, { "input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN", "output": "PKUZYTFYWN" }, { "input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH", "output": "HH" }, { "input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW", "output": "W" }, { "input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP", "output": "XBCP" }, { "input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS", "output": "PYPAS" }, { "input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC", "output": "NC" }, { "input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI", "output": "VOCI" }, { "input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA", "output": "HA" }, { "input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS", "output": "G" }, { "input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL", "output": "WL" } ]

1,600,397,368

2,147,483,647

Python 3

OK

TESTS

34

186

0

import sys def main(): n = int(sys.stdin.readline()) count = {} for _ in range(n): temp = sys.stdin.readline() count[temp] = count.get(temp, 0) + 1 sys.stdout.write(f'{max(count, key=count.get)}') if __name__ == '__main__': main()

Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. Output Specification: Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. Demo Input: ['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n'] Demo Output: ['ABC\n', 'A\n'] Note: none

```python import sys def main(): n = int(sys.stdin.readline()) count = {} for _ in range(n): temp = sys.stdin.readline() count[temp] = count.get(temp, 0) + 1 sys.stdout.write(f'{max(count, key=count.get)}') if __name__ == '__main__': main() ```

3.9535

32

A

Reconnaissance

PROGRAMMING

800

[ "brute force" ]

A. Reconnaissance

2

256

According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment. Ways (1,<=2) and (2,<=1) should be regarded as different.

The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.

Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*.

[ "5 10\n10 20 50 60 65\n", "5 1\n55 30 29 31 55\n" ]

[ "6\n", "6\n" ]

none

500

[ { "input": "5 10\n10 20 50 60 65", "output": "6" }, { "input": "5 1\n55 30 29 31 55", "output": "6" }, { "input": "6 10\n4 6 4 1 9 3", "output": "30" }, { "input": "7 100\n19 1694 261 162 1 234 513", "output": "8" }, { "input": "8 42\n37 53 74 187 568 22 5 65", "output": "20" }, { "input": "10 4\n11 6 76 49 28 20 57 152 5 32", "output": "4" }, { "input": "100 100\n51 93 101 960 2 477 213 129 663 925 254 78 1486 274 160 481 132 156 412 372 5 57 152 298 1771 7 359 468 254 406 202 929 221 366 552 97 555 29 822 118 539 140 992 854 7 163 134 103 940 30 409 1003 398 43 555 79 107 40 23 103 643 171 310 382 770 337 18 189 570 177 29 54 855 171 205 291 299 935 620 180 114 358 88 292 118 400 218 537 369 60 683 192 13 537 59 824 264 191 3 300", "output": "2404" } ]

1,626,275,548

2,147,483,647

Python 3

OK

TESTS

32

686

6,963,200

n,d = map(int, input().split()) count = 0 l = list(map(int, input().split())) for x in l: for y in l: if abs(x - y) <= d: count += 1 print(count - len(l))

Title: Reconnaissance Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment. Ways (1,<=2) and (2,<=1) should be regarded as different. Input Specification: The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109. Output Specification: Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*. Demo Input: ['5 10\n10 20 50 60 65\n', '5 1\n55 30 29 31 55\n'] Demo Output: ['6\n', '6\n'] Note: none

```python n,d = map(int, input().split()) count = 0 l = list(map(int, input().split())) for x in l: for y in l: if abs(x - y) <= d: count += 1 print(count - len(l)) ```

3.81553

568

A

Primes or Palindromes?

PROGRAMMING

1,600

[ "brute force", "implementation", "math", "number theory" ]

null

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this! Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one. Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left. One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(*n*) — the number of primes no larger than *n*, *rub*(*n*) — the number of palindromic numbers no larger than *n*. Rikhail wants to prove that there are a lot more primes than palindromic ones. He asked you to solve the following problem: for a given value of the coefficient *A* find the maximum *n*, such that π(*n*)<=≤<=*A*·*rub*(*n*).

The input consists of two positive integers *p*, *q*, the numerator and denominator of the fraction that is the value of *A* (, ).

If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).

[ "1 1\n", "1 42\n", "6 4\n" ]

[ "40\n", "1\n", "172\n" ]

none

500

[ { "input": "1 1", "output": "40" }, { "input": "1 42", "output": "1" }, { "input": "6 4", "output": "172" }, { "input": "3 1", "output": "2530" }, { "input": "42 1", "output": "1179858" }, { "input": "10000 239", "output": "1168638" }, { "input": "5 8", "output": "16" }, { "input": "7 11", "output": "16" }, { "input": "16 60", "output": "1" }, { "input": "214 210", "output": "40" }, { "input": "620 35", "output": "251262" }, { "input": "940 480", "output": "1372" }, { "input": "1307 3420", "output": "1" }, { "input": "6811 5416", "output": "66" }, { "input": "7 267", "output": "1" }, { "input": "106 6", "output": "250300" }, { "input": "10000 10000", "output": "40" }, { "input": "10000 9999", "output": "40" }, { "input": "9999 9998", "output": "40" }, { "input": "9999 9999", "output": "40" }, { "input": "4 9", "output": "10" }, { "input": "1000 10000", "output": "1" }, { "input": "238 9996", "output": "1" }, { "input": "999 10000", "output": "1" }, { "input": "241 10000", "output": "1" }, { "input": "239 10000", "output": "1" }, { "input": "5858 674", "output": "71118" } ]

1,439,936,696

2,147,483,647

Python 3

OK

TESTS

27

1,746

5,222,400

''' Created on 19 Ağu 2015 @author: enesoncu ''' pr = [0]*1300000 def prime(): pr[1] = 1; for i in range(2,1300000): if pr[i]==1: continue for j in range(2*i,1300000,i): pr[j] = 1 def palindrom(s): s = str(s) return s==s[::-1] (p,q) = map(int,input().split()) res = "Palindromic tree is better than splay tree" rub = 0 pi = 0 prime() for i in range(1,1200000): #print(palindrom(i)) if pr[i]==0: pi += 1 if palindrom(i): rub += 1 #print(i,pi,rub) if pi*q<=rub*p: res = i print( res )

Title: Primes or Palindromes? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this! Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one. Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left. One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(*n*) — the number of primes no larger than *n*, *rub*(*n*) — the number of palindromic numbers no larger than *n*. Rikhail wants to prove that there are a lot more primes than palindromic ones. He asked you to solve the following problem: for a given value of the coefficient *A* find the maximum *n*, such that π(*n*)<=≤<=*A*·*rub*(*n*). Input Specification: The input consists of two positive integers *p*, *q*, the numerator and denominator of the fraction that is the value of *A* (, ). Output Specification: If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes). Demo Input: ['1 1\n', '1 42\n', '6 4\n'] Demo Output: ['40\n', '1\n', '172\n'] Note: none

```python ''' Created on 19 Ağu 2015 @author: enesoncu ''' pr = [0]*1300000 def prime(): pr[1] = 1; for i in range(2,1300000): if pr[i]==1: continue for j in range(2*i,1300000,i): pr[j] = 1 def palindrom(s): s = str(s) return s==s[::-1] (p,q) = map(int,input().split()) res = "Palindromic tree is better than splay tree" rub = 0 pi = 0 prime() for i in range(1,1200000): #print(palindrom(i)) if pr[i]==0: pi += 1 if palindrom(i): rub += 1 #print(i,pi,rub) if pi*q<=rub*p: res = i print( res ) ```

3

339

A

Helpful Maths

PROGRAMMING

800

[ "greedy", "implementation", "sortings", "strings" ]

null

Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation. The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3. You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.

The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.

Print the new sum that Xenia can count.

[ "3+2+1\n", "1+1+3+1+3\n", "2\n" ]

[ "1+2+3\n", "1+1+1+3+3\n", "2\n" ]

none

500

[ { "input": "3+2+1", "output": "1+2+3" }, { "input": "1+1+3+1+3", "output": "1+1+1+3+3" }, { "input": "2", "output": "2" }, { "input": "2+2+1+1+3", "output": "1+1+2+2+3" }, { "input": "2+1+2+2+2+3+1+3+1+2", "output": "1+1+1+2+2+2+2+2+3+3" }, { "input": "1+2+1+2+2+2+2+1+3+3", "output": "1+1+1+2+2+2+2+2+3+3" }, { "input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1", "output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "1", "output": "1" }, { "input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1", "output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2", "output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3", "output": "1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3" }, { "input": "3", "output": "3" }, { "input": "1+1", "output": "1+1" }, { "input": "1+2", "output": "1+2" }, { "input": "1+3", "output": "1+3" }, { "input": "2+1", "output": "1+2" }, { "input": "2+2", "output": "2+2" }, { "input": "2+3", "output": "2+3" }, { "input": "3+1", "output": "1+3" }, { "input": "3+2", "output": "2+3" }, { "input": "3+3", "output": "3+3" } ]

1,698,003,479

2,147,483,647

PyPy 3

OK

TESTS

21

156

0

broj = input() lista = broj.split("+") listaintova=[] for el in lista: listaintova.append(int(el)) listaintova.sort() g = len(listaintova) g-=1 for i in range(len(listaintova)): if i == g: print(f"{listaintova[i]}") break print(f"{listaintova[i]}+",end="")

Title: Helpful Maths Time Limit: None seconds Memory Limit: None megabytes Problem Description: Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation. The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3. You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum. Input Specification: The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long. Output Specification: Print the new sum that Xenia can count. Demo Input: ['3+2+1\n', '1+1+3+1+3\n', '2\n'] Demo Output: ['1+2+3\n', '1+1+1+3+3\n', '2\n'] Note: none

```python broj = input() lista = broj.split("+") listaintova=[] for el in lista: listaintova.append(int(el)) listaintova.sort() g = len(listaintova) g-=1 for i in range(len(listaintova)): if i == g: print(f"{listaintova[i]}") break print(f"{listaintova[i]}+",end="") ```

3

567

A

Lineland Mail

PROGRAMMING

900

[ "greedy", "implementation" ]

null

All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point. Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in). Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city. For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city

The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.

Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.

[ "4\n-5 -2 2 7\n", "2\n-1 1\n" ]

[ "3 12\n3 9\n4 7\n5 12\n", "2 2\n2 2\n" ]

none

500

[ { "input": "4\n-5 -2 2 7", "output": "3 12\n3 9\n4 7\n5 12" }, { "input": "2\n-1 1", "output": "2 2\n2 2" }, { "input": "3\n-1 0 1", "output": "1 2\n1 1\n1 2" }, { "input": "4\n-1 0 1 3", "output": "1 4\n1 3\n1 2\n2 4" }, { "input": "3\n-1000000000 0 1000000000", "output": "1000000000 2000000000\n1000000000 1000000000\n1000000000 2000000000" }, { "input": "2\n-1000000000 1000000000", "output": "2000000000 2000000000\n2000000000 2000000000" }, { "input": "10\n1 10 12 15 59 68 130 912 1239 9123", "output": "9 9122\n2 9113\n2 9111\n3 9108\n9 9064\n9 9055\n62 8993\n327 8211\n327 7884\n7884 9122" }, { "input": "5\n-2 -1 0 1 2", "output": "1 4\n1 3\n1 2\n1 3\n1 4" }, { "input": "5\n-2 -1 0 1 3", "output": "1 5\n1 4\n1 3\n1 3\n2 5" }, { "input": "3\n-10000 1 10000", "output": "10001 20000\n9999 10001\n9999 20000" }, { "input": "5\n-1000000000 -999999999 -999999998 -999999997 -999999996", "output": "1 4\n1 3\n1 2\n1 3\n1 4" }, { "input": "10\n-857422304 -529223472 82412729 145077145 188538640 265299215 527377039 588634631 592896147 702473706", "output": "328198832 1559896010\n328198832 1231697178\n62664416 939835033\n43461495 1002499449\n43461495 1045960944\n76760575 1122721519\n61257592 1384799343\n4261516 1446056935\n4261516 1450318451\n109577559 1559896010" }, { "input": "10\n-876779400 -829849659 -781819137 -570920213 18428128 25280705 121178189 219147240 528386329 923854124", "output": "46929741 1800633524\n46929741 1753703783\n48030522 1705673261\n210898924 1494774337\n6852577 905425996\n6852577 902060105\n95897484 997957589\n97969051 1095926640\n309239089 1405165729\n395467795 1800633524" }, { "input": "30\n-15 1 21 25 30 40 59 60 77 81 97 100 103 123 139 141 157 158 173 183 200 215 226 231 244 256 267 279 289 292", "output": "16 307\n16 291\n4 271\n4 267\n5 262\n10 252\n1 233\n1 232\n4 215\n4 211\n3 195\n3 192\n3 189\n16 169\n2 154\n2 156\n1 172\n1 173\n10 188\n10 198\n15 215\n11 230\n5 241\n5 246\n12 259\n11 271\n11 282\n10 294\n3 304\n3 307" }, { "input": "10\n-1000000000 -999999999 -999999997 -999999996 -999999995 -999999994 -999999992 -999999990 -999999988 -999999986", "output": "1 14\n1 13\n1 11\n1 10\n1 9\n1 8\n2 8\n2 10\n2 12\n2 14" }, { "input": "50\n-50000 -49459 -48875 -48456 -48411 -48096 -47901 -47500 -47150 -46808 -46687 -46679 -46337 -45747 -45604 -45194 -44752 -44242 -44231 -44122 -43636 -43274 -42916 -42881 -42386 -42095 -41830 -41618 -41145 -40897 -40534 -40007 -39761 -39389 -39104 -38909 -38630 -38561 -38364 -38255 -38214 -38084 -37959 -37607 -37202 -36890 -36681 -36136 -36123 -35886", "output": "541 14114\n541 13573\n419 12989\n45 12570\n45 12525\n195 12210\n195 12015\n350 11614\n342 11264\n121 10922\n8 10801\n8 10793\n342 10451\n143 9861\n143 9718\n410 9308\n442 8866\n11 8356\n11 8345\n109 8236\n362 7750\n358 7388\n35 7084\n35 7119\n291 7614\n265 7905\n212 8170\n212 8382\n248 8855\n248 9103\n363 9466\n246 9993\n246 10239\n285 10611\n195 10896\n195 11091\n69 11370\n69 11439\n109 11636\n41 11745\n41 11786\n125 11916\n125 12041\n352 12393\n312 12798\n209 13110\n209 13319\n13 13864\n13 13877\n237 141..." }, { "input": "3\n-1000000000 999999999 1000000000", "output": "1999999999 2000000000\n1 1999999999\n1 2000000000" } ]

1,676,997,444

2,147,483,647

Python 3

OK

TESTS

54

483

8,089,600

n = int(input()) x=[int(i) for i in input().split()] y= len(x) for i in range(y-1): print(min(abs(x[i+1]-x[i]),abs(x[i]-x[i-1])),max(abs(x[y-1]-x[i]),abs(x[i]-x[0]))) print(x[y-1]-x[y-2],x[y-1]-x[0])

Title: Lineland Mail Time Limit: None seconds Memory Limit: None megabytes Problem Description: All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point. Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in). Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city. For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city Input Specification: The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order. Output Specification: Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city. Demo Input: ['4\n-5 -2 2 7\n', '2\n-1 1\n'] Demo Output: ['3 12\n3 9\n4 7\n5 12\n', '2 2\n2 2\n'] Note: none

```python n = int(input()) x=[int(i) for i in input().split()] y= len(x) for i in range(y-1): print(min(abs(x[i+1]-x[i]),abs(x[i]-x[i-1])),max(abs(x[y-1]-x[i]),abs(x[i]-x[0]))) print(x[y-1]-x[y-2],x[y-1]-x[0]) ```

3

794

A

Bank Robbery

PROGRAMMING

800

[ "brute force", "implementation" ]

null

A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes. Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg. The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather.

The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively. The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes. The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct.

Output a single integer: the maximum number of banknotes Oleg can take.

[ "5 3 7\n8\n4 7 5 5 3 6 2 8\n", "6 5 7\n5\n1 5 7 92 3\n" ]

[ "4\n", "0\n" ]

In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes. For the second sample, Oleg can't take any banknotes without bumping into any of the security guards.

500

[ { "input": "5 3 7\n8\n4 7 5 5 3 6 2 8", "output": "4" }, { "input": "6 5 7\n5\n1 5 7 92 3", "output": "0" }, { "input": "3 2 4\n1\n3", "output": "1" }, { "input": "5 3 8\n12\n8 3 4 5 7 6 8 3 5 4 7 6", "output": "8" }, { "input": "7 3 10\n5\n3 3 3 3 3", "output": "0" }, { "input": "3 2 5\n4\n1 3 4 5", "output": "2" }, { "input": "3 2 4\n1\n1", "output": "0" }, { "input": "6 4 8\n1\n4", "output": "0" }, { "input": "2 1 3\n1\n3", "output": "0" } ]

1,658,649,378

2,147,483,647

Python 3

OK

TESTS

46

93

8,294,400

l=list(map(int,input().split())) a=l[0] b=l[1] c=l[2] n=int(input()) m=list(map(int,input().split())) count=0 for i in m: if i>=c or i<=b: continue else: count+=1 print(count)

Title: Bank Robbery Time Limit: None seconds Memory Limit: None megabytes Problem Description: A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes. Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg. The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather. Input Specification: The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively. The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes. The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct. Output Specification: Output a single integer: the maximum number of banknotes Oleg can take. Demo Input: ['5 3 7\n8\n4 7 5 5 3 6 2 8\n', '6 5 7\n5\n1 5 7 92 3\n'] Demo Output: ['4\n', '0\n'] Note: In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes. For the second sample, Oleg can't take any banknotes without bumping into any of the security guards.

```python l=list(map(int,input().split())) a=l[0] b=l[1] c=l[2] n=int(input()) m=list(map(int,input().split())) count=0 for i in m: if i>=c or i<=b: continue else: count+=1 print(count) ```

3

841

A

Generous Kefa

PROGRAMMING

900

[ "brute force", "implementation" ]

null

One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends. Next line contains string *s* — colors of baloons.

Answer to the task — «YES» or «NO» in a single line. You can choose the case (lower or upper) for each letter arbitrary.

[ "4 2\naabb\n", "6 3\naacaab\n" ]

[ "YES\n", "NO\n" ]

In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second. In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

500

[ { "input": "4 2\naabb", "output": "YES" }, { "input": "6 3\naacaab", "output": "NO" }, { "input": "2 2\nlu", "output": "YES" }, { "input": "5 3\novvoo", "output": "YES" }, { "input": "36 13\nbzbzcffczzcbcbzzfzbbfzfzzbfbbcbfccbf", "output": "YES" }, { "input": "81 3\nooycgmvvrophvcvpoupepqllqttwcocuilvyxbyumdmmfapvpnxhjhxfuagpnntonibicaqjvwfhwxhbv", "output": "NO" }, { "input": "100 100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx", "output": "YES" }, { "input": "100 1\nnubcvvjvbjgnjsdkajimdcxvewbcytvfkihunycdrlconddlwgzjasjlsrttlrzsumzpyumpveglfqzmaofbshbojmwuwoxxvrod", "output": "NO" }, { "input": "100 13\nvyldolgryldqrvoldvzvrdrgorlorszddtgqvrlisxxrxdxlqtvtgsrqlzixoyrozxzogqxlsgzdddzqrgitxxritoolzolgrtvl", "output": "YES" }, { "input": "18 6\njzwtnkvmscqhmdlsxy", "output": "YES" }, { "input": "21 2\nfscegcqgzesefghhwcexs", "output": "NO" }, { "input": "32 22\ncduamsptaklqtxlyoutlzepxgyfkvngc", "output": "YES" }, { "input": "49 27\noxyorfnkzwsfllnyvdhdanppuzrnbxehugvmlkgeymqjlmfxd", "output": "YES" }, { "input": "50 24\nxxutzjwbggcwvxztttkmzovtmuwttzcbwoztttohzzxghuuthv", "output": "YES" }, { "input": "57 35\nglxshztrqqfyxthqamagvtmrdparhelnzrqvcwqxjytkbuitovkdxueul", "output": "YES" }, { "input": "75 23\nittttiiuitutuiiuuututiuttiuiuutuuuiuiuuuuttuuttuutuiiuiuiiuiitttuututuiuuii", "output": "NO" }, { "input": "81 66\nfeqevfqfebhvubhuuvfuqheuqhbeeuebehuvhffvbqvqvfbqqvvhevqffbqqhvvqhfeehuhqeqhueuqqq", "output": "YES" }, { "input": "93 42\npqeiafraiavfcteumflpcbpozcomlvpovlzdbldvoopnhdoeqaopzthiuzbzmeieiatthdeqovaqfipqlddllmfcrrnhb", "output": "YES" }, { "input": "100 53\nizszyqyndzwzyzgsdagdwdazadiawizinagqqgczaqqnawgijziziawzszdjdcqjdjqiwgadydcnqisaayjiqqsscwwzjzaycwwc", "output": "YES" }, { "input": "100 14\nvkrdcqbvkwuckpmnbydmczdxoagdsgtqxvhaxntdcxhjcrjyvukhugoglbmyoaqexgtcfdgemmizoniwtmisqqwcwfusmygollab", "output": "YES" }, { "input": "100 42\naaaaaiiiiaiiiaaiaiiaaiiiiiaaaaaiaiiiaiiiiaiiiaaaaaiiiaaaiiaaiiiaiiiaiaaaiaiiiiaaiiiaiiaiaiiaiiiaaaia", "output": "NO" }, { "input": "100 89\ntjbkmydejporbqhcbztkcumxjjgsrvxpuulbhzeeckkbchpbxwhedrlhjsabcexcohgdzouvsgphjdthpuqrlkgzxvqbuhqxdsmf", "output": "YES" }, { "input": "100 100\njhpyiuuzizhubhhpxbbhpyxzhbpjphzppuhiahihiappbhuypyauhizpbibzixjbzxzpbphuiaypyujappuxiyuyaajaxjupbahb", "output": "YES" }, { "input": "100 3\nsszoovvzysavsvzsozzvoozvysozsaszayaszasaysszzzysosyayyvzozovavzoyavsooaoyvoozvvozsaosvayyovazzszzssa", "output": "NO" }, { "input": "100 44\ndluthkxwnorabqsukgnxnvhmsmzilyulpursnxkdsavgemiuizbyzebhyjejgqrvuckhaqtuvdmpziesmpmewpvozdanjyvwcdgo", "output": "YES" }, { "input": "100 90\ntljonbnwnqounictqqctgonktiqoqlocgoblngijqokuquoolciqwnctgoggcbojtwjlculoikbggquqncittwnjbkgkgubnioib", "output": "YES" }, { "input": "100 79\nykxptzgvbqxlregvkvucewtydvnhqhuggdsyqlvcfiuaiddnrrnstityyehiamrggftsqyduwxpuldztyzgmfkehprrneyvtknmf", "output": "YES" }, { "input": "100 79\naagwekyovbviiqeuakbqbqifwavkfkutoriovgfmittulhwojaptacekdirgqoovlleeoqkkdukpadygfwavppohgdrmymmulgci", "output": "YES" }, { "input": "100 93\nearrehrehenaddhdnrdddhdahnadndheeennrearrhraharddreaeraddhehhhrdnredanndneheddrraaneerreedhnadnerhdn", "output": "YES" }, { "input": "100 48\nbmmaebaebmmmbbmxvmammbvvebvaemvbbaxvbvmaxvvmveaxmbbxaaemxmxvxxxvxbmmxaaaevvaxmvamvvmaxaxavexbmmbmmev", "output": "YES" }, { "input": "100 55\nhsavbkehaaesffaeeffakhkhfehbbvbeasahbbbvkesbfvkefeesesevbsvfkbffakvshsbkahfkfakebsvafkbvsskfhfvaasss", "output": "YES" }, { "input": "100 2\ncscffcffsccffsfsfffccssfsscfsfsssffcffsscfccssfffcfscfsscsccccfsssffffcfcfsfffcsfsccffscffcfccccfffs", "output": "NO" }, { "input": "100 3\nzrgznxgdpgfoiifrrrsjfuhvtqxjlgochhyemismjnanfvvpzzvsgajcbsulxyeoepjfwvhkqogiiwqxjkrpsyaqdlwffoockxnc", "output": "NO" }, { "input": "100 5\njbltyyfjakrjeodqepxpkjideulofbhqzxjwlarufwzwsoxhaexpydpqjvhybmvjvntuvhvflokhshpicbnfgsqsmrkrfzcrswwi", "output": "NO" }, { "input": "100 1\nfnslnqktlbmxqpvcvnemxcutebdwepoxikifkzaaixzzydffpdxodmsxjribmxuqhueifdlwzytxkklwhljswqvlejedyrgguvah", "output": "NO" }, { "input": "100 21\nddjenetwgwmdtjbpzssyoqrtirvoygkjlqhhdcjgeurqpunxpupwaepcqkbjjfhnvgpyqnozhhrmhfwararmlcvpgtnopvjqsrka", "output": "YES" }, { "input": "100 100\nnjrhiauqlgkkpkuvciwzivjbbplipvhslqgdkfnmqrxuxnycmpheenmnrglotzuyxycosfediqcuadklsnzjqzfxnbjwvfljnlvq", "output": "YES" }, { "input": "100 100\nbbbbbbbtbbttbtbbbttbttbtbbttttbbbtbttbbbtbttbtbbttttbbbbbtbbttbtbbtbttbbbtbtbtbtbtbtbbbttbbtbtbtbbtb", "output": "YES" }, { "input": "14 5\nfssmmsfffmfmmm", "output": "NO" }, { "input": "2 1\nff", "output": "NO" }, { "input": "2 1\nhw", "output": "YES" }, { "input": "2 2\nss", "output": "YES" }, { "input": "1 1\nl", "output": "YES" }, { "input": "100 50\nfffffttttttjjjuuuvvvvvdddxxxxwwwwgggbsssncccczzyyyyyhhhhhkrreeeeeeaaaaaiiillllllllooooqqqqqqmmpppppp", "output": "YES" }, { "input": "100 50\nbbbbbbbbgggggggggggaaaaaaaahhhhhhhhhhpppppppppsssssssrrrrrrrrllzzzzzzzeeeeeeekkkkkkkwwwwwwwwjjjjjjjj", "output": "YES" }, { "input": "100 50\nwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxzzzzzzzzzzzzzzzzzzbbbbbbbbbbbbbbbbbbbbjjjjjjjjjjjjjjjjjjjjjjjj", "output": "YES" }, { "input": "100 80\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm", "output": "YES" }, { "input": "100 10\nbbttthhhhiiiiiiijjjjjvvvvpppssssseeeeeeewwwwgggkkkkkkkkmmmddddduuuzzzzllllnnnnnxxyyyffffccraaaaooooq", "output": "YES" }, { "input": "100 20\nssssssssssbbbbbbbhhhhhhhyyyyyyyzzzzzzzzzzzzcccccxxxxxxxxxxddddmmmmmmmeeeeeeejjjjjjjjjwwwwwwwtttttttt", "output": "YES" }, { "input": "1 2\na", "output": "YES" }, { "input": "3 1\nabb", "output": "NO" }, { "input": "2 1\naa", "output": "NO" }, { "input": "2 1\nab", "output": "YES" }, { "input": "6 2\naaaaaa", "output": "NO" }, { "input": "8 4\naaaaaaaa", "output": "NO" }, { "input": "4 2\naaaa", "output": "NO" }, { "input": "4 3\naaaa", "output": "NO" }, { "input": "1 3\na", "output": "YES" }, { "input": "4 3\nzzzz", "output": "NO" }, { "input": "4 1\naaaa", "output": "NO" }, { "input": "3 4\nabc", "output": "YES" }, { "input": "2 5\nab", "output": "YES" }, { "input": "2 4\nab", "output": "YES" }, { "input": "1 10\na", "output": "YES" }, { "input": "5 2\nzzzzz", "output": "NO" }, { "input": "53 26\naaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "NO" }, { "input": "4 1\nabab", "output": "NO" }, { "input": "4 1\nabcb", "output": "NO" }, { "input": "4 2\nabbb", "output": "NO" }, { "input": "5 2\nabccc", "output": "NO" }, { "input": "2 3\nab", "output": "YES" }, { "input": "4 3\nbbbs", "output": "YES" }, { "input": "10 2\nazzzzzzzzz", "output": "NO" }, { "input": "1 2\nb", "output": "YES" }, { "input": "1 3\nb", "output": "YES" }, { "input": "4 5\nabcd", "output": "YES" }, { "input": "4 6\naabb", "output": "YES" }, { "input": "5 2\naaaab", "output": "NO" }, { "input": "3 5\naaa", "output": "YES" }, { "input": "5 3\nazzzz", "output": "NO" }, { "input": "4 100\naabb", "output": "YES" }, { "input": "3 10\naaa", "output": "YES" }, { "input": "3 4\naaa", "output": "YES" }, { "input": "12 5\naaaaabbbbbbb", "output": "NO" }, { "input": "5 2\naabbb", "output": "NO" }, { "input": "10 5\nzzzzzzzzzz", "output": "NO" }, { "input": "2 4\naa", "output": "YES" }, { "input": "1 5\na", "output": "YES" }, { "input": "10 5\naaaaaaaaaa", "output": "NO" }, { "input": "6 3\naaaaaa", "output": "NO" }, { "input": "7 1\nabcdeee", "output": "NO" }, { "input": "18 3\naaaaaabbbbbbcccccc", "output": "NO" }, { "input": "8 2\naabbccdd", "output": "YES" }, { "input": "4 2\nzzzz", "output": "NO" }, { "input": "4 2\nabaa", "output": "NO" }, { "input": "3 2\naaa", "output": "NO" }, { "input": "3 1\nzzz", "output": "NO" }, { "input": "5 4\nzzzzz", "output": "NO" }, { "input": "6 2\naabbbc", "output": "NO" }, { "input": "3 6\naaa", "output": "YES" }, { "input": "2 1\nzz", "output": "NO" }, { "input": "10 3\naaaeeeeeee", "output": "NO" }, { "input": "4 5\naabb", "output": "YES" }, { "input": "3 1\naaa", "output": "NO" }, { "input": "5 2\naazzz", "output": "NO" }, { "input": "6 2\nabbbbc", "output": "NO" }, { "input": "4 2\nxxxx", "output": "NO" }, { "input": "6 3\nzzzzzz", "output": "NO" }, { "input": "3 2\nabb", "output": "YES" }, { "input": "3 2\nzzz", "output": "NO" }, { "input": "6 5\nzzzzzz", "output": "NO" }, { "input": "6 3\nbcaaaa", "output": "NO" }, { "input": "100 100\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa", "output": "YES" }, { "input": "3 6\nabc", "output": "YES" } ]

1,545,646,298

2,147,483,647

Python 3

OK

TESTS

114

124

0

n,m=map(int,input().split()) s=input() ans="YES" for i in [s.count(k) for k in set(s)]: if(i>m): ans="NO" break print(ans)

Title: Generous Kefa Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends. Next line contains string *s* — colors of baloons. Output Specification: Answer to the task — «YES» or «NO» in a single line. You can choose the case (lower or upper) for each letter arbitrary. Demo Input: ['4 2\naabb\n', '6 3\naacaab\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second. In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

```python n,m=map(int,input().split()) s=input() ans="YES" for i in [s.count(k) for k in set(s)]: if(i>m): ans="NO" break print(ans) ```

3

43

B

Letter

PROGRAMMING

1,100

[ "implementation", "strings" ]

B. Letter

2

256

Vasya decided to write an anonymous letter cutting the letters out of a newspaper heading. He knows heading *s*1 and text *s*2 that he wants to send. Vasya can use every single heading letter no more than once. Vasya doesn't have to cut the spaces out of the heading — he just leaves some blank space to mark them. Help him; find out if he will manage to compose the needed text.

The first line contains a newspaper heading *s*1. The second line contains the letter text *s*2. *s*1 и *s*2 are non-empty lines consisting of spaces, uppercase and lowercase Latin letters, whose lengths do not exceed 200 symbols. The uppercase and lowercase letters should be differentiated. Vasya does not cut spaces out of the heading.

If Vasya can write the given anonymous letter, print YES, otherwise print NO

[ "Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog\n", "Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears\n", "Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears\n", "abcdefg hijk\nk j i h g f e d c b a\n" ]

[ "NO\n", "YES\n", "NO\n", "YES\n" ]

none

1,000

[ { "input": "Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog", "output": "NO" }, { "input": "Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears", "output": "YES" }, { "input": "Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears", "output": "NO" }, { "input": "abcdefg hijk\nk j i h g f e d c b a", "output": "YES" }, { "input": "HpOKgo\neAtAVB", "output": "NO" }, { "input": "GRZGc\nLPzD", "output": "NO" }, { "input": "GtPXu\nd", "output": "NO" }, { "input": "FVF\nr ", "output": "NO" }, { "input": "HpOKgo\nogK", "output": "YES" }, { "input": "GRZGc\nZG", "output": "YES" }, { "input": "HpOKgoueAtAVBdGffvQheJDejNDHhhwyKJisugiRAH OseK yUwqPPNuThUxTfthqIUeb wS jChGOdFDarNrKRT MlwKecxWNoKEeD BbiHAruE XMlvKYVsJGPP\nAHN XvoaNwV AVBKwKjr u U K wKE D K Jy KiHsR h d W Js IHyMPK Br iSqe E fDA g H", "output": "YES" }, { "input": "GRZGcsLPzDrCSXhhNTaibJqVphhjbcPoZhCDUlzAbDnRWjHvxLKtpGiFWiGbfeDxBwCrdJmJGCGv GebAOinUsFrlqKTILOmxrFjSpEoVGoTdSSstJWVgMLKMPettxHASaQZNdOIObcTxtF qTHWBdNIKwj\nWqrxze Ji x q aT GllLrRV jMpGiMDTwwS JDsPGpAZKACmsFCOS CD Sj bCDgKF jJxa RddtLFAi VGLHH SecObzG q hPF ", "output": "YES" }, { "input": "GtPXuwdAxNhODQbjRslDDKciOALJrCifTjDQurQEBeFUUSZWwCZQPdYwZkYbrduMijFjgodAOrKIuUKwSXageZuOWMIhAMexyLRzFuzuXqBDTEaWMzVdbzhxDGSJC SsIYuYILwpiwwcObEHWpFvHeBkWYNitqYrxqgHReHcKnHbtjcWZuaxPBVPb\nTQIKyqFaewOkY lZUOOuxEw EwuKcArxRQGFYkvVWIAe SuanPeHuDjquurJu aSxwgOSw jYMwjxItNUUArQjO BIujAhSwttLWp", "output": "YES" }, { "input": "FVFSr unvtXbpKWF vPaAgNaoTqklzVqiGYcUcBIcattzBrRuNSnKUtmdGKbjcE\nUzrU K an GFGR Wc zt iBa P c T K v p V In b B c", "output": "YES" }, { "input": "lSwjnYLYtDNIZjxHiTawdh ntSzggZogcIZTuiTMWVgwyloMtEhqkrOxgIcFvwvsboXUPILPIymFAEXnhApewJXJNtFyZ\nAoxe jWZ u yImg o AZ FNI w lpj tNhT g y ZYcb rc J w Dlv", "output": "YES" }, { "input": "kvlekcdJqODUKdsJlXkRaileTmdGwUHWWgvgUokQxRzzbpFnswvNKiDnjfOFGvFcnaaiRnBGQmqoPxDHepgYasLhzjDgmvaFfVNEcSPVQCJKAbSyTGpXsAjIHr\nGjzUllNaGGKXUdYmDFpqFAKIwvTpjmqnyswWRTnxlBnavAGvavxJemrjvRJc", "output": "YES" }, { "input": "kWbvhgvvoYOhwXmgTwOSCDXrtFHhqwvMlCvsuuAUXMmWaYXiqHplFZZemhgkTuvsUtIaUxtyYauBIpjdbyYxjZ ZkaBPzwqPfqF kCqGRmXvWuabnQognnkvdNDtRUsSUvSzgBuxCMBWJifbxWegsknp\nBsH bWHJD n Ca T xq PRCv tatn Wjy sm I q s WCjFqdWe t W XUs Do eb Pfh ii hTbF O Fll", "output": "YES" }, { "input": "OTmLdkMhmDEOMQMiW ZpzEIjyElHFrNCfFQDp SZyoZaEIUIpyCHfwOUqiSkKtFHggrTBGkqfOxkChPztmPrsHoxVwAdrxbZLKxPXHlMnrkgMgiaHFopiFFiUEtKwCjpJtwdwkbJCgA bxeDIscFdmHQJLAMNhWlrZisQrHQpvbALWTwpf jnx\nDbZwrQbydCdkJMCrftiwtPFfpMiwwrfIrKidEChKECxQUBVUEfFirbGWiLkFQkdJiFtkrtkbIAEXCEDkwLpK", "output": "YES" }, { "input": "NwcGaIeSkOva\naIa", "output": "YES" }, { "input": "gSrAcVYgAdbdayzbKGhIzLDjyznLRIJH KyvilAaEddmgkBPCNzpmPNeGEbmmpAyHvUSoPvnaORrPUuafpReEGoDOQsAYnUHYfBqhdcopQfxJuGXgKnbdVMQNhJYkyjiJDKlShqBTtnnDQQzEijOMcYRGMgPGVhfIReYennKBLwDTVvcHMIHMgVpJkvzTrezxqS\nHJerIVvRyfrPgAQMTI AqGNO mQDfDwQHKgeeYmuRmozKHILvehMPOJNMRtPTAfvKvsoGKi xHEeKqDAYmQJPUXRJbIbHrgVOMGMTdvYiLui", "output": "YES" }, { "input": "ReB hksbHqQXxUgpvoNK bFqmNVCEiOyKdKcAJQRkpeohpfuqZabvrLfmpZOMcfyFBJGZwVMxiUPP pbZZtJjxhEwvrAba\nJTCpQnIViIGIdQtLnmkVzmcbBZR CoxAdTtWSYpbOglDFifqIVQ vfGKGtLpxpJHiHSWCMeRcrVOXBGBhoEnVhNTPWGTOErNtSvokcGdgZXbgTEtISUyTwaXUEIlJMmutsdCbiyrPZPJyRdOjnSuAGttLy", "output": "NO" }, { "input": "hrLzRegCuDGxTrhDgVvM KowwyYuXGzIpcXdSMgeQVfVOtJZdkhNYSegwFWWoPqcZoeapbQnyCtojgkcyezUNHGGIZrhzsKrvvcrtokIdcnqXXkCNKjrOjrnEAKBNxyDdiMVeyLvXxUYMZQRFdlcdlcxzKTeYzBlmpNiwWbNAAhWkMoGpRxkCuyqkzXdKWwGH\nJESKDOfnFdxPvUOCkrgSBEPQHJtJHzuNGstRbTCcchRWJvCcveSEAtwtOmZZiW", "output": "NO" }, { "input": "yDBxCtUygQwWqONxQCcuAvVCkMGlqgC zvkfEkwqbhMCQxnkwQIUhucCbVUyOBUcXvTNEGriTBwMDMfdsPZgWRgIUDqM\neptVnORTTyixxmWIBpSTEwOXqGZllBgSxPenYCDlFwckJlWsoVwWLAIbPOmFqcKcTcoQqahetl KLfVSyaLVebzsGwPSVbtQAeUdZAaJtfxlCEvvaRhLlVvRJhKat IaB awdqcDlrrhTbRxjEbzGwcdmdavkhcjHjzmwbxAgw", "output": "NO" }, { "input": "jlMwnnotSdlQMluKWkJwAeCetcqbIEnKeNyLWoKCGONDRBQOjbkGpUvDlmSFUJ bWhohqmmIUWTlDsvelUArAcZJBipMDwUvRfBsYzMdQnPDPAuBaeJmAxVKwUMJrwMDxNtlrtAowVWqWiwFGtmquZAcrpFsLHCrvMSMMlvQUqypAihQWrFMNoaqfs IBg\nNzeWQ bafrmDsYlpNHSGTBBgPl WIcuNhyNaNOEFvL", "output": "NO" }, { "input": "zyWvXBcUZqGqjHwZHQryBtFliLYnweXAoMKNpLaunaOlzaauWmLtywsEvWPiwxJapocAFRMjrqWJXYqfKEbBKnzLO\npsbi bsXpSeJaCkIuPWfSRADXdIClxcDCowwJzGCDTyAl", "output": "NO" }, { "input": "kKhuIwRPLCwPFfcnsyCfBdnsraGeOCcLTfXuGjqFSGPSAeDZJSS bXKFanNqWjpFnvRpWxHJspvisDlADJBioxXNbVoXeUedoPcNEpUyEeYxdJXhGzFAmpAiHotSVwbZQsuWjIVhVaEGgqbZHIoDpiEmjTtFylCwCkWWzUOoUfOHxEZvDwNpXhBWamHn\nK VpJjGhNbwCRhcfmNGVjewBFpEmPlIKeTuWiukDtEWpjgqciqglkyNfWrBLbGAKvlNWxaUelJmSlSoakSpRzePvJsshOsTYrMPXdxKpaShjyVIXGhRIAdtiGpNwtiRmGTBZhkJqIMdxMHX RMxCMYcWjcjhtCHyFnCvjjezGbkRDRiVxkbh", "output": "NO" }, { "input": "AXssNpFKyQmJcBdBdfkhhMUzfqJVgcLBddkwtnFSzSRUCjiDcdtmkzIGkCKSxWUEGhmHmciktJyGMkgCductyHx\nI nYhmJfPnvoKUiXYUBIPIcxNYTtvwPUoXERZvY ahlDpQFNMmVZqEBiYqYlHNqcpSCmhFczBlOAhsYFeqMGfqL EJsDNOgwoJfBzqijKOFcYQ", "output": "NO" }, { "input": "lkhrzDZmkdbjzYKPNMRkiwCFoZsMzBQMnxxdKKVJezSBjnLjPpUYtabcPTIaDJeDEobbWHdKOdVfMQwDXzDDcSrwVenDEYpMqfiOQ xSsqApWnAMoyhQXCKFzHvvzvUvkWwmwZrvZz\nsUzGspYpRFsHRbRgTQuCBgnFgPkisTUfFNwyEEWWRiweWWgjRkVQxgTwxOzdsOwfrGIH O gCXpzvHzfItuEHaihmugEyymSJIogYwX qAwcwIItidfnzZDhZgQHi eRjMAeVkJHceDZuJkmxGowOsmcGYYvk Ajtgi TxwihvjLViNZjvscTWvsaQUelTSivLShhEl", "output": "NO" }, { "input": "BRsVjyNhrqRHVwrJzuzRigEhdpbDmaACSPfed\nlWqKTjlrqOCUbgBBZdZDGCeQJDXawPnnDkQdZDgwrEQk", "output": "NO" }, { "input": "KRmINuyBYPwiTsdlyiNVuylToysJKmOpcLovAtwGPqrgFJQNAYvuAiyQRkeFMECVZvkDEmTauXlyjAaYRnTJXORMZRnTakBaUzSelMilejySDIZjQjzcOIrwXdvDvpeRIkoBgreyFXIyyIZutjiEBtwrmzQtPVUhvvdEtDMbXjBpoPVjGdM EXTAK JbCnw\nXZZqlJvzKKtvdNlzFPDTYxidqlsgufVzyEmO FZuLQ vVQsJESNviUCovCK NwwlbxsmPtOJNmAonCqrOZ bZ LVKAsQGmoLnYjeekvEIECFk", "output": "NO" } ]

1,624,930,707

2,147,483,647

Python 3

OK

TESTS

30

124

204,800

import sys from collections import Counter input = lambda:sys.stdin.readline() int_arr = lambda: list(map(int,input().split())) str_arr = lambda: list(map(str,input().split())) get_str = lambda: map(str,input().split()) get_int = lambda: map(int,input().split()) get_flo = lambda: map(float,input().split()) mod = 1000000007 def solve(h,s): hc = {} for e,t in Counter(h).items(): hc[e] = t for i in s: if i in hc: hc[i] -= 1 if hc[i] < 0: print("NO") return else: print("NO") return print("YES") # for _ in range(int(input())): h = str(input())[:-1] s = str(input())[:-1] h = ''.join(h.split());s = ''.join(s.split()) solve(h,s)

Title: Letter Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya decided to write an anonymous letter cutting the letters out of a newspaper heading. He knows heading *s*1 and text *s*2 that he wants to send. Vasya can use every single heading letter no more than once. Vasya doesn't have to cut the spaces out of the heading — he just leaves some blank space to mark them. Help him; find out if he will manage to compose the needed text. Input Specification: The first line contains a newspaper heading *s*1. The second line contains the letter text *s*2. *s*1 и *s*2 are non-empty lines consisting of spaces, uppercase and lowercase Latin letters, whose lengths do not exceed 200 symbols. The uppercase and lowercase letters should be differentiated. Vasya does not cut spaces out of the heading. Output Specification: If Vasya can write the given anonymous letter, print YES, otherwise print NO Demo Input: ['Instead of dogging Your footsteps it disappears but you dont notice anything\nwhere is your dog\n', 'Instead of dogging Your footsteps it disappears but you dont notice anything\nYour dog is upstears\n', 'Instead of dogging your footsteps it disappears but you dont notice anything\nYour dog is upstears\n', 'abcdefg hijk\nk j i h g f e d c b a\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n', 'YES\n'] Note: none

```python import sys from collections import Counter input = lambda:sys.stdin.readline() int_arr = lambda: list(map(int,input().split())) str_arr = lambda: list(map(str,input().split())) get_str = lambda: map(str,input().split()) get_int = lambda: map(int,input().split()) get_flo = lambda: map(float,input().split()) mod = 1000000007 def solve(h,s): hc = {} for e,t in Counter(h).items(): hc[e] = t for i in s: if i in hc: hc[i] -= 1 if hc[i] < 0: print("NO") return else: print("NO") return print("YES") # for _ in range(int(input())): h = str(input())[:-1] s = str(input())[:-1] h = ''.join(h.split());s = ''.join(s.split()) solve(h,s) ```

3.968619

734

A

Anton and Danik

PROGRAMMING

800

[ "implementation", "strings" ]

null

Anton likes to play chess, and so does his friend Danik. Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie. Now Anton wonders, who won more games, he or Danik? Help him determine this.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played. The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.

If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output. If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output. If Anton and Danik won the same number of games, print "Friendship" (without quotes).

[ "6\nADAAAA\n", "7\nDDDAADA\n", "6\nDADADA\n" ]

[ "Anton\n", "Danik\n", "Friendship\n" ]

In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton". In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik". In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".

500

[ { "input": "6\nADAAAA", "output": "Anton" }, { "input": "7\nDDDAADA", "output": "Danik" }, { "input": "6\nDADADA", "output": "Friendship" }, { "input": "10\nDDDDADDADD", "output": "Danik" }, { "input": "40\nAAAAAAAAADDAAAAAAAAAAADADDAAAAAAAAAAADAA", "output": "Anton" }, { "input": "200\nDDDDDDDADDDDDDAADADAADAAADAADADAAADDDADDDDDDADDDAADDDAADADDDDDADDDAAAADAAADDDDDAAADAADDDAAAADDADADDDAADDAADAAADAADAAAADDAADDADAAAADADDDAAAAAADDAADAADAADADDDAAADAAAADADDADAAAAAADADADDDADDDAADDADDDAAAAD", "output": "Friendship" }, { "input": "1\nA", "output": "Anton" }, { "input": "1\nD", "output": "Danik" }, { "input": "2\nDA", "output": "Friendship" }, { "input": "4\nDADA", "output": "Friendship" }, { "input": "4\nDAAD", "output": "Friendship" }, { "input": "3\nADD", "output": "Danik" }, { "input": "3\nDAD", "output": "Danik" }, { "input": "2\nDA", "output": "Friendship" }, { "input": "379\nAADAAAAAADDAAAAAADAADADADDAAAAADADDAADAAAADDDADAAAAAAADAADAAAAAAADAAAAAAAAADAAAAAAADAAAAAAAAAAADDDADAAAAAAAADAADADAAAADAAAAAAAAAAAAAAAAADAAAADDDAADAAAAAAADAAADAAADAADDDADDAAADAAAAAADDDADDDAAADAAAADAAAAAAAAADAAADAAAAAAAAADAAAAAAAAAAAAAAAAAADADAAAAAAAAAAADAAAAADAAAADAAAAAAAAAAAAADADAADAAAAAAAADAADAAAAAAAADAAAAAAAADDDAAAAAADAAADAAAAAADAADAAAAAADAAAADADAADAAAAAADAAAADAADDAADAADAAA", "output": "Anton" } ]

1,694,200,111

2,147,483,647

PyPy 3-64

OK

TESTS

25

62

1,740,800

x=int (input()) sum_of_A=0 sum_of_D=0 s=input().upper() for i in s: if i=="A": sum_of_A+=1 elif i=="D": sum_of_D+=1 if sum_of_A>sum_of_D: print("Anton") elif sum_of_D>sum_of_A: print("Danik") else: print("Friendship")

Title: Anton and Danik Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anton likes to play chess, and so does his friend Danik. Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie. Now Anton wonders, who won more games, he or Danik? Help him determine this. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played. The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game. Output Specification: If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output. If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output. If Anton and Danik won the same number of games, print "Friendship" (without quotes). Demo Input: ['6\nADAAAA\n', '7\nDDDAADA\n', '6\nDADADA\n'] Demo Output: ['Anton\n', 'Danik\n', 'Friendship\n'] Note: In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton". In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik". In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".

```python x=int (input()) sum_of_A=0 sum_of_D=0 s=input().upper() for i in s: if i=="A": sum_of_A+=1 elif i=="D": sum_of_D+=1 if sum_of_A>sum_of_D: print("Anton") elif sum_of_D>sum_of_A: print("Danik") else: print("Friendship") ```

3

0

none

0

[ "none" ]

null

We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7. The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.

The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.

Print the minimum number of steps modulo 109<=+<=7.

[ "ab\n", "aab\n" ]

[ "1\n", "3\n" ]

The first example: "ab" → "bba". The second example: "aab" → "abba" → "bbaba" → "bbbbaa".

0

[ { "input": "ab", "output": "1" }, { "input": "aab", "output": "3" }, { "input": "aaaaabaabababaaaaaba", "output": "17307" }, { "input": "abaabaaabbabaabab", "output": "1795" }, { "input": "abbaa", "output": "2" }, { "input": "abbaaabaabaaaaabbbbaababaaaaabaabbaaaaabbaabbaaaabbbabbbabb", "output": "690283580" }, { "input": "aababbaaaabbaabbbbbbbbabbababbbaaabbaaabbabbba", "output": "2183418" }, { "input": "aabbaababbabbbaabbaababaaaabbaaaabaaaaaababbaaaabaababbabbbb", "output": "436420225" }, { "input": "aaabaaaabbababbaabbababbbbaaaaaaabbabbba", "output": "8431094" }, { "input": "abbbbababbabbbbbabaabbbaabbbbbbbaaab", "output": "8180" }, { "input": "bbababbbaabaaaaaaaabbabbbb", "output": "40979" }, { "input": "abbbaaabbbbbabaabbaaabbbababbbaabaabababababa", "output": "2065758" }, { "input": "abaaaaaabaaaabbabbaaabbbbabababaaaaabbaabbaaaaabbbaababaaaaaaabbbbbaaaaabaababbabababbabbbbaabbaabbabbbabaabbaabbaaaaaab", "output": "235606597" }, { "input": "abbbbbbbbbbbbbbbbbbbbbbbbbbaababaaaaaaabaabaaababaabaababaaabababaababab", "output": "7" }, { "input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaaaaaaabaabaaababaabaababaaabababaabbbbbbb", "output": "557763786" } ]

1,592,494,619

2,147,483,647

Python 3

OK

TESTS

32

483

3,174,400

s = input() cnt = 0 ans = 0 mod = 1000000007 for i in reversed(range(len(s))): if(s[i] == 'a'): ans = (ans + cnt) % mod cnt = (cnt * 2) % mod else: cnt += 1 print(ans)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7. The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string. Input Specification: The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106. Output Specification: Print the minimum number of steps modulo 109<=+<=7. Demo Input: ['ab\n', 'aab\n'] Demo Output: ['1\n', '3\n'] Note: The first example: "ab" → "bba". The second example: "aab" → "abba" → "bbaba" → "bbbbaa".

```python s = input() cnt = 0 ans = 0 mod = 1000000007 for i in reversed(range(len(s))): if(s[i] == 'a'): ans = (ans + cnt) % mod cnt = (cnt * 2) % mod else: cnt += 1 print(ans) ```

3

760

A

Petr and a calendar

PROGRAMMING

800

[ "implementation", "math" ]

null

Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture: Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.

The only line contain two integers *m* and *d* (1<=≤<=*m*<=≤<=12, 1<=≤<=*d*<=≤<=7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).

Print single integer: the number of columns the table should have.

[ "1 7\n", "1 1\n", "11 6\n" ]

[ "6\n", "5\n", "5\n" ]

The first example corresponds to the January 2017 shown on the picture in the statements. In the second example 1-st January is Monday, so the whole month fits into 5 columns. In the third example 1-st November is Saturday and 5 columns is enough.

500

[ { "input": "1 7", "output": "6" }, { "input": "1 1", "output": "5" }, { "input": "11 6", "output": "5" }, { "input": "2 7", "output": "5" }, { "input": "2 1", "output": "4" }, { "input": "8 6", "output": "6" }, { "input": "1 1", "output": "5" }, { "input": "1 2", "output": "5" }, { "input": "1 3", "output": "5" }, { "input": "1 4", "output": "5" }, { "input": "1 5", "output": "5" }, { "input": "1 6", "output": "6" }, { "input": "1 7", "output": "6" }, { "input": "2 1", "output": "4" }, { "input": "2 2", "output": "5" }, { "input": "2 3", "output": "5" }, { "input": "2 4", "output": "5" }, { "input": "2 5", "output": "5" }, { "input": "2 6", "output": "5" }, { "input": "2 7", "output": "5" }, { "input": "3 1", "output": "5" }, { "input": "3 2", "output": "5" }, { "input": "3 3", "output": "5" }, { "input": "3 4", "output": "5" }, { "input": "3 5", "output": "5" }, { "input": "3 6", "output": "6" }, { "input": "3 7", "output": "6" }, { "input": "4 1", "output": "5" }, { "input": "4 2", "output": "5" }, { "input": "4 3", "output": "5" }, { "input": "4 4", "output": "5" }, { "input": "4 5", "output": "5" }, { "input": "4 6", "output": "5" }, { "input": "4 7", "output": "6" }, { "input": "5 1", "output": "5" }, { "input": "5 2", "output": "5" }, { "input": "5 3", "output": "5" }, { "input": "5 4", "output": "5" }, { "input": "5 5", "output": "5" }, { "input": "5 6", "output": "6" }, { "input": "5 7", "output": "6" }, { "input": "6 1", "output": "5" }, { "input": "6 2", "output": "5" }, { "input": "6 3", "output": "5" }, { "input": "6 4", "output": "5" }, { "input": "6 5", "output": "5" }, { "input": "6 6", "output": "5" }, { "input": "6 7", "output": "6" }, { "input": "7 1", "output": "5" }, { "input": "7 2", "output": "5" }, { "input": "7 3", "output": "5" }, { "input": "7 4", "output": "5" }, { "input": "7 5", "output": "5" }, { "input": "7 6", "output": "6" }, { "input": "7 7", "output": "6" }, { "input": "8 1", "output": "5" }, { "input": "8 2", "output": "5" }, { "input": "8 3", "output": "5" }, { "input": "8 4", "output": "5" }, { "input": "8 5", "output": "5" }, { "input": "8 6", "output": "6" }, { "input": "8 7", "output": "6" }, { "input": "9 1", "output": "5" }, { "input": "9 2", "output": "5" }, { "input": "9 3", "output": "5" }, { "input": "9 4", "output": "5" }, { "input": "9 5", "output": "5" }, { "input": "9 6", "output": "5" }, { "input": "9 7", "output": "6" }, { "input": "10 1", "output": "5" }, { "input": "10 2", "output": "5" }, { "input": "10 3", "output": "5" }, { "input": "10 4", "output": "5" }, { "input": "10 5", "output": "5" }, { "input": "10 6", "output": "6" }, { "input": "10 7", "output": "6" }, { "input": "11 1", "output": "5" }, { "input": "11 2", "output": "5" }, { "input": "11 3", "output": "5" }, { "input": "11 4", "output": "5" }, { "input": "11 5", "output": "5" }, { "input": "11 6", "output": "5" }, { "input": "11 7", "output": "6" }, { "input": "12 1", "output": "5" }, { "input": "12 2", "output": "5" }, { "input": "12 3", "output": "5" }, { "input": "12 4", "output": "5" }, { "input": "12 5", "output": "5" }, { "input": "12 6", "output": "6" }, { "input": "12 7", "output": "6" }, { "input": "1 4", "output": "5" }, { "input": "1 5", "output": "5" }, { "input": "9 7", "output": "6" }, { "input": "2 6", "output": "5" }, { "input": "1 6", "output": "6" }, { "input": "2 2", "output": "5" }, { "input": "4 7", "output": "6" }, { "input": "12 6", "output": "6" }, { "input": "12 3", "output": "5" }, { "input": "3 6", "output": "6" }, { "input": "9 6", "output": "5" }, { "input": "7 6", "output": "6" }, { "input": "11 7", "output": "6" }, { "input": "6 6", "output": "5" } ]

1,485,109,184

284

Python 3

OK

TESTS

104

78

4,710,400

month, day = list(map(int, input().split())) months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] fs = 7 - day + 1 counter = 1 while fs + 7 <= months[month - 1]: counter += 1 fs += 7 if (fs != months[month - 1]): counter += 1 print(counter)

Title: Petr and a calendar Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture: Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap. Input Specification: The only line contain two integers *m* and *d* (1<=≤<=*m*<=≤<=12, 1<=≤<=*d*<=≤<=7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday). Output Specification: Print single integer: the number of columns the table should have. Demo Input: ['1 7\n', '1 1\n', '11 6\n'] Demo Output: ['6\n', '5\n', '5\n'] Note: The first example corresponds to the January 2017 shown on the picture in the statements. In the second example 1-st January is Monday, so the whole month fits into 5 columns. In the third example 1-st November is Saturday and 5 columns is enough.

```python month, day = list(map(int, input().split())) months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] fs = 7 - day + 1 counter = 1 while fs + 7 <= months[month - 1]: counter += 1 fs += 7 if (fs != months[month - 1]): counter += 1 print(counter) ```

3

699

A

Launch of Collider

PROGRAMMING

1,000

[ "implementation" ]

null

There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers. You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time. Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.

The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles. The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right. The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.

In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion. Print the only integer -1, if the collision of particles doesn't happen.

[ "4\nRLRL\n2 4 6 10\n", "3\nLLR\n40 50 60\n" ]

[ "1\n", "-1\n" ]

In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3. In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point.

500

[ { "input": "4\nRLRL\n2 4 6 10", "output": "1" }, { "input": "3\nLLR\n40 50 60", "output": "-1" }, { "input": "4\nRLLR\n46 230 264 470", "output": "92" }, { "input": "6\nLLRLLL\n446 492 650 844 930 970", "output": "97" }, { "input": "8\nRRLLLLLL\n338 478 512 574 594 622 834 922", "output": "17" }, { "input": "10\nLRLRLLRRLR\n82 268 430 598 604 658 670 788 838 1000", "output": "3" }, { "input": "2\nRL\n0 1000000000", "output": "500000000" }, { "input": "12\nLRLLRRRRLRLL\n254 1260 1476 1768 2924 4126 4150 4602 5578 7142 8134 9082", "output": "108" }, { "input": "14\nRLLRRLRLLRLLLR\n698 2900 3476 3724 3772 3948 4320 4798 5680 6578 7754 8034 8300 8418", "output": "88" }, { "input": "16\nRRLLLRLRLLLLRLLR\n222 306 968 1060 1636 1782 2314 2710 3728 4608 5088 6790 6910 7156 7418 7668", "output": "123" }, { "input": "18\nRLRLLRRRLLLRLRRLRL\n1692 2028 2966 3008 3632 4890 5124 5838 6596 6598 6890 8294 8314 8752 8868 9396 9616 9808", "output": "10" }, { "input": "20\nRLLLLLLLRRRRLRRLRRLR\n380 902 1400 1834 2180 2366 2562 2596 2702 2816 3222 3238 3742 5434 6480 7220 7410 8752 9708 9970", "output": "252" }, { "input": "22\nLRRRRRRRRRRRLLRRRRRLRL\n1790 2150 2178 2456 2736 3282 3622 4114 4490 4772 5204 5240 5720 5840 5910 5912 6586 7920 8584 9404 9734 9830", "output": "48" }, { "input": "24\nLLRLRRLLRLRRRRLLRRLRLRRL\n100 360 864 1078 1360 1384 1438 2320 2618 3074 3874 3916 3964 5178 5578 6278 6630 6992 8648 8738 8922 8930 9276 9720", "output": "27" }, { "input": "26\nRLLLLLLLRLRRLRLRLRLRLLLRRR\n908 1826 2472 2474 2728 3654 3716 3718 3810 3928 4058 4418 4700 5024 5768 6006 6128 6386 6968 7040 7452 7774 7822 8726 9338 9402", "output": "59" }, { "input": "28\nRRLRLRRRRRRLLLRRLRRLLLRRLLLR\n156 172 1120 1362 2512 3326 3718 4804 4990 5810 6242 6756 6812 6890 6974 7014 7088 7724 8136 8596 8770 8840 9244 9250 9270 9372 9400 9626", "output": "10" }, { "input": "30\nRLLRLRLLRRRLRRRLLLLLLRRRLRRLRL\n128 610 1680 2436 2896 2994 3008 3358 3392 4020 4298 4582 4712 4728 5136 5900 6088 6232 6282 6858 6934 7186 7224 7256 7614 8802 8872 9170 9384 9794", "output": "7" }, { "input": "10\nLLLLRRRRRR\n0 2 4 6 8 10 12 14 16 18", "output": "-1" }, { "input": "5\nLLLLL\n0 10 20 30 40", "output": "-1" }, { "input": "6\nRRRRRR\n40 50 60 70 80 100", "output": "-1" }, { "input": "1\nR\n0", "output": "-1" }, { "input": "2\nRL\n2 1000000000", "output": "499999999" }, { "input": "2\nRL\n0 400000", "output": "200000" }, { "input": "2\nRL\n0 200002", "output": "100001" }, { "input": "2\nRL\n2 20000000", "output": "9999999" }, { "input": "4\nLLRL\n2 4 10 100", "output": "45" }, { "input": "4\nRLRL\n2 10 12 14", "output": "1" }, { "input": "2\nRL\n0 100000000", "output": "50000000" }, { "input": "2\nRL\n2 600002", "output": "300000" }, { "input": "1\nL\n0", "output": "-1" }, { "input": "2\nRL\n0 600000", "output": "300000" }, { "input": "5\nRRRRR\n0 2 4 6 8", "output": "-1" }, { "input": "2\nRL\n2 200000000", "output": "99999999" }, { "input": "2\nRL\n0 267382766", "output": "133691383" }, { "input": "3\nRRL\n4 8 999999998", "output": "499999995" }, { "input": "2\nRL\n0 2", "output": "1" }, { "input": "2\nRL\n2 400002", "output": "200000" }, { "input": "2\nLL\n2 4", "output": "-1" }, { "input": "2\nLL\n0 2", "output": "-1" }, { "input": "2\nRL\n0 100000", "output": "50000" }, { "input": "2\nRL\n2 200000020", "output": "100000009" }, { "input": "2\nRL\n2000000 4000000", "output": "1000000" }, { "input": "2\nRL\n0 199998", "output": "99999" }, { "input": "3\nLRR\n40 50 60", "output": "-1" }, { "input": "2\nRL\n200 400400", "output": "200100" }, { "input": "2\nRL\n2 400004", "output": "200001" }, { "input": "2\nRL\n0 200000000", "output": "100000000" } ]

1,696,852,513

2,147,483,647

Python 3

OK

TESTS

85

155

28,057,600

n = int(input()) s = input() ans = 1000000000 flag = False arr = list(map(int,input().split())) for i in range(n-1): if s[i] == 'R' and s[i+1] == 'L': ans = min(ans, ((arr[i+1] - arr[i])/2)) flag = True if flag: print(int(ans)) else: print(-1)

Title: Launch of Collider Time Limit: None seconds Memory Limit: None megabytes Problem Description: There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers. You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time. Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point. Input Specification: The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles. The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right. The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order. Output Specification: In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion. Print the only integer -1, if the collision of particles doesn't happen. Demo Input: ['4\nRLRL\n2 4 6 10\n', '3\nLLR\n40 50 60\n'] Demo Output: ['1\n', '-1\n'] Note: In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3. In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point.

```python n = int(input()) s = input() ans = 1000000000 flag = False arr = list(map(int,input().split())) for i in range(n-1): if s[i] == 'R' and s[i+1] == 'L': ans = min(ans, ((arr[i+1] - arr[i])/2)) flag = True if flag: print(int(ans)) else: print(-1) ```

3

61

A

Ultra-Fast Mathematician

PROGRAMMING

800

[ "implementation" ]

A. Ultra-Fast Mathematician

2

256

Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate.

There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.

Write one line — the corresponding answer. Do not omit the leading 0s.

[ "1010100\n0100101\n", "000\n111\n", "1110\n1010\n", "01110\n01100\n" ]

[ "1110001\n", "111\n", "0100\n", "00010\n" ]

none

500

[ { "input": "1010100\n0100101", "output": "1110001" }, { "input": "000\n111", "output": "111" }, { "input": "1110\n1010", "output": "0100" }, { "input": "01110\n01100", "output": "00010" }, { "input": "011101\n000001", "output": "011100" }, { "input": "10\n01", "output": "11" }, { "input": "00111111\n11011101", "output": "11100010" }, { "input": "011001100\n101001010", "output": "110000110" }, { "input": "1100100001\n0110101100", "output": "1010001101" }, { "input": "00011101010\n10010100101", "output": "10001001111" }, { "input": "100000101101\n111010100011", "output": "011010001110" }, { "input": "1000001111010\n1101100110001", "output": "0101101001011" }, { "input": "01011111010111\n10001110111010", "output": "11010001101101" }, { "input": "110010000111100\n001100101011010", "output": "111110101100110" }, { "input": "0010010111110000\n0000000011010110", "output": "0010010100100110" }, { "input": "00111110111110000\n01111100001100000", "output": "01000010110010000" }, { "input": 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1,687,957,466

2,147,483,647

Python 3

OK

TESTS

102

46

0

x = list(input()) y = list(input()) res = [] for i in range(len(x)): if x[i] == y[i]: res.append(0) else: res.append(1) for j in range(len(res)): print(res[j],end='')

Title: Ultra-Fast Mathematician Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second. One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part. In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0. Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length. Now you are going to take part in Shapur's contest. See if you are faster and more accurate. Input Specification: There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. Output Specification: Write one line — the corresponding answer. Do not omit the leading 0s. Demo Input: ['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n'] Demo Output: ['1110001\n', '111\n', '0100\n', '00010\n'] Note: none

```python x = list(input()) y = list(input()) res = [] for i in range(len(x)): if x[i] == y[i]: res.append(0) else: res.append(1) for j in range(len(res)): print(res[j],end='') ```

3.9885

855

B

Marvolo Gaunt's Ring

PROGRAMMING

1,500

[ "brute force", "data structures", "dp" ]

null

Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made. Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative.

First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105). Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).

Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*.

[ "5 1 2 3\n1 2 3 4 5\n", "5 1 2 -3\n-1 -2 -3 -4 -5\n" ]

[ "30\n", "12\n" ]

In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30. In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12.

1,000

[ { "input": "5 1 2 3\n1 2 3 4 5", "output": "30" }, { "input": "5 1 2 -3\n-1 -2 -3 -4 -5", "output": "12" }, { "input": "5 886327859 82309257 -68295239\n-731225382 354766539 -48222231 -474691998 360965777", "output": "376059240645059046" }, { "input": "4 -96405765 -495906217 625385006\n-509961652 392159235 -577128498 -744548876", "output": "547306902373544674" }, { "input": "43 959134961 -868367850 142426380\n921743429 63959718 -797293233 122041422 -407576197 700139744 299598010 168207043 362252658 591926075 941946099 812263640 -76679927 -824267725 89529990 -73303355 83596189 -982699817 -235197848 654773327 125211479 -497091570 -2301804 203486596 -126652024 309810546 -581289415 -740125230 64425927 -501018049 304730559 34930193 -762964086 723645139 -826821494 495947907 816331024 9932423 -876541603 -782692568 322360800 841436938 40787162", "output": "1876641179289775029" }, { "input": "1 0 0 0\n0", "output": "0" }, { "input": "1 1000000000 1000000000 1000000000\n1000000000", "output": "3000000000000000000" }, { "input": "1 -1000000000 -1000000000 1000000000\n1000000000", "output": "-1000000000000000000" }, { "input": "1 -1000000000 -1000000000 -1000000000\n1000000000", "output": "-3000000000000000000" }, { "input": "3 1000000000 1000000000 1000000000\n-1000000000 -1000000000 -1000000000", "output": "-3000000000000000000" }, { "input": "1 1 1 1\n-1", "output": "-3" }, { "input": "1 -1 -1 -1\n1", "output": "-3" }, { "input": "1 1000000000 1000000000 1000000000\n-1000000000", "output": "-3000000000000000000" }, { "input": "1 1 2 3\n-1", "output": "-6" }, { "input": "3 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000", "output": "-3000000000000000000" }, { "input": "2 -1000000000 -1000000000 -1000000000\n1000000000 1000000000", "output": "-3000000000000000000" }, { "input": "3 1 1 1\n-1 -1 -1", "output": "-3" }, { "input": "1 -1000000000 0 0\n1000000000", "output": "-1000000000000000000" }, { "input": "1 -100 -100 -100\n100", "output": "-30000" }, { "input": "5 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000", "output": "-3000000000000000000" }, { "input": "1 999999999 999999999 999999999\n-999999999", "output": "-2999999994000000003" }, { "input": "3 -1000000000 -1000000000 1\n1000000000 1000000000 1000000000", "output": "-1999999999000000000" }, { "input": "3 -2 3 -2\n1 2 1", "output": "2" }, { "input": "2 1 -1 1\n1 -1", "output": "1" }, { "input": "1 -1000000000 1 -1000000000\n1000000000", "output": "-1999999999000000000" }, { "input": "1 1000000000 1000000000 -1000000000\n-1000000000", "output": "-1000000000000000000" }, { "input": "1 -1000000000 -1000000000 0\n1000000000", "output": "-2000000000000000000" } ]

1,581,968,417

2,147,483,647

Python 3

OK

TESTS

82

514

11,571,200

n,p,q,r=map(int,input().split()) arr=[int(i) for i in input().split()] # leftmax=[arr[0]] # for i in range(1,len(arr)): # leftmax.append(max(leftmax[-1],arr[i])) # print(leftmax) # rightmax=[arr[-1]] # for i in range(len(arr)-2,-1,-1): # rightmax.append(max(rightmax[-1],arr[i])) # rightmax.reverse() # print(rightmax) # ans=float('-inf') # for i in range(1,len(arr)-1): # temp=(p*leftmax[i-1])+(q*arr[i])+(r*rightmax[i+1]) # print(temp) # ans=max(ans,temp) parr=[arr[0]*p] for i in range(1,len(arr)): val=max(arr[i]*p,parr[i-1]) parr.append(val) qarr=[arr[0]*q+(p*arr[0])] for i in range(1,len(arr)): val=max(qarr[i-1],(arr[i]*q)+(parr[i])) qarr.append(val) rarr=[(arr[0]*q)+(p*arr[0])+(r*arr[0])] for i in range(1,len(arr)): val=max(rarr[i-1],r*arr[i]+qarr[i]) rarr.append(val) print(rarr[-1])

Title: Marvolo Gaunt's Ring Time Limit: None seconds Memory Limit: None megabytes Problem Description: Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made. Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative. Input Specification: First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105). Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). Output Specification: Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Demo Input: ['5 1 2 3\n1 2 3 4 5\n', '5 1 2 -3\n-1 -2 -3 -4 -5\n'] Demo Output: ['30\n', '12\n'] Note: In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30. In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12.

```python n,p,q,r=map(int,input().split()) arr=[int(i) for i in input().split()] # leftmax=[arr[0]] # for i in range(1,len(arr)): # leftmax.append(max(leftmax[-1],arr[i])) # print(leftmax) # rightmax=[arr[-1]] # for i in range(len(arr)-2,-1,-1): # rightmax.append(max(rightmax[-1],arr[i])) # rightmax.reverse() # print(rightmax) # ans=float('-inf') # for i in range(1,len(arr)-1): # temp=(p*leftmax[i-1])+(q*arr[i])+(r*rightmax[i+1]) # print(temp) # ans=max(ans,temp) parr=[arr[0]*p] for i in range(1,len(arr)): val=max(arr[i]*p,parr[i-1]) parr.append(val) qarr=[arr[0]*q+(p*arr[0])] for i in range(1,len(arr)): val=max(qarr[i-1],(arr[i]*q)+(parr[i])) qarr.append(val) rarr=[(arr[0]*q)+(p*arr[0])+(r*arr[0])] for i in range(1,len(arr)): val=max(rarr[i-1],r*arr[i]+qarr[i]) rarr.append(val) print(rarr[-1]) ```

3

724

A

Checking the Calendar

PROGRAMMING

1,000

[ "implementation" ]

null

You are given names of two days of the week. Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year. In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31. Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".

The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".

Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).

[ "monday\ntuesday\n", "sunday\nsunday\n", "saturday\ntuesday\n" ]

[ "NO\n", "YES\n", "YES\n" ]

In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays. In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday.

500

[ { "input": "monday\ntuesday", "output": "NO" }, { "input": "sunday\nsunday", "output": "YES" }, { "input": "saturday\ntuesday", "output": "YES" }, { "input": "tuesday\nthursday", "output": "YES" }, { "input": "friday\nwednesday", "output": "NO" }, { "input": "sunday\nsaturday", "output": "NO" }, { "input": "monday\nmonday", "output": "YES" }, { "input": "monday\nwednesday", "output": "YES" }, { "input": "monday\nthursday", "output": "YES" }, { "input": "monday\nfriday", "output": "NO" }, { "input": "monday\nsaturday", "output": "NO" }, { "input": "monday\nsunday", "output": "NO" }, { "input": "tuesday\nmonday", "output": "NO" }, { "input": "tuesday\ntuesday", "output": "YES" }, { "input": "tuesday\nwednesday", "output": "NO" }, { "input": "tuesday\nfriday", "output": "YES" }, { "input": "tuesday\nsaturday", "output": "NO" }, { "input": "tuesday\nsunday", "output": "NO" }, { "input": "wednesday\nmonday", "output": "NO" }, { "input": "wednesday\ntuesday", "output": "NO" }, { "input": "wednesday\nwednesday", "output": "YES" }, { "input": "wednesday\nthursday", "output": "NO" }, { "input": "wednesday\nfriday", "output": "YES" }, { "input": "wednesday\nsaturday", "output": "YES" }, { "input": "wednesday\nsunday", "output": "NO" }, { "input": "thursday\nmonday", "output": "NO" }, { "input": "thursday\ntuesday", "output": "NO" }, { "input": "thursday\nwednesday", "output": "NO" }, { "input": "thursday\nthursday", "output": "YES" }, { "input": "thursday\nfriday", "output": "NO" }, { "input": "thursday\nsaturday", "output": "YES" }, { "input": "thursday\nsunday", "output": "YES" }, { "input": "friday\nmonday", "output": "YES" }, { "input": "friday\ntuesday", "output": "NO" }, { "input": "friday\nthursday", "output": "NO" }, { "input": "friday\nsaturday", "output": "NO" }, { "input": "friday\nsunday", "output": "YES" }, { "input": "saturday\nmonday", "output": "YES" }, { "input": "saturday\nwednesday", "output": "NO" }, { "input": "saturday\nthursday", "output": "NO" }, { "input": "saturday\nfriday", "output": "NO" }, { "input": "saturday\nsaturday", "output": "YES" }, { "input": "saturday\nsunday", "output": "NO" }, { "input": "sunday\nmonday", "output": "NO" }, { "input": "sunday\ntuesday", "output": "YES" }, { "input": "sunday\nwednesday", "output": "YES" }, { "input": "sunday\nthursday", "output": "NO" }, { "input": "sunday\nfriday", "output": "NO" }, { "input": "friday\nfriday", "output": "YES" }, { "input": "friday\nsunday", "output": "YES" }, { "input": "monday\nmonday", "output": "YES" }, { "input": "friday\ntuesday", "output": "NO" }, { "input": "thursday\nsaturday", "output": "YES" }, { "input": "tuesday\nfriday", "output": "YES" }, { "input": "sunday\nwednesday", "output": "YES" }, { "input": "monday\nthursday", "output": "YES" }, { "input": "saturday\nsunday", "output": "NO" }, { "input": "friday\nmonday", "output": "YES" }, { "input": "thursday\nthursday", "output": "YES" }, { "input": "wednesday\nfriday", "output": "YES" }, { "input": "thursday\nmonday", "output": "NO" }, { "input": "wednesday\nsunday", "output": "NO" }, { "input": "thursday\nfriday", "output": "NO" }, { "input": "monday\nfriday", "output": "NO" }, { "input": "wednesday\nsaturday", "output": "YES" }, { "input": "thursday\nsunday", "output": "YES" }, { "input": "saturday\nfriday", "output": "NO" }, { "input": "saturday\nmonday", "output": "YES" } ]

1,585,699,776

2,147,483,647

Python 3

OK

TESTS

68

109

307,200

import sys import math #to read string get_string = lambda: sys.stdin.readline().strip() #to read list of integers get_int_list = lambda: list( map(int,sys.stdin.readline().strip().split()) ) #to read integers get_int = lambda: int(sys.stdin.readline()) #--------------------------------WhiteHat010--------------------------------------# a = get_string() b = get_string() days = { "monday": 0, "tuesday": 1, "wednesday": 2, "thursday": 3, "friday": 4, "saturday": 5, "sunday": 6 } valid = {0,2,3} if (days[b] - days[a])%7 in valid: print("YES") else: print("NO")

Title: Checking the Calendar Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given names of two days of the week. Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year. In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31. Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday". Input Specification: The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday". Output Specification: Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes). Demo Input: ['monday\ntuesday\n', 'sunday\nsunday\n', 'saturday\ntuesday\n'] Demo Output: ['NO\n', 'YES\n', 'YES\n'] Note: In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays. In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday.

```python import sys import math #to read string get_string = lambda: sys.stdin.readline().strip() #to read list of integers get_int_list = lambda: list( map(int,sys.stdin.readline().strip().split()) ) #to read integers get_int = lambda: int(sys.stdin.readline()) #--------------------------------WhiteHat010--------------------------------------# a = get_string() b = get_string() days = { "monday": 0, "tuesday": 1, "wednesday": 2, "thursday": 3, "friday": 4, "saturday": 5, "sunday": 6 } valid = {0,2,3} if (days[b] - days[a])%7 in valid: print("YES") else: print("NO") ```

3

1,004

C

Sonya and Robots

PROGRAMMING

1,400

[ "constructive algorithms", "implementation" ]

null

Since Sonya is interested in robotics too, she decided to construct robots that will read and recognize numbers. Sonya has drawn $n$ numbers in a row, $a_i$ is located in the $i$-th position. She also has put a robot at each end of the row (to the left of the first number and to the right of the last number). Sonya will give a number to each robot (they can be either same or different) and run them. When a robot is running, it is moving toward to another robot, reading numbers in the row. When a robot is reading a number that is equal to the number that was given to that robot, it will turn off and stay in the same position. Sonya does not want robots to break, so she will give such numbers that robots will stop before they meet. That is, the girl wants them to stop at different positions so that the first robot is to the left of the second one. For example, if the numbers $[1, 5, 4, 1, 3]$ are written, and Sonya gives the number $1$ to the first robot and the number $4$ to the second one, the first robot will stop in the $1$-st position while the second one in the $3$-rd position. In that case, robots will not meet each other. As a result, robots will not be broken. But if Sonya gives the number $4$ to the first robot and the number $5$ to the second one, they will meet since the first robot will stop in the $3$-rd position while the second one is in the $2$-nd position. Sonya understands that it does not make sense to give a number that is not written in the row because a robot will not find this number and will meet the other robot. Sonya is now interested in finding the number of different pairs that she can give to robots so that they will not meet. In other words, she wants to know the number of pairs ($p$, $q$), where she will give $p$ to the first robot and $q$ to the second one. Pairs ($p_i$, $q_i$) and ($p_j$, $q_j$) are different if $p_i\neq p_j$ or $q_i\neq q_j$. Unfortunately, Sonya is busy fixing robots that broke after a failed launch. That is why she is asking you to find the number of pairs that she can give to robots so that they will not meet.

The first line contains a single integer $n$ ($1\leq n\leq 10^5$) — the number of numbers in a row. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1\leq a_i\leq 10^5$) — the numbers in a row.

Print one number — the number of possible pairs that Sonya can give to robots so that they will not meet.

[ "5\n1 5 4 1 3\n", "7\n1 2 1 1 1 3 2\n" ]

[ "9\n", "7\n" ]

In the first example, Sonya can give pairs ($1$, $1$), ($1$, $3$), ($1$, $4$), ($1$, $5$), ($4$, $1$), ($4$, $3$), ($5$, $1$), ($5$, $3$), and ($5$, $4$). In the second example, Sonya can give pairs ($1$, $1$), ($1$, $2$), ($1$, $3$), ($2$, $1$), ($2$, $2$), ($2$, $3$), and ($3$, $2$).

1,500

[ { "input": "5\n1 5 4 1 3", "output": "9" }, { "input": "7\n1 2 1 1 1 3 2", "output": "7" }, { "input": "10\n2 2 4 4 3 1 1 2 3 2", "output": "14" }, { "input": "15\n1 2 2 1 2 4 2 1 1 6 6 4 2 5 4", "output": "20" }, { "input": "1\n1", "output": "0" } ]

1,671,679,550

2,147,483,647

PyPy 3-64

OK

TESTS

42

124

20,992,000

n = int(input()) l = list(map(int, input().split())) d = dict() s = set() t = [0 for i in range(n)] ans = 0 diff = 0 for i in range(n): if l[i] in d: d[l[i]] += [i] else: d[l[i]] = [i] diff += 1 t[i] = diff # print(t) # print(d) for x in d: i = d[x][-1] # print(i) if i > 0: ans += t[i-1] print(ans)

Title: Sonya and Robots Time Limit: None seconds Memory Limit: None megabytes Problem Description: Since Sonya is interested in robotics too, she decided to construct robots that will read and recognize numbers. Sonya has drawn $n$ numbers in a row, $a_i$ is located in the $i$-th position. She also has put a robot at each end of the row (to the left of the first number and to the right of the last number). Sonya will give a number to each robot (they can be either same or different) and run them. When a robot is running, it is moving toward to another robot, reading numbers in the row. When a robot is reading a number that is equal to the number that was given to that robot, it will turn off and stay in the same position. Sonya does not want robots to break, so she will give such numbers that robots will stop before they meet. That is, the girl wants them to stop at different positions so that the first robot is to the left of the second one. For example, if the numbers $[1, 5, 4, 1, 3]$ are written, and Sonya gives the number $1$ to the first robot and the number $4$ to the second one, the first robot will stop in the $1$-st position while the second one in the $3$-rd position. In that case, robots will not meet each other. As a result, robots will not be broken. But if Sonya gives the number $4$ to the first robot and the number $5$ to the second one, they will meet since the first robot will stop in the $3$-rd position while the second one is in the $2$-nd position. Sonya understands that it does not make sense to give a number that is not written in the row because a robot will not find this number and will meet the other robot. Sonya is now interested in finding the number of different pairs that she can give to robots so that they will not meet. In other words, she wants to know the number of pairs ($p$, $q$), where she will give $p$ to the first robot and $q$ to the second one. Pairs ($p_i$, $q_i$) and ($p_j$, $q_j$) are different if $p_i\neq p_j$ or $q_i\neq q_j$. Unfortunately, Sonya is busy fixing robots that broke after a failed launch. That is why she is asking you to find the number of pairs that she can give to robots so that they will not meet. Input Specification: The first line contains a single integer $n$ ($1\leq n\leq 10^5$) — the number of numbers in a row. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1\leq a_i\leq 10^5$) — the numbers in a row. Output Specification: Print one number — the number of possible pairs that Sonya can give to robots so that they will not meet. Demo Input: ['5\n1 5 4 1 3\n', '7\n1 2 1 1 1 3 2\n'] Demo Output: ['9\n', '7\n'] Note: In the first example, Sonya can give pairs ($1$, $1$), ($1$, $3$), ($1$, $4$), ($1$, $5$), ($4$, $1$), ($4$, $3$), ($5$, $1$), ($5$, $3$), and ($5$, $4$). In the second example, Sonya can give pairs ($1$, $1$), ($1$, $2$), ($1$, $3$), ($2$, $1$), ($2$, $2$), ($2$, $3$), and ($3$, $2$).

```python n = int(input()) l = list(map(int, input().split())) d = dict() s = set() t = [0 for i in range(n)] ans = 0 diff = 0 for i in range(n): if l[i] in d: d[l[i]] += [i] else: d[l[i]] = [i] diff += 1 t[i] = diff # print(t) # print(d) for x in d: i = d[x][-1] # print(i) if i > 0: ans += t[i-1] print(ans) ```

3

550

A

Two Substrings

PROGRAMMING

1,500

[ "brute force", "dp", "greedy", "implementation", "strings" ]

null

You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).

The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters.

Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.

[ "ABA\n", "BACFAB\n", "AXBYBXA\n" ]

[ "NO\n", "YES\n", "NO\n" ]

In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO". In the second sample test there are the following occurrences of the substrings: BACFAB. In the third sample test there is no substring "AB" nor substring "BA".

1,000

[ { "input": "ABA", "output": "NO" }, { "input": "BACFAB", "output": "YES" }, { "input": "AXBYBXA", "output": "NO" }, { "input": "ABABAB", "output": "YES" }, { "input": "BBBBBBBBBB", "output": "NO" }, { "input": "ABBA", "output": "YES" }, { "input": "ABAXXXAB", "output": "YES" }, { "input": "TESTABAXXABTEST", "output": "YES" }, { "input": "A", "output": "NO" }, { "input": "B", "output": "NO" }, { "input": "X", "output": "NO" }, { "input": "BA", "output": "NO" }, { "input": "AB", "output": "NO" }, { "input": "AA", "output": "NO" }, { "input": "BB", "output": "NO" }, { "input": "BAB", "output": "NO" }, { "input": "AAB", "output": "NO" }, { "input": "BAA", "output": "NO" }, { "input": "ABB", "output": "NO" }, { "input": "BBA", "output": "NO" }, { "input": "AAA", "output": "NO" }, { "input": "BBB", "output": "NO" }, { "input": "AXBXBXA", "output": "NO" }, { "input": "SKDSKDJABSDBADKFJDK", "output": "YES" }, { "input": "ABAXXBBXXAA", "output": "NO" }, { "input": "ABAB", "output": "NO" }, { "input": "BABA", "output": "NO" }, { "input": "AAAB", "output": "NO" }, { "input": "AAAA", "output": "NO" }, { "input": "AABA", "output": "NO" }, { "input": "ABAA", "output": "NO" }, { "input": "BAAA", "output": "NO" }, { "input": "AABB", "output": "NO" }, { "input": "BAAB", "output": "YES" }, { "input": "BBAA", "output": "NO" }, { "input": "BBBA", "output": "NO" }, { "input": "BBAB", "output": "NO" }, { "input": "BABB", "output": "NO" }, { "input": "ABBB", "output": "NO" }, { "input": "BBBB", "output": "NO" }, { "input": "BABAB", "output": "YES" }, { "input": "ABABA", "output": "YES" }, { "input": "AAABAAACBBBC", "output": "NO" }, { "input": "AABBBACBBBBBBAACBCCACBBAABBBBBCAACABAACABCACCBCBCCCBCBCABCBBCCBCBBAACBACAABACBBCACCBCCACCABBCBABBAAC", "output": "YES" }, { "input": "CBBABDDBBADAC", "output": "YES" }, { "input": "ABYFAB", "output": "NO" }, { "input": "BABXXAB", "output": "YES" }, { "input": "ABAXAB", "output": "YES" }, { "input": "ABABXAB", "output": "YES" }, { "input": "ABXABXABXABXABXBAXBAXBAXBA", "output": "YES" }, { "input": "QQQQQQQQQABABQQQQQQQQ", "output": "NO" } ]

1,678,480,588

2,147,483,647

PyPy 3-64

OK

TESTS

85

62

3,891,200

def main(): s = input() ab = [] ba = [] for i in range(len(s)-1): if s[i:i+2]=="AB": if any(abs(x-i)>1 for x in ba): print("YES") return ab.append(i) elif s[i:i+2]=="BA": if any(abs(x-i)>1 for x in ab): print("YES") return ba.append(i) print("NO") main()

Title: Two Substrings Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). Input Specification: The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters. Output Specification: Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. Demo Input: ['ABA\n', 'BACFAB\n', 'AXBYBXA\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n'] Note: In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO". In the second sample test there are the following occurrences of the substrings: BACFAB. In the third sample test there is no substring "AB" nor substring "BA".

```python def main(): s = input() ab = [] ba = [] for i in range(len(s)-1): if s[i:i+2]=="AB": if any(abs(x-i)>1 for x in ba): print("YES") return ab.append(i) elif s[i:i+2]=="BA": if any(abs(x-i)>1 for x in ab): print("YES") return ba.append(i) print("NO") main() ```

3

822

A

I'm bored with life

PROGRAMMING

800

[ "implementation", "math", "number theory" ]

null

Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom! Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*. Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?

The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).

Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.

[ "4 3\n" ]

[ "6\n" ]

Consider the sample. 4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.

500

[ { "input": "4 3", "output": "6" }, { "input": "10 399603090", "output": "3628800" }, { "input": "6 973151934", "output": "720" }, { "input": "2 841668075", "output": "2" }, { "input": "7 415216919", "output": "5040" }, { "input": "3 283733059", "output": "6" }, { "input": "11 562314608", "output": "39916800" }, { "input": "3 990639260", "output": "6" }, { "input": "11 859155400", "output": "39916800" }, { "input": "1 1", "output": "1" }, { "input": "5 3", "output": "6" }, { "input": "1 4", "output": "1" }, { "input": "5 4", "output": "24" }, { "input": "1 12", "output": "1" }, { "input": "9 7", "output": "5040" }, { "input": "2 3", "output": "2" }, { "input": "6 11", "output": "720" }, { "input": "6 7", "output": "720" }, { "input": "11 11", "output": "39916800" }, { "input": "4 999832660", "output": "24" }, { "input": "7 999228288", "output": "5040" }, { "input": "11 999257105", "output": "39916800" }, { "input": "11 999286606", "output": "39916800" }, { "input": "3 999279109", "output": "6" }, { "input": "999632727 11", "output": "39916800" }, { "input": "999625230 7", "output": "5040" }, { "input": "999617047 3", "output": "6" }, { "input": "999646548 7", "output": "5040" }, { "input": "999639051 3", "output": "6" }, { "input": "12 12", "output": "479001600" }, { "input": "12 1", "output": "1" }, { "input": "1213 5", "output": "120" }, { "input": "8 9", "output": "40320" }, { "input": "12 9", "output": "362880" }, { "input": "12 1000000000", "output": "479001600" }, { "input": "1000000000 1", "output": "1" }, { "input": "12 13", "output": "479001600" }, { "input": "2 29845", "output": "2" }, { "input": "10 21", "output": "3628800" }, { "input": "12 20", "output": "479001600" }, { "input": "15 12", "output": "479001600" }, { "input": "1 1", "output": "1" }, { "input": "1000000000 12", "output": "479001600" }, { "input": "11 30", "output": "39916800" }, { "input": "17 12", "output": "479001600" }, { "input": "4 19", "output": "24" }, { "input": "12 15", "output": "479001600" }, { "input": "20 6", "output": "720" }, { "input": "10 20", "output": "3628800" }, { "input": "10 10", "output": "3628800" }, { "input": "22 12", "output": "479001600" }, { "input": "20 12", "output": "479001600" }, { "input": "12 23", "output": "479001600" }, { "input": "12 22", "output": "479001600" }, { "input": "18 3", "output": "6" }, { "input": "14 10", "output": "3628800" }, { "input": "14 12", "output": "479001600" }, { "input": "8 3", "output": "6" }, { "input": "5 5", "output": "120" } ]

1,662,568,197

2,147,483,647

Python 3

OK

TESTS

59

46

0

n, m= map(int,input().split()) x = min(n,m) for i in range(1,x): x = x*i print(x)

Title: I'm bored with life Time Limit: None seconds Memory Limit: None megabytes Problem Description: Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom! Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*. Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you? Input Specification: The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12). Output Specification: Print a single integer denoting the greatest common divisor of integers *A*! and *B*!. Demo Input: ['4 3\n'] Demo Output: ['6\n'] Note: Consider the sample. 4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.

```python n, m= map(int,input().split()) x = min(n,m) for i in range(1,x): x = x*i print(x) ```

3

697

B

Barnicle

PROGRAMMING

1,400

[ "brute force", "implementation", "math", "strings" ]

null

Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the distance between him and the girl to make his shot accurate. Barney asked the bar tender Carl about this distance value, but Carl was so busy talking to the customers so he wrote the distance value (it's a real number) on a napkin. The problem is that he wrote it in scientific notation. The scientific notation of some real number *x* is the notation of form *AeB*, where *A* is a real number and *B* is an integer and *x*<==<=*A*<=×<=10*B* is true. In our case *A* is between 0 and 9 and *B* is non-negative. Barney doesn't know anything about scientific notation (as well as anything scientific at all). So he asked you to tell him the distance value in usual decimal representation with minimal number of digits after the decimal point (and no decimal point if it is an integer). See the output format for better understanding.

The first and only line of input contains a single string of form *a*.*deb* where *a*, *d* and *b* are integers and *e* is usual character 'e' (0<=≤<=*a*<=≤<=9,<=0<=≤<=*d*<=<<=10100,<=0<=≤<=*b*<=≤<=100) — the scientific notation of the desired distance value. *a* and *b* contain no leading zeros and *d* contains no trailing zeros (but may be equal to 0). Also, *b* can not be non-zero if *a* is zero.

Print the only real number *x* (the desired distance value) in the only line in its decimal notation. Thus if *x* is an integer, print it's integer value without decimal part and decimal point and without leading zeroes. Otherwise print *x* in a form of *p*.*q* such that *p* is an integer that have no leading zeroes (but may be equal to zero), and *q* is an integer that have no trailing zeroes (and may not be equal to zero).

[ "8.549e2\n", "8.549e3\n", "0.33e0\n" ]

[ "854.9\n", "8549\n", "0.33\n" ]

none

1,000

[ { "input": "8.549e2", "output": "854.9" }, { "input": "8.549e3", "output": "8549" }, { "input": "0.33e0", "output": "0.33" }, { "input": "1.31e1", "output": "13.1" }, { "input": "1.038e0", "output": "1.038" }, { "input": "8.25983e5", "output": "825983" }, { "input": "8.77056e6", "output": "8770560" }, { "input": "4.28522890224373996236468418851564462623381500262405e30", "output": "4285228902243739962364684188515.64462623381500262405" }, { "input": "4.09336275522154223604344399571355118601483591618747e85", "output": "40933627552215422360434439957135511860148359161874700000000000000000000000000000000000" }, { "input": "2.0629094807595491132306264747042243928486303384791951220362096240931158821630792563855724946791054152e85", "output": "20629094807595491132306264747042243928486303384791951220362096240931158821630792563855.724946791054152" }, { "input": "0.7e0", "output": "0.7" }, { "input": "0.75e0", "output": "0.75" }, { "input": "0.3299209894804593859495773277850971828150469972132991597085582244596065712639531451e0", "output": "0.3299209894804593859495773277850971828150469972132991597085582244596065712639531451" }, { "input": "0.1438410315232821898580886049593487999249997483354329425897344341660326482795266134253882860655873197e0", "output": "0.1438410315232821898580886049593487999249997483354329425897344341660326482795266134253882860655873197" }, { "input": "1.7282220592677586155528202123627915992640276211396528871e0", "output": "1.7282220592677586155528202123627915992640276211396528871" }, { "input": "1.91641639840522198229453882518758458881136053577016034847369545687354908120008812644841021662133251e89", "output": "191641639840522198229453882518758458881136053577016034847369545687354908120008812644841021.662133251" }, { "input": "7.0e100", "output": "70000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "1.7390193766535948887334396973270576641602486903095355363287177932797263236084900516267835886881779051e100", "output": "17390193766535948887334396973270576641602486903095355363287177932797263236084900516267835886881779051" }, { "input": "4.6329496401734172195e50", "output": "463294964017341721950000000000000000000000000000000" }, { "input": "2.806303180541991592302230754797823269634e39", "output": "2806303180541991592302230754797823269634" }, { "input": "5.8743505652112692964508303637002e64", "output": "58743505652112692964508303637002000000000000000000000000000000000" }, { "input": "6.8778661934058405217475274375560252344373481358834598914724956711e31", "output": "68778661934058405217475274375560.252344373481358834598914724956711" }, { "input": "9.4e100", "output": "94000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "3.2371070627618799335840070613481911588919091676203766004638236894609230433739617153911544972468224113e50", "output": "323710706276187993358400706134819115889190916762037.66004638236894609230433739617153911544972468224113" }, { "input": "4.8133196117786711780806656271869913331127534865038175322117213586960112955982462632332925275690064929e0", "output": "4.8133196117786711780806656271869913331127534865038175322117213586960112955982462632332925275690064929" }, { "input": "7.7060200967648284035308242369118752594772564843152902469146249303976625961451358536989314351204406625e1", "output": "77.060200967648284035308242369118752594772564843152902469146249303976625961451358536989314351204406625" }, { "input": "8.1089882894234341219420177467603732503076124872188628349726911362800974096687340341040683238197289136e31", "output": "81089882894234341219420177467603.732503076124872188628349726911362800974096687340341040683238197289136" }, { "input": "9.6576660076120385279859051742522204516365367878315639937449558670629833997839913220859648564428655877e99", "output": "9657666007612038527985905174252220451636536787831563993744955867062983399783991322085964856442865587.7" }, { "input": "0.0e0", "output": "0" }, { "input": "1.0e0", "output": "1" }, { "input": "8.0e0", "output": "8" }, { "input": "3.0e0", "output": "3" }, { "input": "4.0e0", "output": "4" }, { "input": "2.0e0", "output": "2" }, { "input": "9.0e0", "output": "9" }, { "input": "0.888888e0", "output": "0.888888" }, { "input": "9.99999999999999999999999999999999999999999999999999999999999999999999999999999999e100", "output": "99999999999999999999999999999999999999999999999999999999999999999999999999999999900000000000000000000" }, { "input": "5.0e0", "output": "5" }, { "input": "1.0e10", "output": "10000000000" }, { "input": "1.0e5", "output": "100000" }, { "input": "6.0e0", "output": "6" }, { "input": "1.1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111e1", "output": "11.111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" } ]

1,509,093,922

2,147,483,647

Python 3

OK

TESTS

42

62

512,000

from decimal import * a = input() b = Decimal(a) if round(b) == b: print("%d" % b) else: print(b)

Title: Barnicle Time Limit: None seconds Memory Limit: None megabytes Problem Description: Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the distance between him and the girl to make his shot accurate. Barney asked the bar tender Carl about this distance value, but Carl was so busy talking to the customers so he wrote the distance value (it's a real number) on a napkin. The problem is that he wrote it in scientific notation. The scientific notation of some real number *x* is the notation of form *AeB*, where *A* is a real number and *B* is an integer and *x*<==<=*A*<=×<=10*B* is true. In our case *A* is between 0 and 9 and *B* is non-negative. Barney doesn't know anything about scientific notation (as well as anything scientific at all). So he asked you to tell him the distance value in usual decimal representation with minimal number of digits after the decimal point (and no decimal point if it is an integer). See the output format for better understanding. Input Specification: The first and only line of input contains a single string of form *a*.*deb* where *a*, *d* and *b* are integers and *e* is usual character 'e' (0<=≤<=*a*<=≤<=9,<=0<=≤<=*d*<=<<=10100,<=0<=≤<=*b*<=≤<=100) — the scientific notation of the desired distance value. *a* and *b* contain no leading zeros and *d* contains no trailing zeros (but may be equal to 0). Also, *b* can not be non-zero if *a* is zero. Output Specification: Print the only real number *x* (the desired distance value) in the only line in its decimal notation. Thus if *x* is an integer, print it's integer value without decimal part and decimal point and without leading zeroes. Otherwise print *x* in a form of *p*.*q* such that *p* is an integer that have no leading zeroes (but may be equal to zero), and *q* is an integer that have no trailing zeroes (and may not be equal to zero). Demo Input: ['8.549e2\n', '8.549e3\n', '0.33e0\n'] Demo Output: ['854.9\n', '8549\n', '0.33\n'] Note: none

```python from decimal import * a = input() b = Decimal(a) if round(b) == b: print("%d" % b) else: print(b) ```

3

770

A

New Password

PROGRAMMING

800

[ "*special", "implementation" ]

null

Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help. Innokentiy decides that new password should satisfy the following conditions: - the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct. Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.

The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it. Pay attention that a desired new password always exists.

Print any password which satisfies all conditions given by Innokentiy.

[ "4 3\n", "6 6\n", "5 2\n" ]

[ "java\n", "python\n", "phphp\n" ]

In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it. In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters. In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it. Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.

500

[ { "input": "4 3", "output": "abca" }, { "input": "6 6", "output": "abcdef" }, { "input": "5 2", "output": "ababa" }, { "input": "3 2", "output": "aba" }, { "input": "10 2", "output": "ababababab" }, { "input": "26 13", "output": "abcdefghijklmabcdefghijklm" }, { "input": "100 2", "output": "abababababababababababababababababababababababababababababababababababababababababababababababababab" }, { "input": "100 10", "output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij" }, { "input": "3 3", "output": "abc" }, { "input": "6 3", "output": "abcabc" }, { "input": "10 3", "output": "abcabcabca" }, { "input": "50 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab" }, { "input": "90 2", "output": "ababababababababababababababababababababababababababababababababababababababababababababab" }, { "input": "6 2", "output": "ababab" }, { "input": "99 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc" }, { "input": "4 2", "output": "abab" }, { "input": "100 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca" }, { "input": "40 22", "output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr" }, { "input": "13 8", "output": "abcdefghabcde" }, { "input": "16 15", "output": "abcdefghijklmnoa" }, { "input": "17 17", "output": "abcdefghijklmnopq" }, { "input": "19 4", "output": "abcdabcdabcdabcdabc" }, { "input": "100 26", "output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv" }, { "input": "100 25", "output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy" }, { "input": "26 26", "output": "abcdefghijklmnopqrstuvwxyz" }, { "input": "27 26", "output": "abcdefghijklmnopqrstuvwxyza" }, { "input": "2 2", "output": "ab" }, { "input": "26 25", "output": "abcdefghijklmnopqrstuvwxya" }, { "input": "99 2", "output": "abababababababababababababababababababababababababababababababababababababababababababababababababa" }, { "input": "99 26", "output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu" }, { "input": "4 4", "output": "abcd" }, { "input": "5 3", "output": "abcab" }, { "input": "5 4", "output": "abcda" }, { "input": "5 5", "output": "abcde" }, { "input": "24 22", "output": "abcdefghijklmnopqrstuvab" }, { "input": "26 14", "output": "abcdefghijklmnabcdefghijkl" }, { "input": "26 15", "output": "abcdefghijklmnoabcdefghijk" }, { "input": "30 12", "output": "abcdefghijklabcdefghijklabcdef" }, { "input": "35 4", "output": "abcdabcdabcdabcdabcdabcdabcdabcdabc" }, { "input": "79 3", "output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca" }, { "input": "79 14", "output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi" }, { "input": "85 13", "output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg" }, { "input": "90 25", "output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno" }, { "input": "90 19", "output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn" }, { "input": "26 24", "output": "abcdefghijklmnopqrstuvwxab" }, { "input": "100 17", "output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno" }, { "input": "26 2", "output": "ababababababababababababab" } ]

1,637,887,061

2,147,483,647

Python 3

OK

TESTS

47

61

0

if __name__ == "__main__": n, k = map(int, input().split()) alphabet = list('abcdefghijklmnopqrstuvwxyz') password = '' i = 0 while k > 0: password= password + alphabet[i] k= k - 1 i= i + 1 i = 0 while len(password) < n: if i == k: i = 0 password= password + password[i] i= i + 1 print(password)

Title: New Password Time Limit: None seconds Memory Limit: None megabytes Problem Description: Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help. Innokentiy decides that new password should satisfy the following conditions: - the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct. Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions. Input Specification: The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it. Pay attention that a desired new password always exists. Output Specification: Print any password which satisfies all conditions given by Innokentiy. Demo Input: ['4 3\n', '6 6\n', '5 2\n'] Demo Output: ['java\n', 'python\n', 'phphp\n'] Note: In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it. In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters. In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it. Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.

```python if __name__ == "__main__": n, k = map(int, input().split()) alphabet = list('abcdefghijklmnopqrstuvwxyz') password = '' i = 0 while k > 0: password= password + alphabet[i] k= k - 1 i= i + 1 i = 0 while len(password) < n: if i == k: i = 0 password= password + password[i] i= i + 1 print(password) ```

3

884

B

Japanese Crosswords Strike Back

PROGRAMMING

1,100

[ "implementation" ]

null

A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect. For example: - If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array. Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!

The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked. The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.

Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.

[ "2 4\n1 3\n", "3 10\n3 3 2\n", "2 10\n1 3\n" ]

[ "NO\n", "YES\n", "NO\n" ]

none

0

[ { "input": "2 4\n1 3", "output": "NO" }, { "input": "3 10\n3 3 2", "output": "YES" }, { "input": "2 10\n1 3", "output": "NO" }, { "input": "1 1\n1", "output": "YES" }, { "input": "1 10\n10", "output": "YES" }, { "input": "1 10000\n10000", "output": "YES" }, { "input": "10 1\n5 78 3 87 4 9 5 8 9 1235", "output": "NO" }, { "input": "3 12\n3 3 3", "output": "NO" }, { "input": "3 9\n2 2 2", "output": "NO" }, { "input": "2 5\n1 1", "output": "NO" }, { "input": "1 2\n1", "output": "NO" }, { "input": "3 13\n3 3 3", "output": "NO" }, { "input": "3 6\n1 1 1", "output": "NO" }, { "input": "1 6\n5", "output": "NO" }, { "input": "3 11\n3 3 2", "output": "NO" }, { "input": "2 6\n1 3", "output": "NO" }, { "input": "3 10\n2 2 2", "output": "NO" }, { "input": "3 8\n2 1 1", "output": "NO" }, { "input": "1 5\n2", "output": "NO" }, { "input": "1 3\n1", "output": "NO" }, { "input": "5 5\n1 1 1 1 1", "output": "NO" }, { "input": "2 10\n4 4", "output": "NO" }, { "input": "2 8\n2 3", "output": "NO" }, { "input": "2 4\n1 1", "output": "NO" }, { "input": "3 10\n1 2 4", "output": "NO" }, { "input": "3 10\n2 1 3", "output": "NO" }, { "input": "2 6\n1 2", "output": "NO" }, { "input": "3 4\n1 1 1", "output": "NO" }, { "input": "3 11\n1 2 4", "output": "NO" }, { "input": "3 12\n3 3 2", "output": "NO" }, { "input": "4 9\n1 1 1 1", "output": "NO" }, { "input": "1 10\n9", "output": "NO" }, { "input": "1 7\n5", "output": "NO" }, { "input": "2 5\n1 2", "output": "NO" }, { "input": "3 8\n1 1 2", "output": "NO" }, { "input": "3 7\n1 1 1", "output": "NO" }, { "input": "1 10\n1", "output": "NO" }, { "input": "3 10\n2 3 4", "output": "NO" }, { "input": "3 9\n1 2 3", "output": "NO" }, { "input": "3 9\n3 3 2", "output": "NO" }, { "input": "3 6\n3 3 2", "output": "NO" }, { "input": "1 1\n3", "output": "NO" }, { "input": "1 3\n2", "output": "NO" }, { "input": "3 10\n3 3 3", "output": "NO" }, { "input": "3 5\n1 1 1", "output": "YES" }, { "input": "2 1\n100 100", "output": "NO" }, { "input": "3 3\n3 3 3", "output": "NO" }, { "input": "4 17\n3 3 9 1", "output": "NO" }, { "input": "4 1660\n1505 13 37 100", "output": "NO" }, { "input": "3 5\n3 3 2", "output": "NO" }, { "input": "4 10\n3 3 2 5", "output": "NO" }, { "input": "5 5\n5 5 5 5 5", "output": "NO" }, { "input": "1 1\n2", "output": "NO" }, { "input": "5 10\n1 2 2 4 5", "output": "NO" }, { "input": "2 1\n1 1", "output": "NO" } ]

1,659,061,738

2,147,483,647

Python 3

OK

TESTS

66

62

6,553,600

###~~~LOTA~~~### a,b=map(int,input().split()) l=list(map(int,input().split())) if sum(l)+(a-1)==b: print('YES') else: print('NO')

Title: Japanese Crosswords Strike Back Time Limit: None seconds Memory Limit: None megabytes Problem Description: A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect. For example: - If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array. Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it! Input Specification: The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked. The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding. Output Specification: Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO. Demo Input: ['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n'] Demo Output: ['NO\n', 'YES\n', 'NO\n'] Note: none

```python ###~~~LOTA~~~### a,b=map(int,input().split()) l=list(map(int,input().split())) if sum(l)+(a-1)==b: print('YES') else: print('NO') ```

3

334

A

Candy Bags

PROGRAMMING

1,000

[ "implementation" ]

null

Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies. Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.

The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers.

Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order. It is guaranteed that the solution exists at the given limits.

[ "2\n" ]

[ "1 4\n2 3\n" ]

The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.

500

[ { "input": "2", "output": "1 4\n2 3" }, { "input": "4", "output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9" }, { "input": "6", "output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19" }, { "input": "8", "output": "1 64 2 63 3 62 4 61\n5 60 6 59 7 58 8 57\n9 56 10 55 11 54 12 53\n13 52 14 51 15 50 16 49\n17 48 18 47 19 46 20 45\n21 44 22 43 23 42 24 41\n25 40 26 39 27 38 28 37\n29 36 30 35 31 34 32 33" }, { "input": "10", "output": "1 100 2 99 3 98 4 97 5 96\n6 95 7 94 8 93 9 92 10 91\n11 90 12 89 13 88 14 87 15 86\n16 85 17 84 18 83 19 82 20 81\n21 80 22 79 23 78 24 77 25 76\n26 75 27 74 28 73 29 72 30 71\n31 70 32 69 33 68 34 67 35 66\n36 65 37 64 38 63 39 62 40 61\n41 60 42 59 43 58 44 57 45 56\n46 55 47 54 48 53 49 52 50 51" }, { "input": "100", "output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951\n51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 993..." }, { "input": "62", "output": "1 3844 2 3843 3 3842 4 3841 5 3840 6 3839 7 3838 8 3837 9 3836 10 3835 11 3834 12 3833 13 3832 14 3831 15 3830 16 3829 17 3828 18 3827 19 3826 20 3825 21 3824 22 3823 23 3822 24 3821 25 3820 26 3819 27 3818 28 3817 29 3816 30 3815 31 3814\n32 3813 33 3812 34 3811 35 3810 36 3809 37 3808 38 3807 39 3806 40 3805 41 3804 42 3803 43 3802 44 3801 45 3800 46 3799 47 3798 48 3797 49 3796 50 3795 51 3794 52 3793 53 3792 54 3791 55 3790 56 3789 57 3788 58 3787 59 3786 60 3785 61 3784 62 3783\n63 3782 64 3781 65 378..." }, { "input": "66", "output": "1 4356 2 4355 3 4354 4 4353 5 4352 6 4351 7 4350 8 4349 9 4348 10 4347 11 4346 12 4345 13 4344 14 4343 15 4342 16 4341 17 4340 18 4339 19 4338 20 4337 21 4336 22 4335 23 4334 24 4333 25 4332 26 4331 27 4330 28 4329 29 4328 30 4327 31 4326 32 4325 33 4324\n34 4323 35 4322 36 4321 37 4320 38 4319 39 4318 40 4317 41 4316 42 4315 43 4314 44 4313 45 4312 46 4311 47 4310 48 4309 49 4308 50 4307 51 4306 52 4305 53 4304 54 4303 55 4302 56 4301 57 4300 58 4299 59 4298 60 4297 61 4296 62 4295 63 4294 64 4293 65 4292..." }, { "input": "18", "output": "1 324 2 323 3 322 4 321 5 320 6 319 7 318 8 317 9 316\n10 315 11 314 12 313 13 312 14 311 15 310 16 309 17 308 18 307\n19 306 20 305 21 304 22 303 23 302 24 301 25 300 26 299 27 298\n28 297 29 296 30 295 31 294 32 293 33 292 34 291 35 290 36 289\n37 288 38 287 39 286 40 285 41 284 42 283 43 282 44 281 45 280\n46 279 47 278 48 277 49 276 50 275 51 274 52 273 53 272 54 271\n55 270 56 269 57 268 58 267 59 266 60 265 61 264 62 263 63 262\n64 261 65 260 66 259 67 258 68 257 69 256 70 255 71 254 72 253\n73 252 7..." }, { "input": "68", "output": "1 4624 2 4623 3 4622 4 4621 5 4620 6 4619 7 4618 8 4617 9 4616 10 4615 11 4614 12 4613 13 4612 14 4611 15 4610 16 4609 17 4608 18 4607 19 4606 20 4605 21 4604 22 4603 23 4602 24 4601 25 4600 26 4599 27 4598 28 4597 29 4596 30 4595 31 4594 32 4593 33 4592 34 4591\n35 4590 36 4589 37 4588 38 4587 39 4586 40 4585 41 4584 42 4583 43 4582 44 4581 45 4580 46 4579 47 4578 48 4577 49 4576 50 4575 51 4574 52 4573 53 4572 54 4571 55 4570 56 4569 57 4568 58 4567 59 4566 60 4565 61 4564 62 4563 63 4562 64 4561 65 4560..." }, { "input": "86", "output": "1 7396 2 7395 3 7394 4 7393 5 7392 6 7391 7 7390 8 7389 9 7388 10 7387 11 7386 12 7385 13 7384 14 7383 15 7382 16 7381 17 7380 18 7379 19 7378 20 7377 21 7376 22 7375 23 7374 24 7373 25 7372 26 7371 27 7370 28 7369 29 7368 30 7367 31 7366 32 7365 33 7364 34 7363 35 7362 36 7361 37 7360 38 7359 39 7358 40 7357 41 7356 42 7355 43 7354\n44 7353 45 7352 46 7351 47 7350 48 7349 49 7348 50 7347 51 7346 52 7345 53 7344 54 7343 55 7342 56 7341 57 7340 58 7339 59 7338 60 7337 61 7336 62 7335 63 7334 64 7333 65 7332..." }, { "input": "96", "output": "1 9216 2 9215 3 9214 4 9213 5 9212 6 9211 7 9210 8 9209 9 9208 10 9207 11 9206 12 9205 13 9204 14 9203 15 9202 16 9201 17 9200 18 9199 19 9198 20 9197 21 9196 22 9195 23 9194 24 9193 25 9192 26 9191 27 9190 28 9189 29 9188 30 9187 31 9186 32 9185 33 9184 34 9183 35 9182 36 9181 37 9180 38 9179 39 9178 40 9177 41 9176 42 9175 43 9174 44 9173 45 9172 46 9171 47 9170 48 9169\n49 9168 50 9167 51 9166 52 9165 53 9164 54 9163 55 9162 56 9161 57 9160 58 9159 59 9158 60 9157 61 9156 62 9155 63 9154 64 9153 65 9152..." }, { "input": "12", "output": "1 144 2 143 3 142 4 141 5 140 6 139\n7 138 8 137 9 136 10 135 11 134 12 133\n13 132 14 131 15 130 16 129 17 128 18 127\n19 126 20 125 21 124 22 123 23 122 24 121\n25 120 26 119 27 118 28 117 29 116 30 115\n31 114 32 113 33 112 34 111 35 110 36 109\n37 108 38 107 39 106 40 105 41 104 42 103\n43 102 44 101 45 100 46 99 47 98 48 97\n49 96 50 95 51 94 52 93 53 92 54 91\n55 90 56 89 57 88 58 87 59 86 60 85\n61 84 62 83 63 82 64 81 65 80 66 79\n67 78 68 77 69 76 70 75 71 74 72 73" }, { "input": "88", "output": "1 7744 2 7743 3 7742 4 7741 5 7740 6 7739 7 7738 8 7737 9 7736 10 7735 11 7734 12 7733 13 7732 14 7731 15 7730 16 7729 17 7728 18 7727 19 7726 20 7725 21 7724 22 7723 23 7722 24 7721 25 7720 26 7719 27 7718 28 7717 29 7716 30 7715 31 7714 32 7713 33 7712 34 7711 35 7710 36 7709 37 7708 38 7707 39 7706 40 7705 41 7704 42 7703 43 7702 44 7701\n45 7700 46 7699 47 7698 48 7697 49 7696 50 7695 51 7694 52 7693 53 7692 54 7691 55 7690 56 7689 57 7688 58 7687 59 7686 60 7685 61 7684 62 7683 63 7682 64 7681 65 7680..." }, { "input": "28", "output": "1 784 2 783 3 782 4 781 5 780 6 779 7 778 8 777 9 776 10 775 11 774 12 773 13 772 14 771\n15 770 16 769 17 768 18 767 19 766 20 765 21 764 22 763 23 762 24 761 25 760 26 759 27 758 28 757\n29 756 30 755 31 754 32 753 33 752 34 751 35 750 36 749 37 748 38 747 39 746 40 745 41 744 42 743\n43 742 44 741 45 740 46 739 47 738 48 737 49 736 50 735 51 734 52 733 53 732 54 731 55 730 56 729\n57 728 58 727 59 726 60 725 61 724 62 723 63 722 64 721 65 720 66 719 67 718 68 717 69 716 70 715\n71 714 72 713 73 712 74 7..." }, { "input": "80", "output": "1 6400 2 6399 3 6398 4 6397 5 6396 6 6395 7 6394 8 6393 9 6392 10 6391 11 6390 12 6389 13 6388 14 6387 15 6386 16 6385 17 6384 18 6383 19 6382 20 6381 21 6380 22 6379 23 6378 24 6377 25 6376 26 6375 27 6374 28 6373 29 6372 30 6371 31 6370 32 6369 33 6368 34 6367 35 6366 36 6365 37 6364 38 6363 39 6362 40 6361\n41 6360 42 6359 43 6358 44 6357 45 6356 46 6355 47 6354 48 6353 49 6352 50 6351 51 6350 52 6349 53 6348 54 6347 55 6346 56 6345 57 6344 58 6343 59 6342 60 6341 61 6340 62 6339 63 6338 64 6337 65 6336..." }, { "input": "48", "output": "1 2304 2 2303 3 2302 4 2301 5 2300 6 2299 7 2298 8 2297 9 2296 10 2295 11 2294 12 2293 13 2292 14 2291 15 2290 16 2289 17 2288 18 2287 19 2286 20 2285 21 2284 22 2283 23 2282 24 2281\n25 2280 26 2279 27 2278 28 2277 29 2276 30 2275 31 2274 32 2273 33 2272 34 2271 35 2270 36 2269 37 2268 38 2267 39 2266 40 2265 41 2264 42 2263 43 2262 44 2261 45 2260 46 2259 47 2258 48 2257\n49 2256 50 2255 51 2254 52 2253 53 2252 54 2251 55 2250 56 2249 57 2248 58 2247 59 2246 60 2245 61 2244 62 2243 63 2242 64 2241 65 224..." }, { "input": "54", "output": "1 2916 2 2915 3 2914 4 2913 5 2912 6 2911 7 2910 8 2909 9 2908 10 2907 11 2906 12 2905 13 2904 14 2903 15 2902 16 2901 17 2900 18 2899 19 2898 20 2897 21 2896 22 2895 23 2894 24 2893 25 2892 26 2891 27 2890\n28 2889 29 2888 30 2887 31 2886 32 2885 33 2884 34 2883 35 2882 36 2881 37 2880 38 2879 39 2878 40 2877 41 2876 42 2875 43 2874 44 2873 45 2872 46 2871 47 2870 48 2869 49 2868 50 2867 51 2866 52 2865 53 2864 54 2863\n55 2862 56 2861 57 2860 58 2859 59 2858 60 2857 61 2856 62 2855 63 2854 64 2853 65 285..." }, { "input": "58", "output": "1 3364 2 3363 3 3362 4 3361 5 3360 6 3359 7 3358 8 3357 9 3356 10 3355 11 3354 12 3353 13 3352 14 3351 15 3350 16 3349 17 3348 18 3347 19 3346 20 3345 21 3344 22 3343 23 3342 24 3341 25 3340 26 3339 27 3338 28 3337 29 3336\n30 3335 31 3334 32 3333 33 3332 34 3331 35 3330 36 3329 37 3328 38 3327 39 3326 40 3325 41 3324 42 3323 43 3322 44 3321 45 3320 46 3319 47 3318 48 3317 49 3316 50 3315 51 3314 52 3313 53 3312 54 3311 55 3310 56 3309 57 3308 58 3307\n59 3306 60 3305 61 3304 62 3303 63 3302 64 3301 65 330..." }, { "input": "64", "output": "1 4096 2 4095 3 4094 4 4093 5 4092 6 4091 7 4090 8 4089 9 4088 10 4087 11 4086 12 4085 13 4084 14 4083 15 4082 16 4081 17 4080 18 4079 19 4078 20 4077 21 4076 22 4075 23 4074 24 4073 25 4072 26 4071 27 4070 28 4069 29 4068 30 4067 31 4066 32 4065\n33 4064 34 4063 35 4062 36 4061 37 4060 38 4059 39 4058 40 4057 41 4056 42 4055 43 4054 44 4053 45 4052 46 4051 47 4050 48 4049 49 4048 50 4047 51 4046 52 4045 53 4044 54 4043 55 4042 56 4041 57 4040 58 4039 59 4038 60 4037 61 4036 62 4035 63 4034 64 4033\n65 403..." }, { "input": "44", "output": "1 1936 2 1935 3 1934 4 1933 5 1932 6 1931 7 1930 8 1929 9 1928 10 1927 11 1926 12 1925 13 1924 14 1923 15 1922 16 1921 17 1920 18 1919 19 1918 20 1917 21 1916 22 1915\n23 1914 24 1913 25 1912 26 1911 27 1910 28 1909 29 1908 30 1907 31 1906 32 1905 33 1904 34 1903 35 1902 36 1901 37 1900 38 1899 39 1898 40 1897 41 1896 42 1895 43 1894 44 1893\n45 1892 46 1891 47 1890 48 1889 49 1888 50 1887 51 1886 52 1885 53 1884 54 1883 55 1882 56 1881 57 1880 58 1879 59 1878 60 1877 61 1876 62 1875 63 1874 64 1873 65 187..." } ]

1,624,341,140

2,147,483,647

PyPy 3

OK

TESTS

21

280

3,993,600

n = int(input()) n_sqr = n**2 r = range(1, n**2+1) for i in range(n_sqr//2): print(r[i], r[n_sqr-1-i])

Title: Candy Bags Time Limit: None seconds Memory Limit: None megabytes Problem Description: Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies. Help him give *n* bags of candies to each brother so that all brothers got the same number of candies. Input Specification: The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers. Output Specification: Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order. It is guaranteed that the solution exists at the given limits. Demo Input: ['2\n'] Demo Output: ['1 4\n2 3\n'] Note: The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.

```python n = int(input()) n_sqr = n**2 r = range(1, n**2+1) for i in range(n_sqr//2): print(r[i], r[n_sqr-1-i]) ```

3

58

A

Chat room

PROGRAMMING

1,000

[ "greedy", "strings" ]

A. Chat room

1

256

Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.

The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.

If Vasya managed to say hello, print "YES", otherwise print "NO".

[ "ahhellllloou\n", "hlelo\n" ]

[ "YES\n", "NO\n" ]

none

500

[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]

1,668,789,162

2,147,483,647

PyPy 3-64

OK

TESTS

40

62

0

s = input() temp = 'hello' ind = 0 for i in s: if i == temp[ind]: ind += 1 if ind == len(temp): break if ind == len(temp): print("YES") else: print("NO")

Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python s = input() temp = 'hello' ind = 0 for i in s: if i == temp[ind]: ind += 1 if ind == len(temp): break if ind == len(temp): print("YES") else: print("NO") ```

3.969

258

A

Little Elephant and Bits

PROGRAMMING

1,100

[ "greedy", "math" ]

null

The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper. To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes). The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.

The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.

In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.

[ "101\n", "110010\n" ]

[ "11\n", "11010\n" ]

In the first sample the best strategy is to delete the second digit. That results in number 112 = 310. In the second sample the best strategy is to delete the third or fourth digits — that results in number 110102 = 2610.

500

[ { "input": "101", "output": "11" }, { "input": "110010", "output": "11010" }, { "input": "10000", "output": "1000" }, { "input": "1111111110", "output": "111111111" }, { "input": "10100101011110101", "output": "1100101011110101" }, { "input": "111010010111", "output": "11110010111" }, { "input": "11110111011100000000", "output": "1111111011100000000" }, { "input": "11110010010100001110110101110011110110100111101", "output": "1111010010100001110110101110011110110100111101" }, { "input": "1001011111010010100111111", "output": "101011111010010100111111" }, { "input": "1111111111", "output": "111111111" }, { "input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101", "output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101" }, { "input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100", "output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100" }, { "input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011", "output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011" }, { "input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011", "output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011" }, { "input": "11", "output": "1" }, { "input": "111", "output": "11" }, { "input": "111111", "output": "11111" }, { "input": "11111", "output": "1111" }, { "input": "1111", "output": "111" } ]

1,670,267,323

2,147,483,647

PyPy 3-64

OK

TESTS

37

124

0

s = input() if '0' in s: e = s.split('0',1) if len(e) == 1: print(e[0]) else: print(e[0]+e[1]) else: print(s[1:])

Title: Little Elephant and Bits Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper. To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes). The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation. Input Specification: The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits. Output Specification: In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem. Demo Input: ['101\n', '110010\n'] Demo Output: ['11\n', '11010\n'] Note: In the first sample the best strategy is to delete the second digit. That results in number 112 = 310. In the second sample the best strategy is to delete the third or fourth digits — that results in number 110102 = 2610.

```python s = input() if '0' in s: e = s.split('0',1) if len(e) == 1: print(e[0]) else: print(e[0]+e[1]) else: print(s[1:]) ```

3

937

A

Olympiad

PROGRAMMING

800

[ "implementation", "sortings" ]

null

The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points. As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria: - At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded. Determine the number of ways to choose a subset of participants that will receive the diplomas.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants. The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores. It's guaranteed that at least one participant has non-zero score.

Print a single integer — the desired number of ways.

[ "4\n1 3 3 2\n", "3\n1 1 1\n", "4\n42 0 0 42\n" ]

[ "3\n", "1\n", "1\n" ]

There are three ways to choose a subset in sample case one. 1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma! The only option in sample case two is to award everyone. Note that in sample case three participants with zero scores cannot get anything.

500

[ { "input": "4\n1 3 3 2", "output": "3" }, { "input": "3\n1 1 1", "output": "1" }, { "input": "4\n42 0 0 42", "output": "1" }, { "input": "10\n1 0 1 0 1 0 0 0 0 1", "output": "1" }, { "input": "10\n572 471 540 163 50 30 561 510 43 200", "output": "10" }, { "input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600", "output": "94" }, { "input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600", "output": "1" }, { "input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12", "output": "13" }, { "input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391", "output": "33" }, { "input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3", "output": "3" }, { "input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55", "output": "93" }, { "input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13", "output": "13" }, { "input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1", "output": "1" }, { "input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37", "output": "61" }, { "input": "2\n0 1", "output": "1" }, { "input": "2\n0 600", "output": "1" }, { "input": "4\n1 1 1 2", "output": "2" }, { "input": "4\n0 0 1 2", "output": "2" }, { "input": "1\n5", "output": "1" }, { "input": "2\n0 5", "output": "1" }, { "input": "5\n1 0 0 1 2", "output": "2" } ]

1,575,397,740

2,147,483,647

Python 3

OK

TESTS

21

124

0

n=int(input()) l=list(map(int,input().split())) s=set(l) if 0 in s:print(len(s)-1) else:print(len(s))

Title: Olympiad Time Limit: None seconds Memory Limit: None megabytes Problem Description: The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points. As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria: - At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded. Determine the number of ways to choose a subset of participants that will receive the diplomas. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants. The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores. It's guaranteed that at least one participant has non-zero score. Output Specification: Print a single integer — the desired number of ways. Demo Input: ['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n'] Demo Output: ['3\n', '1\n', '1\n'] Note: There are three ways to choose a subset in sample case one. 1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma! The only option in sample case two is to award everyone. Note that in sample case three participants with zero scores cannot get anything.

```python n=int(input()) l=list(map(int,input().split())) s=set(l) if 0 in s:print(len(s)-1) else:print(len(s)) ```

3

368

A

Sereja and Coat Rack

PROGRAMMING

1,000

[ "implementation" ]

null

Sereja owns a restaurant for *n* people. The restaurant hall has a coat rack with *n* hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the *i*-th hook costs *a**i* rubles. Only one person can hang clothes on one hook. Tonight Sereja expects *m* guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment a guest arrives the rack has no available hooks, Sereja must pay a *d* ruble fine to the guest. Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides the *m* guests is visiting Sereja's restaurant tonight.

The first line contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100). The third line contains integer *m* (1<=≤<=*m*<=≤<=100).

In a single line print a single integer — the answer to the problem.

[ "2 1\n2 1\n2\n", "2 1\n2 1\n10\n" ]

[ "3\n", "-5\n" ]

In the first test both hooks will be used, so Sereja gets 1 + 2 = 3 rubles. In the second test both hooks will be used but Sereja pays a fine 8 times, so the answer is 3 - 8 = - 5.

500

[ { "input": "2 1\n2 1\n2", "output": "3" }, { "input": "2 1\n2 1\n10", "output": "-5" }, { "input": "1 1\n1\n2", "output": "0" }, { "input": "3 96\n83 22 17\n19", "output": "-1414" }, { "input": "8 4\n27 72 39 70 13 68 100 36\n95", "output": "77" }, { "input": "2 65\n23 34\n74", "output": "-4623" }, { "input": "2 48\n12 54\n69", "output": "-3150" }, { "input": "5 30\n63 58 38 60 24\n42", "output": "-867" }, { "input": "9 47\n17 36 91 43 89 7 41 43 65\n49", "output": "-1448" }, { "input": "6 49\n91 30 71 51 7 2\n94", "output": "-4060" }, { "input": "57 27\n73 51 24 86 57 17 27 58 27 58 38 72 70 62 97 23 18 13 18 97 86 42 24 30 30 66 60 33 97 56 54 63 85 35 55 73 58 70 33 64 8 84 12 36 68 49 76 39 24 43 55 12 42 76 60 26 22\n71", "output": "2454" }, { "input": "35 19\n6 84 51 99 80 2 94 35 38 35 57 94 77 6 63 49 82 1 14 42 56 56 43 63 12 78 25 79 53 44 97 74 41 14 76\n73", "output": "1098" }, { "input": "11 91\n18 33 13 96 70 32 41 89 86 91 98\n90", "output": "-6522" }, { "input": "46 48\n54 15 52 41 45 59 36 60 93 6 65 82 4 30 76 9 93 98 50 57 62 28 68 42 30 41 14 75 2 78 16 84 14 93 25 2 93 60 71 29 28 85 76 87 99 71\n88", "output": "382" }, { "input": "5 72\n4 22 64 7 64\n11", "output": "-271" }, { "input": "90 24\n41 65 43 20 14 92 5 19 33 51 6 76 40 4 23 99 48 85 49 72 65 14 76 46 13 47 79 70 63 20 86 90 45 66 41 46 9 19 71 2 24 33 73 53 88 71 64 2 4 24 28 1 70 16 66 29 44 48 89 44 38 10 64 50 82 89 43 9 61 22 59 55 89 47 91 50 44 31 21 49 68 37 84 36 27 86 39 54 30 25\n49", "output": "1306" }, { "input": "60 63\n58 67 45 56 19 27 12 26 56 2 50 97 85 16 65 43 76 14 43 97 49 73 27 7 74 30 5 6 27 13 76 94 66 37 37 42 15 95 57 53 37 39 83 56 16 32 31 42 26 12 38 87 91 51 63 35 94 54 17 53\n9", "output": "86" }, { "input": "34 79\n55 4 35 4 57 49 25 18 14 10 29 1 81 19 59 51 56 62 65 4 77 44 10 3 62 90 49 83 54 75 21 3 24 32\n70", "output": "-1519" }, { "input": "60 91\n9 20 72 4 46 82 5 93 86 14 99 90 23 39 38 11 62 35 9 62 60 94 16 70 38 70 59 1 72 65 18 16 56 16 31 40 13 89 83 55 86 11 85 75 81 16 52 42 16 80 11 99 74 89 78 33 57 90 14 9\n42", "output": "1406" }, { "input": "24 68\n64 29 85 79 1 72 86 75 72 34 68 54 96 69 26 77 30 51 99 10 94 87 81 17\n50", "output": "-312" }, { "input": "29 19\n80 65 22 6 27 17 17 27 67 88 82 65 41 87 22 63 22 65 10 16 3 74 25 42 46 63 24 32 7\n69", "output": "445" }, { "input": "3 37\n8 8 82\n13", "output": "-272" }, { "input": "31 63\n15 10 85 57 91 94 97 53 55 46 9 49 92 13 32 15 40 59 23 5 96 53 70 80 39 24 19 67 60 99 87\n97", "output": "-2524" }, { "input": "34 30\n59 23 47 93 38 26 48 59 3 8 99 31 93 1 79 100 53 49 83 41 16 76 63 68 37 98 19 98 29 52 17 31 50 26\n59", "output": "963" }, { "input": "21 29\n41 61 48 63 56 76 93 62 55 99 47 15 47 89 70 39 64 76 16 22 76\n16", "output": "782" }, { "input": "35 86\n71 6 65 58 63 62 25 50 70 31 24 51 34 26 11 38 37 38 79 94 37 15 65 92 50 36 6 38 5 38 24 65 71 9 69\n82", "output": "-2489" }, { "input": "53 75\n74 53 95 77 27 97 73 50 41 75 20 44 12 42 90 20 66 6 86 17 51 16 10 65 67 94 75 10 1 96 74 90 62 73 69 59 32 69 27 11 23 75 80 11 53 83 92 96 65 75 65 3 56\n61", "output": "2293" }, { "input": "73 13\n22 23 48 78 90 6 96 95 51 44 55 82 13 73 40 29 13 63 68 9 16 9 24 60 35 5 87 20 59 46 7 67 1 68 93 88 33 57 75 48 22 84 23 32 77 84 49 24 83 19 77 21 12 83 57 91 26 25 87 78 70 44 35 78 69 69 92 97 84 29 28 27 72\n98", "output": "3419" }, { "input": "4 39\n28 9 46 9\n86", "output": "-3106" }, { "input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100\n100", "output": "10000" }, { "input": "1 100\n100\n100", "output": "-9800" }, { "input": "1 1\n1\n1", "output": "1" }, { "input": "5 1\n1 5 2 7 8\n3", "output": "8" }, { "input": "4 44\n3 3 3 3\n1", "output": "3" }, { "input": "3 1\n1 2 3\n1", "output": "1" } ]

1,684,753,752

2,147,483,647

PyPy 3-64

OK

TESTS

35

62

0

# https://codeforces.com/contest/368 import sys input = lambda: sys.stdin.readline().rstrip() # faster! n, d = map(int, input().split()) a = list(map(int, input().split())) m = int(input()) if m >= n: ans = sum(a) - (m - n) * d else: a.sort() ans = sum(a[:m]) print(ans)

Title: Sereja and Coat Rack Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja owns a restaurant for *n* people. The restaurant hall has a coat rack with *n* hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the *i*-th hook costs *a**i* rubles. Only one person can hang clothes on one hook. Tonight Sereja expects *m* guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment a guest arrives the rack has no available hooks, Sereja must pay a *d* ruble fine to the guest. Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides the *m* guests is visiting Sereja's restaurant tonight. Input Specification: The first line contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100). The third line contains integer *m* (1<=≤<=*m*<=≤<=100). Output Specification: In a single line print a single integer — the answer to the problem. Demo Input: ['2 1\n2 1\n2\n', '2 1\n2 1\n10\n'] Demo Output: ['3\n', '-5\n'] Note: In the first test both hooks will be used, so Sereja gets 1 + 2 = 3 rubles. In the second test both hooks will be used but Sereja pays a fine 8 times, so the answer is 3 - 8 = - 5.

```python # https://codeforces.com/contest/368 import sys input = lambda: sys.stdin.readline().rstrip() # faster! n, d = map(int, input().split()) a = list(map(int, input().split())) m = int(input()) if m >= n: ans = sum(a) - (m - n) * d else: a.sort() ans = sum(a[:m]) print(ans) ```

3

844

A

Diversity

PROGRAMMING

1,000

[ "greedy", "implementation", "strings" ]

null

Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible. String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.

First line of input contains string *s*, consisting only of lowercase Latin letters (1<=≤<=|*s*|<=≤<=1000, |*s*| denotes the length of *s*). Second line of input contains integer *k* (1<=≤<=*k*<=≤<=26).

Print single line with a minimum number of necessary changes, or the word «impossible» (without quotes) if it is impossible.

[ "yandex\n6\n", "yahoo\n5\n", "google\n7\n" ]

[ "0\n", "1\n", "impossible\n" ]

In the first test case string contains 6 different letters, so we don't need to change anything. In the second test case string contains 4 different letters: {'*a*', '*h*', '*o*', '*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}. In the third test case, it is impossible to make 7 different letters because the length of the string is 6.

500

[ { "input": "yandex\n6", "output": "0" }, { "input": "yahoo\n5", "output": "1" }, { "input": "google\n7", "output": "impossible" }, { "input": "a\n1", "output": "0" }, { "input": "z\n2", "output": "impossible" }, { "input": "fwgfrwgkuwghfiruhewgirueguhergiqrbvgrgf\n26", "output": "14" }, { "input": "nfevghreuoghrueighoqghbnebvnejbvnbgneluqe\n26", "output": "12" }, { "input": "a\n3", "output": "impossible" }, { "input": "smaxpqplaqqbxuqxalqmbmmgubbpspxhawbxsuqhhegpmmpebqmqpbbeplwaepxmsahuepuhuhwxeqmmlgqubuaxehwuwasgxpqmugbmuawuhwqlswllssueglbxepbmwgs\n1", "output": "0" }, { "input": "cuguccgcugcugucgggggcgcgucgucugcuuuccccuugccg\n4", "output": "1" }, { "input": "fcfccfcfccfcfcffcffffffcfccfccfcffccccfcffffccfccfcffcfcccccffcfffcccffcfccfffffcccfccffffffccfccccf\n20", "output": "18" }, { "input": "swmkwaruyv\n5", "output": "0" }, { "input": "tnbqpsuhkczmejirvyfdolxwga\n22", "output": "0" }, { "input": "abcde\n3", "output": "0" }, { "input": "abb\n1", "output": "0" }, { "input": "aaaa\n1", "output": "0" }, { "input": "abcde\n2", "output": "0" }, { "input": "yandex\n4", "output": "0" }, { "input": "aaabbbccc\n1", "output": "0" }, { "input": "abcd\n2", "output": "0" }, { "input": "asdfgh\n2", "output": "0" }, { "input": "aab\n1", "output": "0" }, { "input": "mynameissako\n5", "output": "0" }, { "input": "abcde\n1", "output": "0" }, { "input": "abcd\n3", "output": "0" }, { "input": "abcdef\n2", "output": "0" }, { "input": "abcdefg\n4", "output": "0" }, { "input": "abc\n1", "output": "0" }, { "input": "asdafjsgljdllgjdgkl\n5", "output": "0" }, { "input": "yaay\n3", "output": "1" }, { "input": "yaay\n4", "output": "2" }, { "input": "zzzzzz\n2", "output": "1" } ]

1,508,265,190

2,147,483,647

Python 3

OK

TESTS

41

62

5,529,600

s = input() k = int(input()) n = 0 a = list() for i in s: test = False for j in a: if i == j: test = True if not test: a.append(i) if len(s) < k: print('impossible') else: print(max(k - len(a),0))

Title: Diversity Time Limit: None seconds Memory Limit: None megabytes Problem Description: Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible. String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too. Input Specification: First line of input contains string *s*, consisting only of lowercase Latin letters (1<=≤<=|*s*|<=≤<=1000, |*s*| denotes the length of *s*). Second line of input contains integer *k* (1<=≤<=*k*<=≤<=26). Output Specification: Print single line with a minimum number of necessary changes, or the word «impossible» (without quotes) if it is impossible. Demo Input: ['yandex\n6\n', 'yahoo\n5\n', 'google\n7\n'] Demo Output: ['0\n', '1\n', 'impossible\n'] Note: In the first test case string contains 6 different letters, so we don't need to change anything. In the second test case string contains 4 different letters: {'*a*', '*h*', '*o*', '*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}. In the third test case, it is impossible to make 7 different letters because the length of the string is 6.

```python s = input() k = int(input()) n = 0 a = list() for i in s: test = False for j in a: if i == j: test = True if not test: a.append(i) if len(s) < k: print('impossible') else: print(max(k - len(a),0)) ```

3

808

D

Array Division

PROGRAMMING

1,900

[ "binary search", "data structures", "implementation" ]

null

Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position). Inserting an element in the same position he was erased from is also considered moving. Can Vasya divide the array after choosing the right element to move and its new position?

The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array. The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.

Print YES if Vasya can divide the array after moving one element. Otherwise print NO.

[ "3\n1 3 2\n", "5\n1 2 3 4 5\n", "5\n2 2 3 4 5\n" ]

[ "YES\n", "NO\n", "YES\n" ]

In the first example Vasya can move the second element to the end of the array. In the second example no move can make the division possible. In the third example Vasya can move the fourth element by one position to the left.

0

[ { "input": "3\n1 3 2", "output": "YES" }, { "input": "5\n1 2 3 4 5", "output": "NO" }, { "input": "5\n2 2 3 4 5", "output": "YES" }, { "input": "5\n72 32 17 46 82", "output": "NO" }, { "input": "6\n26 10 70 11 69 57", "output": "NO" }, { "input": "7\n4 7 10 7 5 5 1", "output": "NO" }, { "input": "8\n9 5 5 10 4 9 5 8", "output": "NO" }, { "input": "10\n9 6 8 5 5 2 8 9 2 2", "output": "YES" }, { "input": "15\n4 8 10 3 1 4 5 9 3 2 1 7 7 3 8", "output": "NO" }, { "input": "20\n71 83 54 6 10 64 91 98 94 49 65 68 14 39 91 60 74 100 17 13", "output": "NO" }, { "input": "20\n2 8 10 4 6 6 4 1 2 2 6 9 5 1 9 1 9 8 10 6", "output": "NO" }, { "input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14", "output": "YES" }, { "input": "100\n89 38 63 73 77 4 99 74 30 5 69 57 97 37 88 71 36 59 19 63 46 20 33 58 61 98 100 31 33 53 99 96 34 17 44 95 54 52 22 77 67 88 20 88 26 43 12 23 96 94 14 7 57 86 56 54 32 8 3 43 97 56 74 22 5 100 12 60 93 12 44 68 31 63 7 71 21 29 19 38 50 47 97 43 50 59 88 40 51 61 20 68 32 66 70 48 19 55 91 53", "output": "NO" }, { "input": "100\n80 100 88 52 25 87 85 8 92 62 35 66 74 39 58 41 55 53 23 73 90 72 36 44 97 67 16 54 3 8 25 34 84 47 77 39 93 19 49 20 29 44 21 48 21 56 82 59 8 31 94 95 84 54 72 20 95 91 85 1 67 19 76 28 31 63 87 98 55 28 16 20 36 91 93 39 94 69 80 97 100 96 68 26 91 45 22 84 20 36 20 92 53 75 58 51 60 26 76 25", "output": "NO" }, { "input": "100\n27 95 57 29 91 85 83 36 72 86 39 5 79 61 78 93 100 97 73 23 82 66 41 92 38 92 100 96 48 56 66 47 5 32 69 13 95 23 46 62 99 83 57 66 98 82 81 57 37 37 81 64 45 76 72 43 99 76 86 22 37 39 93 80 99 36 53 83 3 32 52 9 78 34 47 100 33 72 19 40 29 56 77 32 79 72 15 88 100 98 56 50 22 81 88 92 58 70 21 19", "output": "NO" }, { "input": "100\n35 31 83 11 7 94 57 58 30 26 2 99 33 58 98 6 3 52 13 66 21 53 26 94 100 5 1 3 91 13 97 49 86 25 63 90 88 98 57 57 34 81 32 16 65 94 59 83 44 14 46 18 28 89 75 95 87 57 52 18 46 80 31 43 38 54 69 75 82 9 64 96 75 40 96 52 67 85 86 38 95 55 16 57 17 20 22 7 63 3 12 16 42 87 46 12 51 95 67 80", "output": "NO" }, { "input": "6\n1 4 3 100 100 6", "output": "YES" }, { "input": "6\n6 100 100 3 4 1", "output": "YES" }, { "input": "6\n4 2 3 7 1 1", "output": "YES" }, { "input": "4\n6 1 4 5", "output": "NO" }, { "input": "3\n228 114 114", "output": "YES" }, { "input": "3\n229 232 444", "output": "NO" }, { "input": "3\n322 324 555", "output": "NO" }, { "input": "3\n69 34 5", "output": "NO" }, { "input": "6\n5 4 1 2 2 2", "output": "YES" }, { "input": "3\n545 237 546", "output": "NO" }, { "input": "5\n2 3 1 1 1", "output": "YES" }, { "input": "6\n2 2 10 2 2 2", "output": "YES" }, { "input": "5\n5 4 6 5 6", "output": "NO" }, { "input": "5\n6 1 1 1 1", "output": "NO" }, { "input": "2\n1 3", "output": "NO" }, { "input": "5\n5 2 2 3 4", "output": "YES" }, { "input": "2\n2 2", "output": "YES" }, { "input": "5\n1 2 6 1 2", "output": "YES" }, { "input": "5\n1 1 8 5 1", "output": "YES" }, { "input": "10\n73 67 16 51 56 71 37 49 90 6", "output": "NO" }, { "input": "1\n10", "output": "NO" }, { "input": "1\n1", "output": "NO" }, { "input": "2\n1 1", "output": "YES" }, { "input": "5\n8 2 7 5 4", "output": "YES" }, { "input": "1\n2", "output": "NO" }, { "input": "16\n9 10 2 1 6 7 6 5 8 3 2 10 8 4 9 2", "output": "YES" }, { "input": "4\n8 2 2 4", "output": "YES" }, { "input": "19\n9 9 3 2 4 5 5 7 8 10 8 10 1 2 2 6 5 3 3", "output": "NO" }, { "input": "11\n7 2 1 8 8 2 4 10 8 7 1", "output": "YES" }, { "input": "6\n10 20 30 40 99 1", "output": "YES" }, { "input": "10\n3 7 9 2 10 1 9 6 4 1", "output": "NO" }, { "input": "3\n3 1 2", "output": "YES" }, { "input": "2\n9 3", "output": "NO" }, { "input": "7\n1 2 3 12 1 2 3", "output": "YES" }, { "input": "6\n2 4 4 5 8 5", "output": "YES" }, { "input": "18\n2 10 3 6 6 6 10 8 8 1 10 9 9 3 1 9 7 4", "output": "YES" }, { "input": "20\n9 6 6 10 4 4 8 7 4 10 10 2 10 5 9 5 3 10 1 9", "output": "NO" }, { "input": "12\n3 8 10 2 4 4 6 9 5 10 10 3", "output": "YES" }, { "input": "11\n9 2 7 7 7 3 7 5 4 10 7", "output": "NO" }, { "input": "5\n1 1 4 1 1", "output": "YES" }, { "input": "2\n4 4", "output": "YES" }, { "input": "2\n7 1", "output": "NO" }, { "input": "5\n10 5 6 7 6", "output": "YES" }, { "input": "11\n4 3 10 3 7 8 4 9 2 1 1", "output": "YES" }, { "input": "6\n705032704 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "NO" }, { "input": "8\n1 5 6 8 3 1 7 3", "output": "YES" }, { "input": "20\n8 6 3 6 3 5 10 2 6 1 7 6 9 10 8 3 5 9 3 8", "output": "YES" }, { "input": "11\n2 4 8 3 4 7 9 10 5 3 3", "output": "YES" }, { "input": "7\n6 4 2 24 6 4 2", "output": "YES" }, { "input": "17\n7 1 1 1 8 9 1 10 8 8 7 9 7 9 1 6 5", "output": "NO" }, { "input": "7\n7 10 1 2 6 2 2", "output": "NO" }, { "input": "5\n10 10 40 10 10", "output": "YES" }, { "input": "3\n4 3 13", "output": "NO" }, { "input": "5\n5 2 10 2 1", "output": "YES" }, { "input": "7\n7 4 5 62 20 20 6", "output": "YES" }, { "input": "6\n1 5 2 20 10 2", "output": "YES" }, { "input": "2\n5 6", "output": "NO" }, { "input": "14\n5 2 9 7 5 8 3 2 2 4 9 1 3 10", "output": "YES" }, { "input": "5\n1 2 3 4 2", "output": "YES" }, { "input": "5\n2 2 2 5 5", "output": "NO" }, { "input": "11\n1 1 1 1 1 10 1 1 1 1 1", "output": "YES" }, { "input": "9\n8 4 13 19 11 1 8 2 8", "output": "YES" }, { "input": "6\n14 16 14 14 15 11", "output": "YES" }, { "input": "9\n14 19 1 13 11 3 1 1 7", "output": "YES" }, { "input": "6\n16 13 3 7 4 15", "output": "YES" }, { "input": "4\n11 7 12 14", "output": "NO" }, { "input": "3\n3 2 1", "output": "YES" }, { "input": "5\n2 1 3 6 4", "output": "YES" }, { "input": "5\n3 4 8 11 2", "output": "YES" }, { "input": "5\n1 2 10 3 4", "output": "YES" }, { "input": "6\n8 15 12 14 15 4", "output": "YES" }, { "input": "5\n1 2 4 4 5", "output": "YES" }, { "input": "3\n2 4 2", "output": "YES" }, { "input": "5\n2 3 1 6 4", "output": "YES" }, { "input": "7\n1 2 3 12 3 2 1", "output": "YES" }, { "input": "3\n3 4 13", "output": "NO" }, { "input": "6\n1 1 1 1 1000000000 1000000000", "output": "YES" }, { "input": "6\n19 6 5 13 6 13", "output": "YES" }, { "input": "8\n2 2 2 5 1 2 3 3", "output": "YES" } ]

1,565,187,757

2,147,483,647

PyPy 3

OK

TESTS

115

327

15,462,400

def possible(a): if len(a)==1: return False if len(a)==2: return a[0]==a[1] total = sum(a) vals = set([a[0]]) running = a[0] for el in a[1:]: running += el vals.add(el) if running == total-running: return True if (running - (total-running))%2==0 and (running - (total-running))//2 in vals: return True return False n = int(input()) a = list(map(int, input().split())) if possible(a) or possible(a[::-1]): print("YES") else: print("NO")

Title: Array Division Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position). Inserting an element in the same position he was erased from is also considered moving. Can Vasya divide the array after choosing the right element to move and its new position? Input Specification: The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array. The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. Output Specification: Print YES if Vasya can divide the array after moving one element. Otherwise print NO. Demo Input: ['3\n1 3 2\n', '5\n1 2 3 4 5\n', '5\n2 2 3 4 5\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: In the first example Vasya can move the second element to the end of the array. In the second example no move can make the division possible. In the third example Vasya can move the fourth element by one position to the left.

```python def possible(a): if len(a)==1: return False if len(a)==2: return a[0]==a[1] total = sum(a) vals = set([a[0]]) running = a[0] for el in a[1:]: running += el vals.add(el) if running == total-running: return True if (running - (total-running))%2==0 and (running - (total-running))//2 in vals: return True return False n = int(input()) a = list(map(int, input().split())) if possible(a) or possible(a[::-1]): print("YES") else: print("NO") ```

3

104

A

Blackjack

PROGRAMMING

800

[ "implementation" ]

A. Blackjack

2

256

One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one! Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture. In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses. The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.

The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.

Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.

[ "12\n", "20\n", "10\n" ]

[ "4", "15", "0" ]

In the first sample only four two's of different suits can earn the required sum of points. In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use. In the third sample there is no card, that would add a zero to the current ten points.

500

[ { "input": "12", "output": "4" }, { "input": "20", "output": "15" }, { "input": "10", "output": "0" }, { "input": "11", "output": "4" }, { "input": "15", "output": "4" }, { "input": "18", "output": "4" }, { "input": "25", "output": "0" }, { "input": "22", "output": "0" }, { "input": "1", "output": "0" }, { "input": "2", "output": "0" }, { "input": "3", "output": "0" }, { "input": "4", "output": "0" }, { "input": "5", "output": "0" }, { "input": "6", "output": "0" }, { "input": "7", "output": "0" }, { "input": "8", "output": "0" }, { "input": "9", "output": "0" }, { "input": "13", "output": "4" }, { "input": "14", "output": "4" }, { "input": "16", "output": "4" }, { "input": "17", "output": "4" }, { "input": "19", "output": "4" }, { "input": "21", "output": "4" }, { "input": "23", "output": "0" }, { "input": "24", "output": "0" } ]

1,626,073,614

2,147,483,647

Python 3

OK

TESTS

25

154

6,758,400

n = int(input()) reqNumber = n - 10 if reqNumber in [2,3,4,5,6,7,8,9] : print(4) elif reqNumber == 10: print(15) elif reqNumber == 11: print(4) elif reqNumber == 1 : print(4) else : print(0)

Title: Blackjack Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one! Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture. In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses. The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. Input Specification: The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. Output Specification: Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. Demo Input: ['12\n', '20\n', '10\n'] Demo Output: ['4', '15', '0'] Note: In the first sample only four two's of different suits can earn the required sum of points. In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use. In the third sample there is no card, that would add a zero to the current ten points.

```python n = int(input()) reqNumber = n - 10 if reqNumber in [2,3,4,5,6,7,8,9] : print(4) elif reqNumber == 10: print(15) elif reqNumber == 11: print(4) elif reqNumber == 1 : print(4) else : print(0) ```

3.948911

214

A

System of Equations

PROGRAMMING

800

[ "brute force" ]

null

Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you? You are given a system of equations: You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.

A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.

On a single line print the answer to the problem.

[ "9 3\n", "14 28\n", "4 20\n" ]

[ "1\n", "1\n", "0\n" ]

In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.

500

[ { "input": "9 3", "output": "1" }, { "input": "14 28", "output": "1" }, { "input": "4 20", "output": "0" }, { "input": "18 198", "output": "1" }, { "input": "22 326", "output": "1" }, { "input": "26 104", "output": "1" }, { "input": "14 10", "output": "0" }, { "input": "8 20", "output": "0" }, { "input": "2 8", "output": "0" }, { "input": "20 11", "output": "0" }, { "input": "57 447", "output": "1" }, { "input": "1 1", "output": "2" }, { "input": "66 296", "output": "1" }, { "input": "75 683", "output": "1" }, { "input": "227 975", "output": "1" }, { "input": "247 499", "output": "1" }, { "input": "266 116", "output": "1" }, { "input": "286 916", "output": "1" }, { "input": "307 341", "output": "1" }, { "input": "451 121", "output": "1" }, { "input": "471 921", "output": "1" }, { "input": "502 346", "output": "1" }, { "input": "535 59", "output": "1" }, { "input": "555 699", "output": "1" }, { "input": "747 351", "output": "1" }, { "input": "790 64", "output": "1" }, { "input": "810 704", "output": "1" }, { "input": "855 225", "output": "1" }, { "input": "902 34", "output": "1" }, { "input": "922 514", "output": "1" }, { "input": "971 131", "output": "1" }, { "input": "991 931", "output": "1" }, { "input": "840 780", "output": "0" }, { "input": "102 595", "output": "0" }, { "input": "139 433", "output": "0" }, { "input": "968 288", "output": "0" }, { "input": "563 354", "output": "0" }, { "input": "994 975", "output": "0" }, { "input": "456 221", "output": "0" }, { "input": "205 210", "output": "0" }, { "input": "1 11", "output": "0" }, { "input": "1000 1000", "output": "0" }, { "input": "3 3", "output": "0" }, { "input": "11 99", "output": "0" }, { "input": "2 2", "output": "1" }, { "input": "11 1", "output": "0" }, { "input": "6 6", "output": "1" }, { "input": "100 452", "output": "0" }, { "input": "420 380", "output": "0" }, { "input": "31 31", "output": "0" }, { "input": "2 6", "output": "0" }, { "input": "112 134", "output": "0" }, { "input": "13 13", "output": "0" }, { "input": "1 571", "output": "0" } ]

1,667,490,038

2,147,483,647

PyPy 3-64

OK

TESTS

54

124

0

def solve(n, m): valid_pairs_count = 0 valid_pairs = {} a = 0 while (a ** 2) <= n: b = n - (a ** 2) if int(b) == b: # checks to see if b is an integer valid_pairs[(a, b)] = True a += 1 b = 0 while (b ** 2) <= m: a = m - (b ** 2) if int(a) == a: if (a, b) in valid_pairs: valid_pairs_count += 1 b += 1 return valid_pairs_count if __name__ == "__main__": n, m = [int(_) for _ in input().split(sep=" ")] result = solve(n, m) print(result)

Title: System of Equations Time Limit: None seconds Memory Limit: None megabytes Problem Description: Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you? You are given a system of equations: You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system. Input Specification: A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space. Output Specification: On a single line print the answer to the problem. Demo Input: ['9 3\n', '14 28\n', '4 20\n'] Demo Output: ['1\n', '1\n', '0\n'] Note: In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.

```python def solve(n, m): valid_pairs_count = 0 valid_pairs = {} a = 0 while (a ** 2) <= n: b = n - (a ** 2) if int(b) == b: # checks to see if b is an integer valid_pairs[(a, b)] = True a += 1 b = 0 while (b ** 2) <= m: a = m - (b ** 2) if int(a) == a: if (a, b) in valid_pairs: valid_pairs_count += 1 b += 1 return valid_pairs_count if __name__ == "__main__": n, m = [int(_) for _ in input().split(sep=" ")] result = solve(n, m) print(result) ```

3

999

B

Reversing Encryption

PROGRAMMING

900

[ "implementation" ]

null

A string $s$ of length $n$ can be encrypted by the following algorithm: - iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$). For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$). You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique.

The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters.

Print a string $s$ such that the above algorithm results in $t$.

[ "10\nrocesfedoc\n", "16\nplmaetwoxesisiht\n", "1\nz\n" ]

[ "codeforces\n", "thisisexampletwo\n", "z\n" ]

The first example is described in the problem statement.

0

[ { "input": "10\nrocesfedoc", "output": "codeforces" }, { "input": "16\nplmaetwoxesisiht", "output": "thisisexampletwo" }, { "input": "1\nz", "output": "z" }, { "input": "2\nir", "output": "ri" }, { "input": "3\nilj", "output": "jli" }, { "input": "4\njfyy", "output": "yyjf" }, { "input": "6\nkrdych", "output": "hcyrkd" }, { "input": "60\nfnebsopcvmlaoecpzmakqigyuutueuozjxutlwwiochekmhjgwxsgfbcrpqj", "output": "jqprcbfgsxwgjhmkehcoiwwltuxjzokamzpalobnfespcvmoecqigyuutueu" }, { "input": "64\nhnlzzhrvqnldswxfsrowfhmyzbxtyoxhogudasgywxycyhzgiseerbislcncvnwy", "output": "ywnvcnclsibreesigzhycyxwygsadugofxwsdlnqzlhnzhrvsrowfhmyzbxtyoxh" }, { "input": "97\nqnqrmdhmbubaijtwsecbidqouhlecladwgwcuxbigckrfzasnbfbslukoayhcgquuacygakhxoubibxtqkpyyhzjipylujgrc", "output": "crgjulypijzhyypkqtxbibuoxhkagycauuqgchyaokulsbfbnsazfrkcgibxucwgwdalcelhuoqdibceswtjiabubmhdmrqnq" }, { "input": "100\nedykhvzcntljuuoqghptioetqnfllwekzohiuaxelgecabvsbibgqodqxvyfkbyjwtgbyhvssntinkwsinwsmalusiwnjmtcoovf", "output": "fvooctmjnwisulamswniswknitnssvhybgtwjybkfyvxqdoqgbqteoitnczvkyedhljuuoqghptnfllwekzohiuaxelgecabvsbi" }, { "input": "96\nqtbcksuvxonzbkokhqlgkrvimzqmqnrvqlihrmksldyydacbtckfphenxszcnzhfjmpeykrvshgiboivkvabhrpphgavvprz", "output": "zrpvvaghpprhbavkviobighsvrkyepmjfhznczsxnehpfkctvrnqmqzmkokbvuctqbksxonzhqlgkrviqlihrmksldyydacb" }, { "input": "90\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm", "output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm" }, { "input": "89\nwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww", "output": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww" }, { "input": "99\nqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq", "output": "qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq" }, { "input": "100\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo", "output": "oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo" }, { "input": "60\nwwwwwxwwwwwwfhwwhwwwwwwawwwwwwwwwwwwwnwwwwwwwwwwwwwwwwwwwwww", "output": "wwwwwwwwwwwwwwwwwwwwwwnwwwwwwwwwwhwwwxwwwwwwwwwfhwwwwawwwwww" }, { "input": "90\ncccchccccccccccccccccccccccccccwcccccccccgcccccchccccccccccccccccccccccxccccccncccccccuccc", "output": "cccucccccccnccccccxcccccccccccccccccccccchccccccccccccccccccccccchccccccccccwcccccccccgccc" }, { "input": "97\nfwffffffffffffffffffffffffrffffffffffffffzfffffffffffffffftfcfffffffqffffffffffffffffffffffyfffff", "output": "fffffyffffffffffffffffffffffqfffffffcftffffffffffffffffzffffffffffffffrffffffffffffffffffffffffwf" }, { "input": "100\ndjjjjjjjjjjgjjjjjjjjjjjjjjsvjjjjjjjjjjmjjjjjjjjjjjjjajjjjjjajjjjjjrjjjjjjjjjjjjrjjtjjjjjjjjjjjjjojjj", "output": "jjjojjjjjjjjjjjjjtjjrjjjjjjjjjjjjrjjjjjjajjjjjjajjjjjjjjjjjjjjdjjjgjjjjjjjjjsvjjjjjjjjjjmjjjjjjjjjjj" } ]

1,613,939,704

2,147,483,647

Python 3

OK

TESTS

20

62

0

n=int(input()) s=input() for i in range(2, n+1): if n%i==0: z=str() for j in range(i-1, -1, -1): z=z+s[j] for j in range(i, n): z=z+s[j] s=z print(s)

Title: Reversing Encryption Time Limit: None seconds Memory Limit: None megabytes Problem Description: A string $s$ of length $n$ can be encrypted by the following algorithm: - iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$). For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$). You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique. Input Specification: The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters. Output Specification: Print a string $s$ such that the above algorithm results in $t$. Demo Input: ['10\nrocesfedoc\n', '16\nplmaetwoxesisiht\n', '1\nz\n'] Demo Output: ['codeforces\n', 'thisisexampletwo\n', 'z\n'] Note: The first example is described in the problem statement.

```python n=int(input()) s=input() for i in range(2, n+1): if n%i==0: z=str() for j in range(i-1, -1, -1): z=z+s[j] for j in range(i, n): z=z+s[j] s=z print(s) ```

3

515

B

Drazil and His Happy Friends

PROGRAMMING

1,300

[ "brute force", "dsu", "meet-in-the-middle", "number theory" ]

null

Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan. There are *n* boys and *m* girls among his friends. Let's number them from 0 to *n*<=-<=1 and 0 to *m*<=-<=1 separately. In *i*-th day, Drazil invites -th boy and -th girl to have dinner together (as Drazil is programmer, *i* starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever. Drazil wants to know whether he can use this plan to make all his friends become happy at some moment.

The first line contains two integer *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains integer *b* (0<=≤<=*b*<=≤<=*n*), denoting the number of happy boys among friends of Drazil, and then follow *b* distinct integers *x*1,<=*x*2,<=...,<=*x**b* (0<=≤<=*x**i*<=<<=*n*), denoting the list of indices of happy boys. The third line conatins integer *g* (0<=≤<=*g*<=≤<=*m*), denoting the number of happy girls among friends of Drazil, and then follow *g* distinct integers *y*1,<=*y*2,<=... ,<=*y**g* (0<=≤<=*y**j*<=<<=*m*), denoting the list of indices of happy girls. It is guaranteed that there is at least one person that is unhappy among his friends.

If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No".

[ "2 3\n0\n1 0\n", "2 4\n1 0\n1 2\n", "2 3\n1 0\n1 1\n" ]

[ "Yes\n", "No\n", "Yes\n" ]

By <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/169ade208e6eb4f9263c57aaff716529d59c3288.png" style="max-width: 100.0%;max-height: 100.0%;"/> we define the remainder of integer division of *i* by *k*. In first sample case: - On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. - On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes at this day. - On the 2-nd day, Drazil invites 0-th boy and 2-nd girl. Because 0-th boy is already happy he makes 2-nd girl become happy at this day. - On the 3-rd day, Drazil invites 1-st boy and 0-th girl. 0-th girl is happy, so she makes 1-st boy happy. - On the 4-th day, Drazil invites 0-th boy and 1-st girl. 0-th boy is happy, so he makes the 1-st girl happy. So, all friends become happy at this moment.

1,000

[ { "input": "2 3\n0\n1 0", "output": "Yes" }, { "input": "2 4\n1 0\n1 2", "output": "No" }, { "input": "2 3\n1 0\n1 1", "output": "Yes" }, { "input": "16 88\n6 5 14 2 0 12 7\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82", "output": "Yes" }, { "input": "52 91\n13 26 1 3 43 17 19 32 46 33 48 23 37 50\n25 78 26 1 40 2 67 42 4 56 30 70 84 32 20 85 59 8 86 34 73 23 10 88 24 11", "output": "No" }, { "input": "26 52\n8 0 14 16 17 7 9 10 11\n15 39 15 2 41 42 30 17 18 31 6 21 35 48 50 51", "output": "No" }, { "input": "50 50\n0\n0", "output": "No" }, { "input": "27 31\n4 25 5 19 20\n26 5 28 17 2 1 0 26 23 12 29 6 4 25 19 15 13 20 24 8 27 22 30 3 10 9 7", "output": "Yes" }, { "input": "55 79\n5 51 27 36 45 53\n30 15 28 0 5 38 3 34 30 35 1 32 12 27 42 39 69 33 10 63 16 29 76 19 60 70 67 31 78 68 45", "output": "Yes" }, { "input": "79 23\n35 31 62 14 9 46 18 68 69 42 13 50 77 23 76 5 53 40 16 32 74 54 38 25 45 39 26 37 66 78 3 48 10 17 56 59\n13 16 0 8 6 18 14 21 11 20 4 15 13 22", "output": "Yes" }, { "input": "7 72\n1 4\n3 49 32 28", "output": "Yes" }, { "input": "100 50\n31 52 54 8 60 61 62 63 64 16 19 21 73 25 76 77 79 30 81 32 33 34 37 88 39 40 91 42 94 95 96 98\n18 0 1 3 5 6 7 9 15 18 20 22 24 28 35 36 43 47 49", "output": "No" }, { "input": "98 49\n33 0 51 52 6 57 10 12 63 15 16 19 20 21 72 73 74 76 77 78 30 31 81 33 83 37 38 39 40 92 44 45 95 97\n15 4 5 7 9 11 13 17 18 22 26 35 36 41 42 47", "output": "No" }, { "input": "50 50\n14 7 8 12 16 18 22 23 24 28 30 35 40 46 49\n35 0 1 2 3 4 5 6 9 10 11 13 14 15 17 19 20 21 25 26 27 29 31 32 33 34 36 37 38 39 41 43 44 45 47 48", "output": "No" }, { "input": "30 44\n3 8 26 28\n6 2 30 38 26 8 6", "output": "No" }, { "input": "69 72\n18 58 46 52 43 1 55 16 7 4 38 68 14 32 53 41 29 2 59\n21 22 43 55 13 70 4 7 31 10 23 56 44 62 17 50 53 5 41 11 65 32", "output": "No" }, { "input": "76 28\n10 24 13 61 45 29 57 41 21 37 11\n2 12 9", "output": "No" }, { "input": "65 75\n15 25 60 12 62 37 22 47 52 3 63 58 13 14 49 34\n18 70 10 2 52 22 47 72 57 38 48 13 73 3 19 4 74 49 34", "output": "No" }, { "input": "6 54\n1 5\n14 13 49 31 37 44 2 15 51 52 22 28 10 35 47", "output": "No" }, { "input": "96 36\n34 84 24 0 48 85 13 61 37 62 38 86 75 3 16 64 40 28 76 53 5 17 42 6 7 91 67 55 68 92 57 11 71 35 59\n9 1 14 15 17 18 30 6 8 35", "output": "No" }, { "input": "40 40\n23 0 2 3 4 5 7 11 15 16 17 18 19 22 25 28 29 30 31 32 34 35 36 37\n16 1 6 8 9 10 12 13 14 20 21 23 24 26 27 38 39", "output": "No" }, { "input": "66 66\n24 2 35 3 36 4 5 10 45 14 48 18 51 19 21 55 22 23 24 25 26 63 31 65 32\n21 0 1 37 6 40 7 8 42 45 13 15 16 50 53 23 24 60 28 62 63 31", "output": "No" }, { "input": "20 20\n9 0 3 4 6 7 8 10 12 13\n10 1 2 5 9 11 14 15 16 18 19", "output": "No" }, { "input": "75 30\n18 46 47 32 33 3 34 35 21 51 7 9 54 39 72 42 59 29 14\n8 0 17 5 6 23 26 27 13", "output": "No" }, { "input": "100 50\n30 50 54 7 8 59 60 61 62 63 64 15 16 18 19 20 22 73 27 79 83 86 87 89 42 93 94 45 46 97 98\n20 1 2 3 5 6 17 21 24 25 26 28 30 31 32 34 35 38 40 41 49", "output": "Yes" }, { "input": "98 98\n43 49 1 51 3 53 4 55 56 8 9 10 60 11 12 61 64 16 65 17 19 20 21 72 24 74 25 77 78 31 34 35 36 37 87 88 89 42 92 43 44 94 46 96\n34 50 2 52 5 54 9 62 63 15 18 68 70 22 72 75 26 27 77 30 81 82 83 35 36 37 87 88 89 90 41 93 95 96 48", "output": "No" }, { "input": "100 100\n45 50 1 4 5 55 7 8 10 60 61 62 63 14 65 66 17 18 20 21 22 24 25 27 78 28 29 30 31 82 83 33 84 36 37 38 39 40 41 42 44 45 46 48 98 49\n34 50 1 2 52 3 54 56 7 9 59 61 14 16 67 18 69 22 73 24 76 79 81 82 84 35 36 38 39 90 43 44 45 47 49", "output": "No" }, { "input": "76 72\n29 4 64 68 20 8 12 50 42 46 0 70 11 37 75 47 45 29 17 19 73 9 41 31 35 67 65 39 51 55\n25 60 32 48 42 8 6 9 7 31 19 25 5 33 51 61 67 55 49 27 29 53 39 65 35 13", "output": "Yes" }, { "input": "39 87\n16 18 15 30 33 21 9 3 31 16 10 34 20 35 8 26 23\n36 33 75 81 24 42 54 78 39 57 60 30 36 63 4 76 25 1 40 73 22 58 49 85 31 74 59 20 44 83 65 23 41 71 47 14 35", "output": "Yes" }, { "input": "36 100\n10 0 32 4 5 33 30 18 14 35 7\n29 60 32 20 4 16 69 5 38 50 46 74 94 18 82 2 66 22 42 55 51 91 67 75 35 95 43 79 3 27", "output": "Yes" }, { "input": "90 25\n26 55 30 35 20 15 26 6 1 41 81 76 46 57 17 12 67 77 27 47 62 8 43 63 3 48 19\n9 10 16 21 7 17 12 13 19 9", "output": "Yes" }, { "input": "66 66\n26 0 54 6 37 43 13 25 38 2 32 56 20 50 39 27 51 9 64 4 16 17 65 11 5 47 23\n15 6 24 43 49 25 20 14 63 27 3 58 52 53 11 41", "output": "No" }, { "input": "24 60\n4 0 2 19 23\n15 12 24 49 2 14 3 52 28 5 6 19 32 33 34 35", "output": "Yes" }, { "input": "80 40\n27 0 41 44 45 6 47 8 10 52 13 14 16 17 18 59 21 62 23 64 26 68 29 32 75 37 78 39\n13 2 3 9 11 15 20 25 27 30 31 33 34 36", "output": "Yes" }, { "input": "66 99\n23 33 35 36 38 8 10 44 11 45 46 47 50 19 54 22 55 23 58 59 27 61 30 65\n32 33 67 69 4 70 38 6 39 7 74 42 9 43 12 13 14 15 81 82 84 85 20 87 89 90 24 58 59 27 95 97 31", "output": "Yes" }, { "input": "100 40\n25 61 42 2 3 25 46 66 68 69 49 9 10 50 91 72 92 33 73 53 14 15 55 96 36 39\n12 0 22 3 23 4 6 27 11 35 37 38 39", "output": "Yes" }, { "input": "90 30\n27 15 16 2 32 78 49 64 65 50 6 66 21 22 82 23 39 84 85 10 86 56 27 87 13 58 44 74\n7 19 4 20 24 25 12 27", "output": "No" }, { "input": "75 75\n33 30 74 57 23 19 42 71 11 44 29 58 43 48 61 63 13 27 50 17 18 70 64 39 12 32 36 10 40 51 49 1 54 73\n8 43 23 0 7 63 47 74 28", "output": "No" }, { "input": "98 98\n23 6 81 90 28 38 51 23 69 13 95 15 16 88 58 10 26 42 44 54 92 27 45 39\n18 20 70 38 82 72 61 37 78 74 23 15 56 59 35 93 64 28 57", "output": "No" }, { "input": "75 75\n19 48 3 5 67 23 8 70 45 63 36 38 56 15 10 37 52 11 9 27\n21 13 9 45 28 59 36 30 43 5 38 27 40 50 17 41 71 8 51 63 1 33", "output": "No" }, { "input": "3 20\n0\n1 19", "output": "Yes" }, { "input": "41 2\n1 33\n0", "output": "Yes" }, { "input": "50 49\n1 49\n0", "output": "Yes" }, { "input": "3 50\n0\n1 49", "output": "Yes" }, { "input": "100 100\n50 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49\n49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98", "output": "No" }, { "input": "100 100\n50 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49\n50 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99", "output": "Yes" }, { "input": "91 98\n78 0 1 2 3 4 5 7 8 9 10 11 12 14 15 16 17 18 19 21 22 23 24 25 26 28 29 30 31 32 33 35 36 37 38 39 40 42 43 44 45 46 47 49 50 51 52 53 54 56 57 58 59 60 61 63 64 65 66 67 68 70 71 72 73 74 75 77 78 79 80 81 82 84 85 86 87 88 89\n84 0 1 2 3 4 5 7 8 9 10 11 12 14 15 16 17 18 19 21 22 23 24 25 26 28 29 30 31 32 33 35 36 37 38 39 40 42 43 44 45 46 47 49 50 51 52 53 54 56 57 58 59 60 61 63 64 65 66 67 68 70 71 72 73 74 75 77 78 79 80 81 82 84 85 86 87 88 89 91 92 93 94 95 96", "output": "No" }, { "input": "99 84\n66 0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 26 27 29 30 32 33 35 36 38 39 41 42 44 45 47 48 50 51 53 54 56 57 59 60 62 63 65 66 68 69 71 72 74 75 77 78 80 81 83 84 86 87 89 90 92 93 95 96 98\n56 0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 26 27 29 30 32 33 35 36 38 39 41 42 44 45 47 48 50 51 53 54 56 57 59 60 62 63 65 66 68 69 71 72 74 75 77 78 80 81 83", "output": "No" }, { "input": "75 90\n60 0 2 3 4 5 7 8 9 10 12 13 14 15 17 18 19 20 22 23 24 25 27 28 29 30 32 33 34 35 37 38 39 40 42 43 44 45 47 48 49 50 52 53 54 55 57 58 59 60 62 63 64 65 67 68 69 70 72 73 74\n72 0 2 3 4 5 7 8 9 10 12 13 14 15 17 18 19 20 22 23 24 25 27 28 29 30 32 33 34 35 37 38 39 40 42 43 44 45 47 48 49 50 52 53 54 55 57 58 59 60 62 63 64 65 67 68 69 70 72 73 74 75 77 78 79 80 82 83 84 85 87 88 89", "output": "No" }, { "input": "5 7\n1 0\n1 0", "output": "Yes" }, { "input": "100 1\n1 99\n0", "output": "Yes" }, { "input": "4 1\n1 3\n0", "output": "Yes" }, { "input": "4 5\n3 0 1 3\n4 0 1 3 4", "output": "Yes" }, { "input": "100 99\n1 99\n0", "output": "Yes" }, { "input": "2 3\n1 0\n2 0 2", "output": "Yes" } ]

1,599,329,127

2,147,483,647

PyPy 3

OK

TESTS

56

140

1,331,200

n, m = [int(p) for p in input().split()] arrb = [int(p) for p in input().split()] arrg = [int(p) for p in input().split()] arrb.pop(0) arrg.pop(0) cb = [False]*n cg = [False]*m for i in arrb: cb[i] = True for i in arrg: cg[i] = True i = 0 while i < 10**5: b = cb[i%n] g = cg[i%m] if b and g: i += 1 continue elif b and not g: cg[i%m] = True elif not b and g: cb[i%n] = True else: i += 1 continue i += 1 if all(cb) and all(cg): print('Yes') else: print('No')

Title: Drazil and His Happy Friends Time Limit: None seconds Memory Limit: None megabytes Problem Description: Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan. There are *n* boys and *m* girls among his friends. Let's number them from 0 to *n*<=-<=1 and 0 to *m*<=-<=1 separately. In *i*-th day, Drazil invites -th boy and -th girl to have dinner together (as Drazil is programmer, *i* starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever. Drazil wants to know whether he can use this plan to make all his friends become happy at some moment. Input Specification: The first line contains two integer *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains integer *b* (0<=≤<=*b*<=≤<=*n*), denoting the number of happy boys among friends of Drazil, and then follow *b* distinct integers *x*1,<=*x*2,<=...,<=*x**b* (0<=≤<=*x**i*<=<<=*n*), denoting the list of indices of happy boys. The third line conatins integer *g* (0<=≤<=*g*<=≤<=*m*), denoting the number of happy girls among friends of Drazil, and then follow *g* distinct integers *y*1,<=*y*2,<=... ,<=*y**g* (0<=≤<=*y**j*<=<<=*m*), denoting the list of indices of happy girls. It is guaranteed that there is at least one person that is unhappy among his friends. Output Specification: If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". Demo Input: ['2 3\n0\n1 0\n', '2 4\n1 0\n1 2\n', '2 3\n1 0\n1 1\n'] Demo Output: ['Yes\n', 'No\n', 'Yes\n'] Note: By <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/169ade208e6eb4f9263c57aaff716529d59c3288.png" style="max-width: 100.0%;max-height: 100.0%;"/> we define the remainder of integer division of *i* by *k*. In first sample case: - On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. - On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes at this day. - On the 2-nd day, Drazil invites 0-th boy and 2-nd girl. Because 0-th boy is already happy he makes 2-nd girl become happy at this day. - On the 3-rd day, Drazil invites 1-st boy and 0-th girl. 0-th girl is happy, so she makes 1-st boy happy. - On the 4-th day, Drazil invites 0-th boy and 1-st girl. 0-th boy is happy, so he makes the 1-st girl happy. So, all friends become happy at this moment.

```python n, m = [int(p) for p in input().split()] arrb = [int(p) for p in input().split()] arrg = [int(p) for p in input().split()] arrb.pop(0) arrg.pop(0) cb = [False]*n cg = [False]*m for i in arrb: cb[i] = True for i in arrg: cg[i] = True i = 0 while i < 10**5: b = cb[i%n] g = cg[i%m] if b and g: i += 1 continue elif b and not g: cg[i%m] = True elif not b and g: cb[i%n] = True else: i += 1 continue i += 1 if all(cb) and all(cg): print('Yes') else: print('No') ```

3

817

C

Really Big Numbers

PROGRAMMING

1,600

[ "binary search", "brute force", "dp", "math" ]

null

Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*. Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.

The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018).

Print one integer — the quantity of really big numbers that are not greater than *n*.

[ "12 1\n", "25 20\n", "10 9\n" ]

[ "3\n", "0\n", "1\n" ]

In the first example numbers 10, 11 and 12 are really big. In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20). In the third example 10 is the only really big number (10 - 1 ≥ 9).

0

[ { "input": "12 1", "output": "3" }, { "input": "25 20", "output": "0" }, { "input": "10 9", "output": "1" }, { "input": "300 1000", "output": "0" }, { "input": "500 1000", "output": "0" }, { "input": "1000 2000", "output": "0" }, { "input": "10000 1000", "output": "8991" }, { "input": "1000000000000000000 1000000000000000000", "output": "0" }, { "input": "1000000000000000000 100000000000000000", "output": "899999999999999991" }, { "input": "1000000000000000000 10000000000000000", "output": "989999999999999991" }, { "input": "1000000000000000000 1000000000000000", "output": "998999999999999991" }, { "input": "1000000000000000000 100000000000000", "output": "999899999999999991" }, { "input": "1000000000000000000 200000000000000000", "output": "799999999999999991" }, { "input": "10 5", "output": "1" }, { "input": "20 5", "output": "11" }, { "input": "20 9", "output": "11" }, { "input": "100 9", "output": "91" }, { "input": "1 1", "output": "0" }, { "input": "130 118", "output": "1" }, { "input": "190 181", "output": "0" }, { "input": "1999 1971", "output": "10" }, { "input": "100 99", "output": "1" }, { "input": "6909094398 719694282", "output": "6189400069" }, { "input": "260 258", "output": "0" }, { "input": "35 19", "output": "6" }, { "input": "100 87", "output": "1" }, { "input": "91 89", "output": "0" }, { "input": "109 89", "output": "10" }, { "input": "109 91", "output": "10" }, { "input": "20331 11580", "output": "8732" }, { "input": "405487470 255750281", "output": "149737161" }, { "input": "17382 12863", "output": "4493" }, { "input": "19725 14457", "output": "5246" }, { "input": "24848 15384", "output": "9449" }, { "input": "25727 15982", "output": "9728" }, { "input": "109 90", "output": "10" }, { "input": "1000000000000000000 999999999999999999", "output": "1" }, { "input": "1000000000000000000 999999999999999998", "output": "1" }, { "input": "1009 980", "output": "10" }, { "input": "999999999999999999 999999999999999838", "output": "0" }, { "input": "1000000000000000000 99999999999999800", "output": "900000000000000061" }, { "input": "8785369357 3377262261", "output": "5408107058" }, { "input": "110 109", "output": "0" }, { "input": "999 777", "output": "200" }, { "input": "327170000015578 77230000029054", "output": "249939999986479" }, { "input": "12515000022229 1791000022317", "output": "10723999999880" }, { "input": "9999999999999 9999999999882", "output": "10" }, { "input": "213 196", "output": "14" }, { "input": "92 82", "output": "0" }, { "input": "148 136", "output": "0" }, { "input": "8 9", "output": "0" }, { "input": "309 299", "output": "0" }, { "input": "9999 9963", "output": "10" }, { "input": "82 81", "output": "0" }, { "input": "9999999 9999936", "output": "10" }, { "input": "171 155", "output": "2" }, { "input": "999 972", "output": "10" }, { "input": "999999999999 999999999891", "output": "10" }, { "input": "9 9", "output": "0" }, { "input": "6900 6885", "output": "1" }, { "input": "96 57", "output": "27" }, { "input": "5 4", "output": "0" }, { "input": "17386 5814", "output": "11557" }, { "input": "493679757404593 316259583979965", "output": "177420173424564" }, { "input": "18474 9478", "output": "8975" }, { "input": "270091571496186 250931112649966", "output": "19160458846177" }, { "input": "565751690089037 381448507916936", "output": "184303182172038" }, { "input": "19 10", "output": "0" } ]

1,497,543,367

4,267

Python 3

OK

TESTS

68

62

0

def f(x): ans = x while x > 0: ans -= x % 10 x //= 10 return ans n, s = map(int, input().split()) if f(n) < s: print(0) exit(0) l, r = 1, n while l < r: m = (l + r) // 2 if f(m) >= s: r = (m // 10) * 10 else: l = m + 1 print(n - l + 1)

Title: Really Big Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*. Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations. Input Specification: The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018). Output Specification: Print one integer — the quantity of really big numbers that are not greater than *n*. Demo Input: ['12 1\n', '25 20\n', '10 9\n'] Demo Output: ['3\n', '0\n', '1\n'] Note: In the first example numbers 10, 11 and 12 are really big. In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20). In the third example 10 is the only really big number (10 - 1 ≥ 9).

```python def f(x): ans = x while x > 0: ans -= x % 10 x //= 10 return ans n, s = map(int, input().split()) if f(n) < s: print(0) exit(0) l, r = 1, n while l < r: m = (l + r) // 2 if f(m) >= s: r = (m // 10) * 10 else: l = m + 1 print(n - l + 1) ```

3

807

A

Is it rated?

PROGRAMMING

900

[ "implementation", "sortings" ]

null

Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.

If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".

[ "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n", "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n", "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n" ]

[ "rated\n", "unrated\n", "maybe\n" ]

In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

500

[ { "input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated" }, { "input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n1 1\n1 1", "output": "maybe" }, { "input": "2\n4126 4126\n4126 4126", "output": "maybe" }, { "input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423", "output": "rated" }, { "input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110", "output": "unrated" }, { "input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143", "output": "maybe" }, { "input": "2\n3936 3936\n2967 2967", "output": "maybe" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 1\n1 2", "output": "rated" }, { "input": "2\n2967 2967\n3936 3936", "output": "unrated" }, { "input": "3\n1200 1200\n1200 1200\n1300 1300", "output": "unrated" }, { "input": "3\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "3\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "2\n3 2\n3 2", "output": "rated" }, { "input": "3\n5 5\n4 4\n3 4", "output": "rated" }, { "input": "3\n200 200\n200 200\n300 300", "output": "unrated" }, { "input": "3\n1 1\n2 2\n3 3", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699", "output": "maybe" }, { "input": "2\n10 10\n8 8", "output": "maybe" }, { "input": "3\n1500 1500\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "3\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n100 100\n100 100\n70 70\n80 80", "output": "unrated" }, { "input": "2\n1 2\n2 1", "output": "rated" }, { "input": "3\n5 5\n4 3\n3 3", "output": "rated" }, { "input": "3\n1600 1650\n1500 1550\n1400 1450", "output": "rated" }, { "input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700", "output": "unrated" }, { "input": "2\n1600 1600\n1400 1400", "output": "maybe" }, { "input": "2\n3 1\n9 8", "output": "rated" }, { "input": "2\n2 1\n1 1", "output": "rated" }, { "input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670", "output": "unrated" }, { "input": "2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n10 11\n5 4", "output": "rated" }, { "input": "2\n15 14\n13 12", "output": "rated" }, { "input": "2\n2 1\n2 2", "output": "rated" }, { "input": "3\n2670 2670\n3670 3670\n4106 4106", "output": "unrated" }, { "input": "3\n4 5\n3 3\n2 2", "output": "rated" }, { "input": "2\n10 9\n10 10", "output": "rated" }, { "input": "3\n1011 1011\n1011 999\n2200 2100", "output": "rated" }, { "input": "2\n3 3\n5 5", "output": "unrated" }, { "input": "2\n1500 1500\n3000 2000", "output": "rated" }, { "input": "2\n5 6\n5 5", "output": "rated" }, { "input": "3\n2000 2000\n1500 1501\n500 500", "output": "rated" }, { "input": "2\n2 3\n2 2", "output": "rated" }, { "input": "2\n3 3\n2 2", "output": "maybe" }, { "input": "2\n1 2\n1 1", "output": "rated" }, { "input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n15 14\n14 13", "output": "rated" }, { "input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900", "output": "unrated" }, { "input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884", "output": "rated" }, { "input": "2\n100 99\n100 100", "output": "rated" }, { "input": "4\n2 2\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "3\n100 101\n100 100\n100 100", "output": "rated" }, { "input": "4\n1000 1001\n900 900\n950 950\n890 890", "output": "rated" }, { "input": "2\n2 3\n1 1", "output": "rated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n3 2\n2 2", "output": "rated" }, { "input": "2\n3 2\n3 3", "output": "rated" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "3\n3 2\n3 3\n3 3", "output": "rated" }, { "input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "3\n1000 1000\n500 500\n400 300", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000", "output": "unrated" }, { "input": "2\n1 1\n2 3", "output": "rated" }, { "input": "2\n6 2\n6 2", "output": "rated" }, { "input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246", "output": "unrated" }, { "input": "2\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699", "output": "maybe" }, { "input": "2\n20 30\n10 5", "output": "rated" }, { "input": "3\n1 1\n2 2\n1 1", "output": "unrated" }, { "input": "2\n1 2\n3 3", "output": "rated" }, { "input": "5\n5 5\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 2\n2 1", "output": "rated" }, { "input": "2\n100 100\n90 89", "output": "rated" }, { "input": "2\n1000 900\n2000 2000", "output": "rated" }, { "input": "2\n50 10\n10 50", "output": "rated" }, { "input": "2\n200 200\n100 100", "output": "maybe" }, { "input": "3\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "3\n1000 1000\n300 300\n100 100", "output": "maybe" }, { "input": "4\n2 2\n2 2\n3 3\n4 4", "output": "unrated" }, { "input": "2\n5 3\n6 3", "output": "rated" }, { "input": "2\n1200 1100\n1200 1000", "output": "rated" }, { "input": "2\n5 5\n4 4", "output": "maybe" }, { "input": "2\n5 5\n3 3", "output": "maybe" }, { "input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100", "output": "unrated" }, { "input": "5\n10 10\n9 9\n8 8\n7 7\n6 6", "output": "maybe" }, { "input": "3\n1000 1000\n300 300\n10 10", "output": "maybe" }, { "input": "5\n6 6\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "2\n3 3\n1 1", "output": "maybe" }, { "input": "4\n2 2\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n1000 1000\n700 700", "output": "maybe" }, { "input": "2\n4 3\n5 3", "output": "rated" }, { "input": "2\n1000 1000\n1100 1100", "output": "unrated" }, { "input": "4\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "3\n1 1\n2 3\n2 2", "output": "rated" }, { "input": "2\n1 2\n1 3", "output": "rated" }, { "input": "2\n3 3\n1 2", "output": "rated" }, { "input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "5\n1 1\n2 2\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "2\n10 10\n1 2", "output": "rated" }, { "input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900", "output": "unrated" }, { "input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699", "output": "unrated" }, { "input": "2\n100 100\n110 110", "output": "unrated" }, { "input": "3\n3 3\n3 3\n4 4", "output": "unrated" }, { "input": "3\n3 3\n3 2\n4 4", "output": "rated" }, { "input": "3\n5 2\n4 4\n3 3", "output": "rated" }, { "input": "4\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n1 1\n3 2", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699", "output": "unrated" }, { "input": "3\n3 3\n3 3\n3 4", "output": "rated" }, { "input": "3\n1 2\n2 2\n3 3", "output": "rated" }, { "input": "3\n1 2\n1 2\n1 2", "output": "rated" }, { "input": "2\n2 1\n2 1", "output": "rated" }, { "input": "2\n1 2\n3 4", "output": "rated" }, { "input": "2\n3 2\n2 3", "output": "rated" }, { "input": "3\n1500 1500\n1600 1600\n1600 1600", "output": "unrated" }, { "input": "3\n1 1\n3 3\n4 4", "output": "unrated" }, { "input": "3\n1 1\n2 2\n2 2", "output": "unrated" }, { "input": "2\n10 12\n8 8", "output": "rated" }, { "input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500", "output": "unrated" }, { "input": "2\n1 2\n2 2", "output": "rated" }, { "input": "3\n1500 1400\n1200 1200\n1100 1100", "output": "rated" }, { "input": "2\n10 12\n10 10", "output": "rated" }, { "input": "3\n1500 1500\n1400 1400\n1300 1300", "output": "maybe" }, { "input": "3\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "3\n2 6\n3 5\n4 4", "output": "rated" }, { "input": "2\n5 6\n4 6", "output": "rated" }, { "input": "4\n10 10\n10 10\n7 7\n8 8", "output": "unrated" }, { "input": "2\n4 4\n3 3", "output": "maybe" } ]

1,598,376,402

2,147,483,647

Python 3

OK

TESTS

150

124

307,200

n = int(input()) a=[[int(i) for i in input().split()] for i in range(n)] ans = 'maybe' for i in range(n): if a[i][0] != a[i][1]: ans = 'rated' break else: for i in range(n-1): if a[i][0] < a[i+1][0]: ans = 'unrated' break print(ans)

Title: Is it rated? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. Output Specification: If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". Demo Input: ['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n'] Demo Output: ['rated\n', 'unrated\n', 'maybe\n'] Note: In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

```python n = int(input()) a=[[int(i) for i in input().split()] for i in range(n)] ans = 'maybe' for i in range(n): if a[i][0] != a[i][1]: ans = 'rated' break else: for i in range(n-1): if a[i][0] < a[i+1][0]: ans = 'unrated' break print(ans) ```

3

570

A

Elections

PROGRAMMING

1,100

[ "implementation" ]

null

The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate. The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index. At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index. Determine who will win the elections.

The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively. Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*. It is guaranteed that the total number of people in all the cities does not exceed 109.

Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.

[ "3 3\n1 2 3\n2 3 1\n1 2 1\n", "3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n" ]

[ "2", "1" ]

Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes. Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.

500

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1,472,980,234

2,147,483,647

Python 3

OK

TESTS

62

102,400

lineInput = list(map(int,str(input()).split())) m = lineInput[0] n = lineInput[1] city = [0] * 110 for i in range(0,n): cnt = list(map(int,str(input()).split())) winner = 0 for j in range(0,m): if(cnt[j] > cnt[winner]): winner = j city[winner] += 1 answer = int(0) for i in range(0,m): if(city[i] > city[answer]): answer = i answer += 1 print(answer)

Title: Elections Time Limit: None seconds Memory Limit: None megabytes Problem Description: The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate. The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index. At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index. Determine who will win the elections. Input Specification: The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively. Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*. It is guaranteed that the total number of people in all the cities does not exceed 109. Output Specification: Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one. Demo Input: ['3 3\n1 2 3\n2 3 1\n1 2 1\n', '3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n'] Demo Output: ['2', '1'] Note: Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes. Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.

```python lineInput = list(map(int,str(input()).split())) m = lineInput[0] n = lineInput[1] city = [0] * 110 for i in range(0,n): cnt = list(map(int,str(input()).split())) winner = 0 for j in range(0,m): if(cnt[j] > cnt[winner]): winner = j city[winner] += 1 answer = int(0) for i in range(0,m): if(city[i] > city[answer]): answer = i answer += 1 print(answer) ```

3

735

B

Urbanization

PROGRAMMING

1,100

[ "greedy", "number theory", "sortings" ]

null

Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home. To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding. Please, help authorities find the optimal way to pick residents for two cities.

The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.

Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .

[ "2 1 1\n1 5\n", "4 2 1\n1 4 2 3\n" ]

[ "6.00000000\n", "6.50000000\n" ]

In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second. In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*3 + *a*4) / 2 + *a*2 = (3 + 2) / 2 + 4 = 6.5

1,000

[ { "input": "2 1 1\n1 5", "output": "6.00000000" }, { "input": "4 2 1\n1 4 2 3", "output": "6.50000000" }, { "input": "3 1 2\n1 2 3", "output": "4.50000000" }, { "input": "10 4 6\n3 5 7 9 12 25 67 69 83 96", "output": "88.91666667" }, { "input": "19 7 12\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 100000 100000", "output": "47052.10714286" }, { "input": "100 9 6\n109 711 40 95 935 48 228 253 308 726 816 534 252 8 966 363 162 508 84 83 807 506 748 178 45 30 106 108 764 698 825 198 336 353 158 790 64 262 403 334 577 571 742 541 946 602 279 621 910 776 421 886 29 133 114 394 762 965 339 263 750 530 49 80 124 31 322 292 27 590 960 278 111 932 849 491 561 744 469 511 106 271 156 160 836 363 149 473 457 543 976 809 490 29 85 626 265 88 995 946", "output": "1849.66666667" }, { "input": "69 6 63\n53475 22876 79144 6335 33763 79104 65441 45527 65847 94406 74670 43529 75330 19403 67629 56187 57949 23071 64910 54409 55348 18056 855 24961 50565 6622 26467 33989 22660 79469 41246 13965 79706 14422 16075 93378 81313 48173 13470 97348 2346 27452 59427 29925 29847 73823 32021 10988 24609 98855 90919 45939 17203 8439 43007 40138 55693 30314 71734 33458 66850 4011 20089 20546 92090 50842 78859 62756 40177", "output": "135712.88888889" }, { "input": "69 6 9\n2612 17461 69001 33130 10662 85485 88195 45974 16712 81365 67119 87797 15559 20197 74716 92979 97268 49466 68603 48351 99905 35606 54242 98603 68232 54398 82637 49647 38979 46171 54680 23334 15892 92186 69670 29711 67999 2220 32317 717 70667 68262 86760 55720 97158 61122 7251 138 21022 27197 12691 59331 13576 66999 38332 13574 83484 66646 17704 33065 98583 80259 64631 16745 69431 40747 82089 82788 32739", "output": "183129.44444444" } ]

1,614,342,549

2,147,483,647

PyPy 3

OK

TESTS

37

171

10,649,600

n,n1,n2=map(int,input().split()) n1,n2=min(n1,n2),max(n1,n2) l=sorted(list(map(int,input().split()))) l.reverse() print(sum(l[:n1])/n1+sum(l[n1:n1+n2])/n2)

Title: Urbanization Time Limit: None seconds Memory Limit: None megabytes Problem Description: Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home. To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding. Please, help authorities find the optimal way to pick residents for two cities. Input Specification: The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate. Output Specification: Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . Demo Input: ['2 1 1\n1 5\n', '4 2 1\n1 4 2 3\n'] Demo Output: ['6.00000000\n', '6.50000000\n'] Note: In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second. In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*3 + *a*4) / 2 + *a*2 = (3 + 2) / 2 + 4 = 6.5

```python n,n1,n2=map(int,input().split()) n1,n2=min(n1,n2),max(n1,n2) l=sorted(list(map(int,input().split()))) l.reverse() print(sum(l[:n1])/n1+sum(l[n1:n1+n2])/n2) ```

3

0

none

0

[ "none" ]

null

Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen.

The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors. Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000). The total number of balls doesn't exceed 1000.

A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007.

[ "3\n2\n2\n1\n", "4\n1\n2\n3\n4\n" ]

[ "3\n", "1680\n" ]

In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:

0

[ { "input": "3\n2\n2\n1", "output": "3" }, { "input": "4\n1\n2\n3\n4", "output": "1680" }, { "input": "10\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100", "output": "12520708" }, { "input": "5\n10\n10\n10\n10\n10", "output": "425711769" }, { "input": "11\n291\n381\n126\n39\n19\n20\n3\n1\n20\n45\n2", "output": "902382672" }, { "input": "1\n1", "output": "1" }, { "input": "13\n67\n75\n76\n80\n69\n86\n75\n86\n81\n84\n73\n72\n76", "output": "232242896" }, { "input": "25\n35\n43\n38\n33\n47\n44\n40\n36\n41\n42\n33\n30\n49\n42\n62\n39\n40\n35\n43\n31\n42\n46\n42\n34\n33", "output": "362689152" }, { "input": "47\n20\n21\n16\n18\n24\n20\n25\n13\n20\n22\n26\n24\n17\n18\n21\n22\n21\n23\n17\n15\n24\n19\n18\n21\n20\n19\n26\n25\n20\n17\n17\n17\n26\n32\n20\n21\n25\n28\n24\n21\n21\n17\n28\n20\n20\n31\n19", "output": "295545118" }, { "input": "3\n343\n317\n337", "output": "691446102" }, { "input": "1\n5", "output": "1" } ]

1,533,046,157

2,147,483,647

PyPy 3

OK

TESTS

27

171

0

import math MOD = 1000000007 MAXN = 1005 def fac(n, k): res = 1 for i in range(n + 1 - k, n + 1): res *= i return res k = int(input()) sum = 0 ans = 1 for i in range(0, k): num = int(input()) sum += num ans = ans * fac(sum - 1, num - 1) // fac(num - 1, num - 1) print (ans % MOD)

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen. Input Specification: The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors. Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000). The total number of balls doesn't exceed 1000. Output Specification: A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007. Demo Input: ['3\n2\n2\n1\n', '4\n1\n2\n3\n4\n'] Demo Output: ['3\n', '1680\n'] Note: In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:

```python import math MOD = 1000000007 MAXN = 1005 def fac(n, k): res = 1 for i in range(n + 1 - k, n + 1): res *= i return res k = int(input()) sum = 0 ans = 1 for i in range(0, k): num = int(input()) sum += num ans = ans * fac(sum - 1, num - 1) // fac(num - 1, num - 1) print (ans % MOD) ```

3

145

A

Lucky Conversion

PROGRAMMING

1,200

[ "greedy", "implementation" ]

null

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types: - replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*. Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task.

The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.

Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*.

[ "47\n74\n", "774\n744\n", "777\n444\n" ]

[ "1\n", "1\n", "3\n" ]

In the first sample it is enough simply to swap the first and the second digit. In the second sample we should replace the second digit with its opposite. In the third number we should replace all three digits with their opposites.

500

[ { "input": "47\n74", "output": "1" }, { "input": "774\n744", "output": "1" }, { "input": "777\n444", "output": "3" }, { "input": "74747474\n77777777", "output": "4" }, { "input": "444444444444\n777777777777", "output": "12" }, { "input": "4744744447774474447474774\n4477774777444444444777447", "output": "8" }, { "input": "7\n4", "output": "1" }, { "input": "4\n7", "output": "1" }, { "input": "7777777777\n7777777774", "output": "1" }, { "input": "47777777777\n77777777774", "output": "1" }, { "input": "47747477747744447774774444444777444747474747777774\n44777444774477447777444774477777477774444477447777", "output": "14" }, { "input": "44447777447744444777777747477444777444447744444\n47444747774774744474747744447744477747777777447", "output": "13" }, { "input": "4447744774744774744747744774474474444447477477444747477444\n7477477444744774744744774774744474744447744774744477744477", "output": "14" }, { "input": "44747744777777444\n47774747747744777", "output": "6" }, { "input": "44447774444474477747774774477777474774744744477444447777477477744747477774744444744777777777747777477447744774744444747477744744\n77777474477477747774777777474474477444474777477747747777477747747744474474747774747747444777474444744744444477477777747744747477", "output": "37" }, { "input": "774774747744474477447477777447477747477474777477744744747444774474477477747474477447774444774744777\n744477444747477447477777774477447444447747477747477747774477474447474477477474444777444444447474747", "output": "27" }, { "input": "4747447477\n4747444744", "output": "3" }, { "input": "47744447444\n74477447744", "output": "4" }, { "input": "447444777744\n777747744477", "output": "6" }, { "input": "474777477774444\n774747777774477", "output": "4" }, { "input": "47744474447747744777777447\n44744747477474777744777477", "output": "7" }, { "input": "77447447444777777744744747744747774747477774777774447447777474477477774774777\n74777777444744447447474474477747747444444447447774444444747777444747474777447", "output": "28" }, { "input": "7\n7", "output": "0" }, { "input": "444\n444", "output": "0" }, { "input": "77747\n47474", "output": "3" } ]

1,617,813,459

2,147,483,647

Python 3

OK

TESTS

51

248

1,945,600

a = list(input()) b = list(input()) a_copy = a[:] a_copy.sort() b_copy = b[:] b_copy.sort() replace_counter = 0 swap_counter = 0 for i in range(len(a_copy)): if(a_copy[i] != b_copy[i]): replace_counter += 1 for i in range(len(a)): if (a[i] != b[i]): swap_counter += 1 swap_counter = (swap_counter - replace_counter) // 2 print(replace_counter + swap_counter)

Title: Lucky Conversion Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types: - replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*. Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task. Input Specification: The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105. Output Specification: Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*. Demo Input: ['47\n74\n', '774\n744\n', '777\n444\n'] Demo Output: ['1\n', '1\n', '3\n'] Note: In the first sample it is enough simply to swap the first and the second digit. In the second sample we should replace the second digit with its opposite. In the third number we should replace all three digits with their opposites.

```python a = list(input()) b = list(input()) a_copy = a[:] a_copy.sort() b_copy = b[:] b_copy.sort() replace_counter = 0 swap_counter = 0 for i in range(len(a_copy)): if(a_copy[i] != b_copy[i]): replace_counter += 1 for i in range(len(a)): if (a[i] != b[i]): swap_counter += 1 swap_counter = (swap_counter - replace_counter) // 2 print(replace_counter + swap_counter) ```

3

855

A

Tom Riddle's Diary

PROGRAMMING

800

[ "brute force", "implementation", "strings" ]

null

Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence. He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not. Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes).

First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list. Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100.

Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not. You can print each letter in any case (upper or lower).

[ "6\ntom\nlucius\nginny\nharry\nginny\nharry\n", "3\na\na\na\n" ]

[ "NO\nNO\nNO\nNO\nYES\nYES\n", "NO\nYES\nYES\n" ]

In test case 1, for *i* = 5 there exists *j* = 3 such that *s**i* = *s**j* and *j* < *i*, which means that answer for *i* = 5 is "YES".

500

[ { "input": "6\ntom\nlucius\nginny\nharry\nginny\nharry", "output": "NO\nNO\nNO\nNO\nYES\nYES" }, { "input": "3\na\na\na", "output": "NO\nYES\nYES" }, { "input": "1\nzn", "output": "NO" }, { "input": "9\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nhrtm\nssjqvixduertmotgagizamvfucfwtxqnhuowbqbzctgznivehelpcyigwrbbdsxnewfqvcf\nhyrtxvozpbveexfkgalmguozzakitjiwsduqxonb\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nbdbivqzvhggth", "output": "NO\nYES\nYES\nNO\nNO\nNO\nNO\nYES\nNO" }, { "input": "10\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nmvutw\nqooeqoxzxwetlpecqiwgdbogiqqulttysyohwhzxzphvsfmnplizxoebzcvvfyppqbhxjksuzepuezqqzxlfmdanoeaoqmor\nmvutw\nvchawxjoreboqzuklifv\nvchawxjoreboqzuklifv\nnivijte\nrflybruq\nvchawxjoreboqzuklifv", "output": "NO\nYES\nNO\nNO\nYES\nNO\nYES\nNO\nNO\nYES" }, { "input": "1\nz", "output": "NO" }, { "input": "9\nl\ny\nm\nj\nn\nr\nj\nk\nf", "output": "NO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO" }, { "input": "14\nw\na\nh\np\nk\nw\ny\nv\ns\nf\nx\nd\nk\nr", "output": "NO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO" }, { "input": "25\np\nk\nu\nl\nf\nt\nc\ns\nq\nd\nb\nq\no\ni\ni\nd\ni\nw\nn\ng\nw\nt\na\ne\ni", "output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nYES\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nYES" }, { "input": "20\nd\nh\ng\no\np\ne\nt\nj\nv\ni\nt\nh\ns\ni\nw\nf\nx\na\nl\ni", "output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES" }, { "input": "3\nbbbbbbb\nbbbbbbbbb\nbbbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "NO\nNO\nNO" }, { "input": "2\nab\nba", "output": "NO\nNO" }, { "input": "6\ntom\nlucius\nginnys\nharpy\nginny\nharry", "output": "NO\nNO\nNO\nNO\nNO\nNO" }, { "input": "2\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde", "output": "NO\nYES" }, { "input": "42\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na", "output": "NO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES" } ]

1,646,909,372

2,147,483,647

Python 3

OK

TESTS

55

46

0

memory=dict() for _ in [0]*int(input()): s=input() try:memory[s]+=1;print('YES') except:memory[s]=0;print('NO')

Title: Tom Riddle's Diary Time Limit: None seconds Memory Limit: None megabytes Problem Description: Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence. He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not. Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes). Input Specification: First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list. Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100. Output Specification: Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not. You can print each letter in any case (upper or lower). Demo Input: ['6\ntom\nlucius\nginny\nharry\nginny\nharry\n', '3\na\na\na\n'] Demo Output: ['NO\nNO\nNO\nNO\nYES\nYES\n', 'NO\nYES\nYES\n'] Note: In test case 1, for *i* = 5 there exists *j* = 3 such that *s**i* = *s**j* and *j* < *i*, which means that answer for *i* = 5 is "YES".

```python memory=dict() for _ in [0]*int(input()): s=input() try:memory[s]+=1;print('YES') except:memory[s]=0;print('NO') ```

3

104

A

Blackjack

PROGRAMMING

800

[ "implementation" ]

A. Blackjack

2

256

One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one! Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture. In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses. The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.

The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.

Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.

[ "12\n", "20\n", "10\n" ]

[ "4", "15", "0" ]

In the first sample only four two's of different suits can earn the required sum of points. In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use. In the third sample there is no card, that would add a zero to the current ten points.

500

[ { "input": "12", "output": "4" }, { "input": "20", "output": "15" }, { "input": "10", "output": "0" }, { "input": "11", "output": "4" }, { "input": "15", "output": "4" }, { "input": "18", "output": "4" }, { "input": "25", "output": "0" }, { "input": "22", "output": "0" }, { "input": "1", "output": "0" }, { "input": "2", "output": "0" }, { "input": "3", "output": "0" }, { "input": "4", "output": "0" }, { "input": "5", "output": "0" }, { "input": "6", "output": "0" }, { "input": "7", "output": "0" }, { "input": "8", "output": "0" }, { "input": "9", "output": "0" }, { "input": "13", "output": "4" }, { "input": "14", "output": "4" }, { "input": "16", "output": "4" }, { "input": "17", "output": "4" }, { "input": "19", "output": "4" }, { "input": "21", "output": "4" }, { "input": "23", "output": "0" }, { "input": "24", "output": "0" } ]

1,545,760,108

2,147,483,647

Python 3

OK

TESTS

25

218

0

c = int(input()) c = c-10 if (c >= 1 and c < 10) or (c is 11): print(4) elif c is 10: print(15) else: print(0)

Title: Blackjack Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one! Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture. In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses. The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. Input Specification: The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. Output Specification: Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. Demo Input: ['12\n', '20\n', '10\n'] Demo Output: ['4', '15', '0'] Note: In the first sample only four two's of different suits can earn the required sum of points. In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use. In the third sample there is no card, that would add a zero to the current ten points.

```python c = int(input()) c = c-10 if (c >= 1 and c < 10) or (c is 11): print(4) elif c is 10: print(15) else: print(0) ```

3.9455

930

D

Game with Tokens

PROGRAMMING

2,500

[ "data structures", "games", "implementation" ]

null

Consider the following game for two players. There is one white token and some number of black tokens. Each token is placed on a plane in a point with integer coordinates *x* and *y*. The players take turn making moves, white starts. On each turn, a player moves all tokens of their color by 1 to up, down, left or right. Black player can choose directions for each token independently. After a turn of the white player the white token can not be in a point where a black token is located. There are no other constraints on locations of the tokens: positions of black tokens can coincide, after a turn of the black player and initially the white token can be in the same point with some black point. If at some moment the white player can't make a move, he loses. If the white player makes 10100500 moves, he wins. You are to solve the following problem. You are given initial positions of all black tokens. It is guaranteed that initially all these positions are distinct. In how many places can the white token be located initially so that if both players play optimally, the black player wins?

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of black points. The (*i*<=+<=1)-th line contains two integers *x**i*, *y**i* (<=-<=105<=≤<=*x**i*,<=*y**i*,<=<=≤<=105) — the coordinates of the point where the *i*-th black token is initially located. It is guaranteed that initial positions of black tokens are distinct.

Print the number of points where the white token can be located initially, such that if both players play optimally, the black player wins.

[ "4\n-2 -1\n0 1\n0 -3\n2 -1\n", "4\n-2 0\n-1 1\n0 -2\n1 -1\n", "16\n2 1\n1 2\n-1 1\n0 1\n0 0\n1 1\n2 -1\n2 0\n1 0\n-1 -1\n1 -1\n2 2\n0 -1\n-1 0\n0 2\n-1 2\n" ]

[ "4\n", "2\n", "4\n" ]

In the first and second examples initial positions of black tokens are shown with black points, possible positions of the white token (such that the black player wins) are shown with white points. The first example: <img class="tex-graphics" src="https://espresso.codeforces.com/5054b8d2df2fac92c92f96fae82d21c365d12983.png" style="max-width: 100.0%;max-height: 100.0%;"/> The second example: <img class="tex-graphics" src="https://espresso.codeforces.com/eb795dd6abb95cfafb1d1cb7d8c8798825dcc180.png" style="max-width: 100.0%;max-height: 100.0%;"/> In the third example the white tokens should be located in the inner square 2 × 2, to make the black player win. <img class="tex-graphics" src="https://espresso.codeforces.com/6dfd863f649b92860dfd6b446ea004abc01b71a6.png" style="max-width: 100.0%;max-height: 100.0%;"/>

2,000

[ { "input": "4\n-2 -1\n0 1\n0 -3\n2 -1", "output": "4" }, { "input": "4\n-2 0\n-1 1\n0 -2\n1 -1", "output": "2" }, { "input": "16\n2 1\n1 2\n-1 1\n0 1\n0 0\n1 1\n2 -1\n2 0\n1 0\n-1 -1\n1 -1\n2 2\n0 -1\n-1 0\n0 2\n-1 2", "output": "4" }, { "input": "1\n1 2", "output": "0" }, { "input": "4\n0 99999\n-99999 0\n99999 0\n0 -99999", "output": "9999800001" }, { "input": "10\n-1 3\n-1 7\n-8 -4\n5 14\n-6 -7\n11 -8\n-11 0\n5 -1\n9 4\n-2 -14", "output": "110" }, { "input": "50\n-15 -80\n-80 21\n90 38\n-100 27\n-64 -75\n-10 59\n38 44\n-31 -91\n97 76\n87 43\n5 43\n-73 74\n-45 42\n31 -100\n-87 19\n-21 -13\n-71 38\n-54 -39\n-89 -32\n-18 99\n-44 -78\n9 76\n-69 -40\n-29 23\n-88 42\n-95 86\n45 15\n-39 100\n17 -33\n5 -48\n-4 -22\n-19 54\n-13 -64\n-86 68\n-52 -95\n-73 -29\n-24 -93\n-60 96\n41 57\n55 43\n-64 15\n-43 9\n29 88\n44 -2\n67 -94\n-20 -81\n-75 -74\n-80 -44\n-49 -7\n39 -59", "output": "17145" }, { "input": "2\n-3 0\n3 2", "output": "0" }, { "input": "3\n-5 3\n4 -5\n-3 2", "output": "0" }, { "input": "4\n-5 1\n0 -3\n-1 4\n5 -5", "output": "0" }, { "input": "4\n-1 4\n-3 2\n-2 1\n-5 3", "output": "0" }, { "input": "5\n-3 -5\n5 2\n-4 1\n-2 0\n1 2", "output": "0" }, { "input": "5\n2 3\n1 -1\n0 2\n0 5\n3 2", "output": "0" }, { "input": "6\n-1 2\n5 -4\n0 4\n3 0\n-4 -1\n-3 -2", "output": "3" }, { "input": "6\n-3 4\n-1 -3\n1 -4\n1 -1\n-5 -1\n1 4", "output": "0" }, { "input": "7\n0 4\n0 3\n-1 3\n4 3\n1 3\n-4 4\n5 4", "output": "0" }, { "input": "7\n4 4\n2 3\n5 -2\n-1 1\n2 2\n-2 -2\n-1 2", "output": "0" }, { "input": "8\n2 -4\n-4 -2\n-3 3\n-3 -1\n4 -4\n2 3\n4 -5\n0 0", "output": "4" }, { "input": "8\n4 -4\n5 -5\n3 2\n-2 5\n-4 -2\n2 5\n-5 5\n5 4", "output": "2" }, { "input": "9\n4 -5\n-4 -3\n4 -4\n1 0\n5 -1\n-3 1\n5 -4\n2 -4\n4 -3", "output": "4" }, { "input": "9\n-2 0\n-4 -4\n0 4\n2 2\n-3 -2\n1 3\n-5 5\n-3 -3\n-1 1", "output": "5" }, { "input": "10\n-1 -2\n-5 2\n-5 0\n-1 0\n4 0\n4 5\n0 -3\n-3 -3\n-5 5\n3 -4", "output": "2" }, { "input": "10\n2 -3\n-3 3\n-1 -1\n3 -5\n5 -3\n0 5\n-4 -4\n2 -4\n-2 -5\n-2 4", "output": "11" }, { "input": "10\n2 -4\n0 -3\n3 2\n-1 3\n1 -1\n4 -5\n-4 2\n1 0\n-2 -5\n-2 2", "output": "3" }, { "input": "10\n-4 -2\n-1 0\n1 -3\n2 5\n3 1\n3 -3\n2 4\n-2 -1\n-3 3\n5 2", "output": "7" }, { "input": "10\n2 0\n1 2\n4 0\n3 -1\n4 3\n-5 4\n-4 -1\n1 -1\n2 -1\n-5 -4", "output": "3" }, { "input": "10\n3 4\n-2 -3\n-2 5\n-2 1\n5 4\n2 -1\n5 -4\n0 1\n4 4\n3 -1", "output": "6" }, { "input": "10\n-1 3\n2 3\n3 2\n-4 -3\n-2 -5\n5 -5\n-4 -4\n0 1\n4 -1\n3 3", "output": "11" }, { "input": "10\n1 -3\n0 4\n-1 3\n-2 3\n4 1\n-1 5\n5 4\n-5 5\n-4 -2\n-5 1", "output": "5" }, { "input": "10\n5 -1\n-2 5\n-5 -1\n-3 -3\n-5 -4\n-3 -2\n-1 -4\n2 5\n4 -5\n1 -4", "output": "5" }, { "input": "10\n-1 0\n5 -1\n-4 1\n-3 0\n-5 -1\n-3 -4\n3 3\n-2 2\n-3 -2\n3 -1", "output": "2" }, { "input": "20\n-16 24\n9 13\n-1 -3\n5 7\n-20 17\n21 5\n-10 8\n0 -14\n17 -5\n7 1\n-6 16\n-18 -9\n-7 -8\n-13 -23\n4 4\n10 -3\n2 -5\n-18 24\n19 -19\n12 -25", "output": "268" }, { "input": "20\n-4 23\n-10 3\n20 25\n24 -23\n1 18\n-23 -24\n-20 -6\n7 22\n11 -18\n-25 -19\n7 -6\n-9 22\n-24 -2\n-9 -17\n-1 12\n-20 -21\n-19 -24\n10 -20\n20 8\n25 -14", "output": "312" }, { "input": "20\n21 20\n23 -21\n-22 24\n-18 -2\n-6 -15\n-20 -10\n-15 21\n-18 5\n13 10\n-11 15\n-6 -1\n17 6\n-13 -23\n8 -9\n-24 21\n8 11\n21 9\n22 12\n-2 -21\n-12 -10", "output": "459" }, { "input": "20\n-5 -7\n-17 22\n13 -4\n19 8\n2 6\n-4 1\n7 -15\n-5 -15\n-14 -13\n14 8\n-13 -23\n8 4\n-13 18\n-17 3\n9 3\n7 -11\n6 -16\n-15 9\n-24 -17\n-20 -18", "output": "227" }, { "input": "20\n-9 5\n-25 -4\n14 -22\n-17 23\n-20 -8\n19 22\n23 -3\n-23 -11\n-2 -15\n22 -4\n-10 -16\n16 22\n9 9\n-18 16\n-25 6\n8 -10\n-2 -17\n-12 6\n20 -10\n17 -6", "output": "529" }, { "input": "20\n-13 15\n1 14\n-12 7\n-18 -15\n-19 -11\n-7 6\n7 -15\n4 18\n-4 10\n-23 16\n-8 -15\n-3 14\n-8 1\n17 19\n15 19\n-3 -12\n-25 16\n-7 -1\n-14 1\n18 3", "output": "295" }, { "input": "20\n-17 7\n0 -21\n15 -10\n-5 12\n18 -12\n-19 11\n24 -19\n-25 -1\n-5 -25\n20 -23\n-4 9\n7 -15\n8 -9\n23 -15\n-2 5\n10 -4\n12 -24\n25 2\n5 -6\n2 25", "output": "577" }, { "input": "20\n17 23\n-7 8\n3 9\n9 -22\n-9 -14\n-18 -10\n-4 2\n10 -3\n-9 19\n-7 9\n-22 4\n6 14\n-9 -18\n2 0\n-17 4\n6 20\n24 13\n22 4\n-14 -1\n-6 -14", "output": "454" }, { "input": "20\n-10 16\n24 18\n-1 -22\n1 4\n4 -19\n-22 8\n-20 -20\n25 24\n-4 8\n7 -11\n-17 -14\n25 -12\n24 23\n-18 15\n23 -1\n-11 -14\n-4 -6\n-14 18\n-10 18\n2 -17", "output": "487" }, { "input": "20\n11 1\n-15 23\n5 24\n7 -13\n-13 -13\n-25 20\n22 -16\n-23 -2\n11 -21\n12 1\n2 3\n-3 -17\n4 21\n-17 12\n13 -14\n1 4\n23 -22\n-18 9\n14 5\n-23 -3", "output": "502" } ]

1,632,748,661

2,147,483,647

PyPy 3

OK

TESTS

62

795

121,548,800

#FB HACKERCUP import os,sys,math from io import BytesIO, IOBase if sys.version_info[0] < 3: from _builtin_ import xrange as range from future_builtins import ascii, filter, hex, map, oct, zip from bisect import bisect_left as lower_bound, bisect_right as upper_bound def so(): return int(input()) def st(): return input() def mj(): return map(int,input().strip().split(" ")) def msj(): return map(str,input().strip().split(" ")) def le(): return list(map(int,input().split())) def lebe():return list(map(int, input())) # def dmain(): # sys.setrecursionlimit(1000000) # threading.stack_size(1024000) # thread = threading.Thread(target=main) # thread.start() def joro(L): return(''.join(map(str, L))) def cheems(c,d): a=1 while(d!=0): if(d%2==1): a*=c c=c*c d=d//2 return a def decimalToBinary(n): return bin(n).replace("0b","") def isprime(n): for i in range(2,int(n**0.5)+1): if n%i==0: return False return True # def npr(n, r): # return factorial(n) // factorial(n - r) if n >= r else 0 # def ncr(n, r): # return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0 def lower_bound(li, num): answer = -1 start = 0 end = len(li) - 1 while (start <= end): middle = (end + start) // 2 if li[middle] >= num: answer = middle end = middle - 1 else: start = middle + 1 return answer # min index where x is not less than num def upper_bound(li, num): answer = -1 start = 0 end = len(li) - 1 while (start <= end): middle = (end + start) // 2 if li[middle] <= num: answer = middle start = middle + 1 else: end = middle - 1 return answer # max index where x is not greater than num def tir(a,b,c): if(0==c): return 1 if(len(a)<=b): return 0 if(c!=-1): return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c)) else: return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1)) def abs(x): return x if x >= 0 else -x doi=int(1e9+7) boi=int(1e9-7) koi=int(5+2e6) moi=int(4e6+5+100) def binary_search(li, val, lb, ub): # print(lb, ub, li) ans = -1 while (lb <= ub): mid = (lb + ub) // 2 # print('mid is',mid, li[mid]) if li[mid] > val: ub = mid - 1 elif val > li[mid]: lb = mid + 1 else: ans = mid # return index break return ans def kadane(x): # maximum sum contiguous subarray sum_so_far = 0 current_sum = 0 for i in x: current_sum += i if current_sum < 0: current_sum = 0 else: sum_so_far = max(sum_so_far, current_sum) return sum_so_far def wubu(m): import math as my d=0 while(not m%2): m=m//2 d=1+d for i in range(3,int(my.sqrt(m))+1,2): while(not m%i): m=m//i d=1+d return int(m>1)+d def pref(li): pref_sum = [0] for i in li: pref_sum.append(pref_sum[-1] + i) return pref_sum def SieveOfEratosthenes(n): prime = [True for i in range(n + 1)] p = 2 li = [] while (p * p <= n): if (prime[p] == True): for i in range(p * p, n + 1, p): prime[i] = False p += 1 for p in range(2, len(prime)): if prime[p]: li.append(p) return li def primefactors(n): import math as my factors = [] while (n % 2 == 0): factors.append(2) n //= 2 for i in range(3, int(my.sqrt(n)) + 1, 2): # only odd factors left while n % i == 0: factors.append(i) n //= i if n > 2: # incase of prime factors.append(n) return factors def read(): sys.stdin = open('input.txt', 'r') sys.stdout = open('output.txt', 'w') def tr(n): return n*(n+1)//2 def bro(q,r,c): bec,pot,h=0,0,0 ds=[] bs=[] es=[] for i in range(c): while(h<len(r) and r[h]<=q[i][0]): pot=0 ds.clear() bs.clear() h=1+h for i in range(c): ds.append(q[i][1]) es.append(q[i][1]) ds=list(set(ds)) for i in ds: bs.append(es.count(i)) print(bs) for i in range(c): if(q[i][0]>r[h-1]): bec=bec+pot-bs[i] pot=1+pot return bec def iu(): import sys input = sys.stdin.readline import math as my import bisect as by m=so() bec=0 A=[0]*moi B=[0]*moi C=[0]*moi D=[0]*moi E=[0]*moi F=[0]*moi for i in range(koi*2+1): A[i]=sys.maxsize B[i]=-sys.maxsize for i in range(1,m+1): p,q=mj() r=p+q+koi s=p-q A[r]=min(A[r],s) B[r]=max(B[r],s) for i in range(2): C[i]=A[i] D[i]=B[i] for i in range(2,1+2*koi): C[i]=min(C[i-2],A[i]) D[i]=max(B[i],D[i-2]) for i in range(2*koi-2,2*koi+1): E[i]=A[i] F[i]=B[i] for i in range(2*koi-2,-1,-1): E[i]=min(E[2+i],A[i]) F[i]=max(B[i],F[2+i]) for i in range(1,2*koi+1,1): se=min(D[i-1],F[1+i])-max(C[i-1],E[1+i]) bec=max(0,se//2)+bec print(bec) def main(): for i in range(1): iu() # region fastio # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") def print(*args, **kwargs): """Prints the values to a stream, or to sys.stdout by default.""" sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout) at_start = True for x in args: if not at_start: file.write(sep) file.write(str(x)) at_start = False file.write(kwargs.pop("end", "\n")) if kwargs.pop("flush", False): file.flush() if sys.version_info[0] < 3: sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout) else: sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # endregion if __name__ == "__main__": #read() main() #dmain() # Comment Read()

Title: Game with Tokens Time Limit: None seconds Memory Limit: None megabytes Problem Description: Consider the following game for two players. There is one white token and some number of black tokens. Each token is placed on a plane in a point with integer coordinates *x* and *y*. The players take turn making moves, white starts. On each turn, a player moves all tokens of their color by 1 to up, down, left or right. Black player can choose directions for each token independently. After a turn of the white player the white token can not be in a point where a black token is located. There are no other constraints on locations of the tokens: positions of black tokens can coincide, after a turn of the black player and initially the white token can be in the same point with some black point. If at some moment the white player can't make a move, he loses. If the white player makes 10100500 moves, he wins. You are to solve the following problem. You are given initial positions of all black tokens. It is guaranteed that initially all these positions are distinct. In how many places can the white token be located initially so that if both players play optimally, the black player wins? Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of black points. The (*i*<=+<=1)-th line contains two integers *x**i*, *y**i* (<=-<=105<=≤<=*x**i*,<=*y**i*,<=<=≤<=105) — the coordinates of the point where the *i*-th black token is initially located. It is guaranteed that initial positions of black tokens are distinct. Output Specification: Print the number of points where the white token can be located initially, such that if both players play optimally, the black player wins. Demo Input: ['4\n-2 -1\n0 1\n0 -3\n2 -1\n', '4\n-2 0\n-1 1\n0 -2\n1 -1\n', '16\n2 1\n1 2\n-1 1\n0 1\n0 0\n1 1\n2 -1\n2 0\n1 0\n-1 -1\n1 -1\n2 2\n0 -1\n-1 0\n0 2\n-1 2\n'] Demo Output: ['4\n', '2\n', '4\n'] Note: In the first and second examples initial positions of black tokens are shown with black points, possible positions of the white token (such that the black player wins) are shown with white points. The first example: <img class="tex-graphics" src="https://espresso.codeforces.com/5054b8d2df2fac92c92f96fae82d21c365d12983.png" style="max-width: 100.0%;max-height: 100.0%;"/> The second example: <img class="tex-graphics" src="https://espresso.codeforces.com/eb795dd6abb95cfafb1d1cb7d8c8798825dcc180.png" style="max-width: 100.0%;max-height: 100.0%;"/> In the third example the white tokens should be located in the inner square 2 × 2, to make the black player win. <img class="tex-graphics" src="https://espresso.codeforces.com/6dfd863f649b92860dfd6b446ea004abc01b71a6.png" style="max-width: 100.0%;max-height: 100.0%;"/>

```python #FB HACKERCUP import os,sys,math from io import BytesIO, IOBase if sys.version_info[0] < 3: from _builtin_ import xrange as range from future_builtins import ascii, filter, hex, map, oct, zip from bisect import bisect_left as lower_bound, bisect_right as upper_bound def so(): return int(input()) def st(): return input() def mj(): return map(int,input().strip().split(" ")) def msj(): return map(str,input().strip().split(" ")) def le(): return list(map(int,input().split())) def lebe():return list(map(int, input())) # def dmain(): # sys.setrecursionlimit(1000000) # threading.stack_size(1024000) # thread = threading.Thread(target=main) # thread.start() def joro(L): return(''.join(map(str, L))) def cheems(c,d): a=1 while(d!=0): if(d%2==1): a*=c c=c*c d=d//2 return a def decimalToBinary(n): return bin(n).replace("0b","") def isprime(n): for i in range(2,int(n**0.5)+1): if n%i==0: return False return True # def npr(n, r): # return factorial(n) // factorial(n - r) if n >= r else 0 # def ncr(n, r): # return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0 def lower_bound(li, num): answer = -1 start = 0 end = len(li) - 1 while (start <= end): middle = (end + start) // 2 if li[middle] >= num: answer = middle end = middle - 1 else: start = middle + 1 return answer # min index where x is not less than num def upper_bound(li, num): answer = -1 start = 0 end = len(li) - 1 while (start <= end): middle = (end + start) // 2 if li[middle] <= num: answer = middle start = middle + 1 else: end = middle - 1 return answer # max index where x is not greater than num def tir(a,b,c): if(0==c): return 1 if(len(a)<=b): return 0 if(c!=-1): return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c)) else: return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1)) def abs(x): return x if x >= 0 else -x doi=int(1e9+7) boi=int(1e9-7) koi=int(5+2e6) moi=int(4e6+5+100) def binary_search(li, val, lb, ub): # print(lb, ub, li) ans = -1 while (lb <= ub): mid = (lb + ub) // 2 # print('mid is',mid, li[mid]) if li[mid] > val: ub = mid - 1 elif val > li[mid]: lb = mid + 1 else: ans = mid # return index break return ans def kadane(x): # maximum sum contiguous subarray sum_so_far = 0 current_sum = 0 for i in x: current_sum += i if current_sum < 0: current_sum = 0 else: sum_so_far = max(sum_so_far, current_sum) return sum_so_far def wubu(m): import math as my d=0 while(not m%2): m=m//2 d=1+d for i in range(3,int(my.sqrt(m))+1,2): while(not m%i): m=m//i d=1+d return int(m>1)+d def pref(li): pref_sum = [0] for i in li: pref_sum.append(pref_sum[-1] + i) return pref_sum def SieveOfEratosthenes(n): prime = [True for i in range(n + 1)] p = 2 li = [] while (p * p <= n): if (prime[p] == True): for i in range(p * p, n + 1, p): prime[i] = False p += 1 for p in range(2, len(prime)): if prime[p]: li.append(p) return li def primefactors(n): import math as my factors = [] while (n % 2 == 0): factors.append(2) n //= 2 for i in range(3, int(my.sqrt(n)) + 1, 2): # only odd factors left while n % i == 0: factors.append(i) n //= i if n > 2: # incase of prime factors.append(n) return factors def read(): sys.stdin = open('input.txt', 'r') sys.stdout = open('output.txt', 'w') def tr(n): return n*(n+1)//2 def bro(q,r,c): bec,pot,h=0,0,0 ds=[] bs=[] es=[] for i in range(c): while(h<len(r) and r[h]<=q[i][0]): pot=0 ds.clear() bs.clear() h=1+h for i in range(c): ds.append(q[i][1]) es.append(q[i][1]) ds=list(set(ds)) for i in ds: bs.append(es.count(i)) print(bs) for i in range(c): if(q[i][0]>r[h-1]): bec=bec+pot-bs[i] pot=1+pot return bec def iu(): import sys input = sys.stdin.readline import math as my import bisect as by m=so() bec=0 A=[0]*moi B=[0]*moi C=[0]*moi D=[0]*moi E=[0]*moi F=[0]*moi for i in range(koi*2+1): A[i]=sys.maxsize B[i]=-sys.maxsize for i in range(1,m+1): p,q=mj() r=p+q+koi s=p-q A[r]=min(A[r],s) B[r]=max(B[r],s) for i in range(2): C[i]=A[i] D[i]=B[i] for i in range(2,1+2*koi): C[i]=min(C[i-2],A[i]) D[i]=max(B[i],D[i-2]) for i in range(2*koi-2,2*koi+1): E[i]=A[i] F[i]=B[i] for i in range(2*koi-2,-1,-1): E[i]=min(E[2+i],A[i]) F[i]=max(B[i],F[2+i]) for i in range(1,2*koi+1,1): se=min(D[i-1],F[1+i])-max(C[i-1],E[1+i]) bec=max(0,se//2)+bec print(bec) def main(): for i in range(1): iu() # region fastio # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") def print(*args, **kwargs): """Prints the values to a stream, or to sys.stdout by default.""" sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout) at_start = True for x in args: if not at_start: file.write(sep) file.write(str(x)) at_start = False file.write(kwargs.pop("end", "\n")) if kwargs.pop("flush", False): file.flush() if sys.version_info[0] < 3: sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout) else: sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # endregion if __name__ == "__main__": #read() main() #dmain() # Comment Read() ```

3

991

A

If at first you don't succeed...

PROGRAMMING

1,000

[ "implementation" ]

null

Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam. Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group. Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?

The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).

If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam. If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.

[ "10 10 5 20\n", "2 2 0 4\n", "2 2 2 1\n" ]

[ "5", "-1", "-1" ]

The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam. In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible. The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible.

500

[ { "input": "10 10 5 20", "output": "5" }, { "input": "2 2 0 4", "output": "-1" }, { "input": "2 2 2 1", "output": "-1" }, { "input": "98 98 97 100", "output": "1" }, { "input": "1 5 2 10", "output": "-1" }, { "input": "5 1 2 10", "output": "-1" }, { "input": "6 7 5 8", "output": "-1" }, { "input": "6 7 5 9", "output": "1" }, { "input": "6 7 5 7", "output": "-1" }, { "input": "50 50 1 100", "output": "1" }, { "input": "8 3 2 12", "output": "3" }, { "input": "10 19 6 25", "output": "2" }, { "input": "1 0 0 99", "output": "98" }, { "input": "0 1 0 98", "output": "97" }, { "input": "1 1 0 97", "output": "95" }, { "input": "1 1 1 96", "output": "95" }, { "input": "0 0 0 0", "output": "-1" }, { "input": "100 0 0 0", "output": "-1" }, { "input": "0 100 0 0", "output": "-1" }, { "input": "100 100 0 0", "output": "-1" }, { "input": "0 0 100 0", "output": "-1" }, { "input": "100 0 100 0", "output": "-1" }, { "input": "0 100 100 0", "output": "-1" }, { "input": "100 100 100 0", "output": "-1" }, { "input": "0 0 0 100", "output": "100" }, { "input": "100 0 0 100", "output": "-1" }, { "input": "0 100 0 100", "output": "-1" }, { "input": "100 100 0 100", "output": "-1" }, { "input": "0 0 100 100", "output": "-1" }, { "input": "100 0 100 100", "output": "-1" }, { "input": "0 100 100 100", "output": "-1" }, { "input": "100 100 100 100", "output": "-1" }, { "input": "10 45 7 52", "output": "4" }, { "input": "38 1 1 68", "output": "30" }, { "input": "8 45 2 67", "output": "16" }, { "input": "36 36 18 65", "output": "11" }, { "input": "10 30 8 59", "output": "27" }, { "input": "38 20 12 49", "output": "3" }, { "input": "8 19 4 38", "output": "15" }, { "input": "36 21 17 72", "output": "32" }, { "input": "14 12 12 89", "output": "75" }, { "input": "38 6 1 44", "output": "1" }, { "input": "13 4 6 82", "output": "-1" }, { "input": "5 3 17 56", "output": "-1" }, { "input": "38 5 29 90", "output": "-1" }, { "input": "22 36 18 55", "output": "15" }, { "input": "13 0 19 75", "output": "-1" }, { "input": "62 65 10 89", "output": "-1" }, { "input": "2 29 31 72", "output": "-1" }, { "input": "1 31 19 55", "output": "-1" }, { "input": "1 25 28 88", "output": "-1" }, { "input": "34 32 28 33", "output": "-1" }, { "input": "43 36 1 100", "output": "22" }, { "input": "16 39 55 70", "output": "-1" }, { "input": "2 3 0 91", "output": "86" }, { "input": "55 29 12 48", "output": "-1" }, { "input": "7 33 20 88", "output": "-1" }, { "input": "40 38 27 99", "output": "48" }, { "input": "18 28 14 84", "output": "52" }, { "input": "34 25 25 92", "output": "58" }, { "input": "4 24 5 76", "output": "-1" }, { "input": "5 22 16 96", "output": "-1" }, { "input": "1 1 0 4", "output": "2" }, { "input": "5 5 3 1", "output": "-1" }, { "input": "0 0 0 1", "output": "1" }, { "input": "2 3 0 8", "output": "3" }, { "input": "5 5 2 5", "output": "-1" }, { "input": "1 2 1 3", "output": "1" }, { "input": "3 0 0 4", "output": "1" }, { "input": "0 0 0 5", "output": "5" }, { "input": "5 5 0 3", "output": "-1" }, { "input": "5 6 1 7", "output": "-1" }, { "input": "10 10 10 11", "output": "1" }, { "input": "0 0 0 10", "output": "10" }, { "input": "5 15 5 30", "output": "15" }, { "input": "3 2 0 7", "output": "2" } ]

1,606,661,164

2,147,483,647

Python 3

OK

TESTS

76

124

0

a,b,c,d=list(map(int,input().split())) print(-1 if d-a-b+c<=0 or a<c or b<c else d-a-b+c)

Title: If at first you don't succeed... Time Limit: None seconds Memory Limit: None megabytes Problem Description: Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam. Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group. Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time? Input Specification: The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$). Output Specification: If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam. If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$. Demo Input: ['10 10 5 20\n', '2 2 0 4\n', '2 2 2 1\n'] Demo Output: ['5', '-1', '-1'] Note: The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam. In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible. The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible.

```python a,b,c,d=list(map(int,input().split())) print(-1 if d-a-b+c<=0 or a<c or b<c else d-a-b+c) ```

3

343

B

Alternating Current

PROGRAMMING

1,600

[ "data structures", "greedy", "implementation" ]

null

Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again. The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view): Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut. To understand the problem better please read the notes to the test samples.

The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.

Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.

[ "-++-\n", "+-\n", "++\n", "-\n" ]

[ "Yes\n", "No\n", "Yes\n", "No\n" ]

The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses. In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled: In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher: In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:

1,000

[ { "input": "-++-", "output": "Yes" }, { "input": "+-", "output": "No" }, { "input": "++", "output": "Yes" }, { "input": "-", "output": "No" }, { "input": "+-+-", "output": "No" }, { "input": "-+-", "output": "No" }, { "input": "-++-+--+", "output": "Yes" }, { "input": "+", "output": "No" }, { "input": "-+", "output": "No" }, { "input": "--", "output": "Yes" }, { "input": "+++", "output": "No" }, { "input": "--+", "output": "No" }, { "input": "++--++", "output": "Yes" }, { "input": "+-++-+", "output": "Yes" }, { "input": "+-+--+", "output": "No" }, { "input": "--++-+", "output": "No" }, { "input": "-+-+--", "output": "No" }, { "input": "+-+++-", "output": "No" }, { "input": "-+-+-+", "output": "No" }, { "input": "-++-+--++--+-++-", "output": "Yes" }, { "input": "+-----+-++---+------+++-++++", "output": "No" }, { "input": "-+-++--+++-++++---+--+----+--+-+-+++-+++-+---++-++++-+--+--+--+-+-++-+-+-++++++---++--+++++-+--++--+-+--++-----+--+-++---+++---++----+++-++++--++-++-", "output": "No" }, { "input": "-+-----++++--++-+-++", "output": "Yes" }, { "input": "+--+--+------+++++++-+-+++--++---+--+-+---+--+++-+++-------+++++-+-++++--+-+-+++++++----+----+++----+-+++-+++-----+++-+-++-+-+++++-+--++----+--+-++-----+-+-++++---+++---+-+-+-++++--+--+++---+++++-+---+-----+++-++--+++---++-++-+-+++-+-+-+---+++--+--++++-+-+--++-------+--+---++-----+++--+-+++--++-+-+++-++--+++-++++++++++-++-++++++-+++--+--++-+++--+++-++++----+++---+-+----++++-+-+", "output": "Yes" }, { "input": "-+-+-++-+-+-", "output": "Yes" }, { "input": "-+-++-+-", "output": "Yes" }, { "input": "-+-++-+-+-", "output": "No" }, { "input": "++-+-+-+-+--+", "output": "No" }, { "input": "+++---", "output": "No" }, { "input": "+-+-+-+-+--+-+-+-+-++--++--+", "output": "Yes" }, { "input": "+-+-++", "output": "No" }, { "input": "-++--+--+++-+-+-+-+-", "output": "No" }, { "input": "+---+-+-", "output": "No" }, { "input": "+-+--+-+", "output": "Yes" }, { "input": "+++---+++---", "output": "No" }, { "input": "-+++++", "output": "No" }, { "input": "-+-+-+-+-+-+-++-+-+-+-+-+-+-", "output": "Yes" }, { "input": "-+++--", "output": "No" }, { "input": "+---+", "output": "No" }, { "input": "-++", "output": "No" }, { "input": "-+--+-", "output": "Yes" }, { "input": "+---++--++", "output": "No" }, { "input": "+++-", "output": "No" }, { "input": "--+++", "output": "No" }, { "input": "++-+", "output": "No" } ]

1,602,432,838

2,147,483,647

PyPy 3

OK

TESTS

62

280

5,017,600

s = input() ss = [] for i in s: if len(ss) > 0 and ss[-1] == i: ss.pop() else: ss.append(i) if len(ss) == 0: print("Yes") else: print("No")

Title: Alternating Current Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again. The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view): Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut. To understand the problem better please read the notes to the test samples. Input Specification: The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise. Output Specification: Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled. Demo Input: ['-++-\n', '+-\n', '++\n', '-\n'] Demo Output: ['Yes\n', 'No\n', 'Yes\n', 'No\n'] Note: The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses. In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled: In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher: In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:

```python s = input() ss = [] for i in s: if len(ss) > 0 and ss[-1] == i: ss.pop() else: ss.append(i) if len(ss) == 0: print("Yes") else: print("No") ```

3

447

B

DZY Loves Strings

PROGRAMMING

1,000

[ "greedy", "implementation" ]

null

DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?

The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103). The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103). The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.

Print a single integer — the largest possible value of the resulting string DZY could get.

[ "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n" ]

[ "41\n" ]

In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

1,000

[ { "input": "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "41" }, { "input": "mmzhr\n3\n443 497 867 471 195 670 453 413 579 466 553 881 847 642 269 996 666 702 487 209 257 741 974 133 519 453", "output": "29978" }, { "input": "ajeeseerqnpaujubmajpibxrccazaawetywxmifzehojf\n23\n359 813 772 413 733 654 33 87 890 433 395 311 801 852 376 148 914 420 636 695 583 733 664 394 407 314", "output": "1762894" }, { "input": "uahngxejpomhbsebcxvelfsojbaouynnlsogjyvktpwwtcyddkcdqcqs\n34\n530 709 150 660 947 830 487 142 208 276 885 542 138 214 76 184 273 753 30 195 722 236 82 691 572 585", "output": "2960349" }, { "input": "xnzeqmouqyzvblcidmhbkqmtusszuczadpooslqxegldanwopilmdwzbczvrwgnwaireykwpugvpnpafbxlyggkgawghysufuegvmzvpgcqyjkoadcreaguzepbendwnowsuekxxivkziibxvxfoilofxcgnxvfefyezfhevfvtetsuhwtyxdlkccdkvqjl\n282\n170 117 627 886 751 147 414 187 150 960 410 70 576 681 641 729 798 877 611 108 772 643 683 166 305 933", "output": "99140444" }, { "input": "pplkqmluhfympkjfjnfdkwrkpumgdmbkfbbldpepicbbmdgafttpopzdxsevlqbtywzkoxyviglbbxsohycbdqksrhlumsldiwzjmednbkcjishkiekfrchzuztkcxnvuykhuenqojrmzaxlaoxnljnvqgnabtmcftisaazzgbmubmpsorygyusmeonrhrgphnfhlaxrvyhuxsnnezjxmdoklpquzpvjbxgbywppmegzxknhfzyygrmejleesoqfwheulmqhonqaukyuejtwxskjldplripyihbfpookxkuehiwqthbfafyrgmykuxglpplozycgydyecqkgfjljfqvigqhuxssqqtfanwszduwbsoytnrtgc\n464\n838 95 473 955 690 84 436 19 179 437 674 626 377 365 781 4 733 776 462 203 119 256 381 668 855 686", "output": "301124161" }, { "input": "qkautnuilwlhjsldfcuwhiqtgtoihifszlyvfaygrnivzgvwthkrzzdtfjcirrjjlrmjtbjlzmjeqmuffsjorjyggzefwgvmblvotvzffnwjhqxorpowzdcnfksdibezdtfjjxfozaghieksbmowrbeehuxlesmvqjsphlvauxiijm\n98\n121 622 0 691 616 959 838 161 581 862 876 830 267 812 598 106 337 73 588 323 999 17 522 399 657 495", "output": "30125295" }, { "input": "tghyxqfmhz\n8\n191 893 426 203 780 326 148 259 182 140 847 636 778 97 167 773 219 891 758 993 695 603 223 779 368 165", "output": "136422" }, { "input": "nyawbfjxnxjiyhwkydaruozobpphgjqdpfdqzezcsoyvurnapu\n30\n65 682 543 533 990 148 815 821 315 916 632 771 332 513 472 864 12 73 548 687 660 572 507 192 226 348", "output": "2578628" }, { "input": "pylrnkrbcjgoytvdnhmlvnkknijkdgdhworlvtwuonrkhrilkewcnofodaumgvnsisxooswgrgtvdeauyxhkipfoxrrtysuepjcf\n60\n894 206 704 179 272 337 413 828 119 182 330 46 440 102 250 191 242 539 678 783 843 431 612 567 33 338", "output": "9168707" }, { "input": "vhjnkrxbyhjhnjrxvwxmhxwoxttbtqosfxtcuvhfjlkyfspeypthsdkkwnqdpxdlnxsgtzvkrgqosgfjrwetqbxgoarkjhrjbspzgblsapifltkfxbfdbxqwoohlgyzijmiwnpmveybyzvasoctxsmgjehpyysmqblwnmkappbecklqjfmxhlyceordroflnposohfplrvijxbwvqdtvzhobtrumiujnyrfbwthvciinuveoizkccelxtaveiiagryqnyvsgfnipnavrtmdqlcnldepocbpzmqnarkdvykds\n276\n364 244 798 82 582 9 309 950 286 547 892 371 569 159 705 975 740 845 655 179 130 993 255 552 882 657", "output": "144901921" }, { "input": "gsaddmezrnttfalbwlqbnedumvikplfosw\n12\n290 850 872 361 483 895 152 118 974 619 701 154 899 285 328 712 669 984 407 340 851 775 324 892 554 860", "output": "809931" }, { "input": "a\n0\n5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "5" }, { "input": "lol\n3\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "21" } ]

1,405,259,264

2,864

Python 3

OK

TESTS

24

77

0

st=input() k=int(input()) w=list(map(int,input().split())) w2={chr(ord('a')+i):w[i] for i in range(26)} m=max(w) print(sum([w2[st[i]]*(i+1) for i in range(len(st))])+sum([i*m for i in range(len(st)+1,len(st)+1+k)]))

Title: DZY Loves Strings Time Limit: None seconds Memory Limit: None megabytes Problem Description: DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get? Input Specification: The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103). The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103). The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000. Output Specification: Print a single integer — the largest possible value of the resulting string DZY could get. Demo Input: ['abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n'] Demo Output: ['41\n'] Note: In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.

```python st=input() k=int(input()) w=list(map(int,input().split())) w2={chr(ord('a')+i):w[i] for i in range(26)} m=max(w) print(sum([w2[st[i]]*(i+1) for i in range(len(st))])+sum([i*m for i in range(len(st)+1,len(st)+1+k)])) ```

3

478

C

Table Decorations

PROGRAMMING

1,800

[ "greedy" ]

null

You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color? Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner.

The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.

Print a single integer *t* — the maximum number of tables that can be decorated in the required manner.

[ "5 4 3\n", "1 1 1\n", "2 3 3\n" ]

[ "4\n", "1\n", "2\n" ]

In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively.

1,500

[ { "input": "5 4 3", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 3 3", "output": "2" }, { "input": "0 1 0", "output": "0" }, { "input": "0 3 3", "output": "2" }, { "input": "4 0 4", "output": "2" }, { "input": "1000000000 1000000000 1000000000", "output": "1000000000" }, { "input": "100 99 56", "output": "85" }, { "input": "1000 1000 1002", "output": "1000" }, { "input": "0 1 1000000000", "output": "1" }, { "input": "500000000 1000000000 500000000", "output": "666666666" }, { "input": "1000000000 2000000000 1000000000", "output": "1333333333" }, { "input": "2000000000 2000000000 2000000000", "output": "2000000000" }, { "input": "0 0 0", "output": "0" }, { "input": "1 2000000000 1000000000", "output": "1000000000" }, { "input": "1585222789 1889821127 2000000000", "output": "1825014638" }, { "input": "10000 7500 7500", "output": "8333" }, { "input": "150000 75000 75000", "output": "100000" }, { "input": "999288131 55884921 109298382", "output": "165183303" }, { "input": "100500 100500 3", "output": "67001" }, { "input": "1463615122 1988383731 837331500", "output": "1429776784" }, { "input": "1938 8999 1882", "output": "3820" }, { "input": "45 33 76", "output": "51" }, { "input": "100000 1 2", "output": "3" }, { "input": "198488 50 18", "output": "68" }, { "input": "82728372 939848 100139442", "output": "61269220" }, { "input": "99 5747 5298", "output": "3714" }, { "input": "3 5 2", "output": "3" }, { "input": "7511 7512 7513", "output": "7512" }, { "input": "1234567890 123456789 987654321", "output": "781893000" }, { "input": "500000000 2000000000 500000000", "output": "1000000000" }, { "input": "500000002 2000000000 500000001", "output": "1000000001" }, { "input": "520000000 1000000033 501000000", "output": "673666677" }, { "input": "10000 1000 100000", "output": "11000" }, { "input": "2000000000 500000000 499999999", "output": "999999999" }, { "input": "1999999999 500000000 500000000", "output": "999999999" }, { "input": "1 1 9", "output": "2" }, { "input": "3 0 0", "output": "0" }, { "input": "6 1 1", "output": "2" }, { "input": "2000000000 1999999999 1999999999", "output": "1999999999" }, { "input": "3 4 9", "output": "5" }, { "input": "3 3 6", "output": "4" } ]

1,668,599,366

2,147,483,647

PyPy 3-64

OK

TESTS

42

77

0

r,g,b = map(int,input().split()) print(min(r+g,r+b,b+g,(r+b+g)//3))

Title: Table Decorations Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color? Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner. Input Specification: The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space. Output Specification: Print a single integer *t* — the maximum number of tables that can be decorated in the required manner. Demo Input: ['5 4 3\n', '1 1 1\n', '2 3 3\n'] Demo Output: ['4\n', '1\n', '2\n'] Note: In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively.

```python r,g,b = map(int,input().split()) print(min(r+g,r+b,b+g,(r+b+g)//3)) ```

3

879

A

Borya's Diagnosis

PROGRAMMING

900

[ "implementation" ]

null

It seems that Borya is seriously sick. He is going visit *n* doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor. Doctors have a strange working schedule. The doctor *i* goes to work on the *s**i*-th day and works every *d**i* day. So, he works on days *s**i*,<=*s**i*<=+<=*d**i*,<=*s**i*<=+<=2*d**i*,<=.... The doctor's appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors?

First line contains an integer *n* — number of doctors (1<=≤<=*n*<=≤<=1000). Next *n* lines contain two numbers *s**i* and *d**i* (1<=≤<=*s**i*,<=*d**i*<=≤<=1000).

Output a single integer — the minimum day at which Borya can visit the last doctor.

[ "3\n2 2\n1 2\n2 2\n", "2\n10 1\n6 5\n" ]

[ "4\n", "11\n" ]

In the first sample case, Borya can visit all doctors on days 2, 3 and 4. In the second sample case, Borya can visit all doctors on days 10 and 11.

500

[ { "input": "3\n2 2\n1 2\n2 2", "output": "4" }, { "input": "2\n10 1\n6 5", "output": "11" }, { "input": "3\n6 10\n3 3\n8 2", "output": "10" }, { "input": "4\n4 8\n10 10\n4 2\n8 2", "output": "14" }, { "input": "5\n7 1\n5 1\n6 1\n1 6\n6 8", "output": "14" }, { "input": "6\n1 3\n2 5\n4 7\n7 5\n6 8\n8 8", "output": "16" }, { "input": "10\n4 10\n8 7\n6 5\n2 1\n2 3\n8 8\n2 4\n2 2\n6 7\n7 9", "output": "34" }, { "input": "1\n1 1", "output": "1" }, { "input": "1\n1000 1000", "output": "1000" }, { "input": "42\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2", "output": "83" }, { "input": "2\n5 5\n5 1", "output": "6" }, { "input": "2\n5 5\n5 5", "output": "10" }, { "input": "2\n1 1\n1 1", "output": "2" }, { "input": "2\n1 6\n7 1", "output": "7" }, { "input": "2\n4 3\n4 5", "output": "9" }, { "input": "2\n1 2\n1 3", "output": "4" }, { "input": "3\n2 3\n5 1\n2 1", "output": "6" }, { "input": "2\n2 1\n6 3", "output": "6" }, { "input": "3\n10 1\n4 4\n12 1", "output": "13" }, { "input": "2\n2 2\n10 2", "output": "10" }, { "input": "2\n1 1\n1000 2", "output": "1000" }, { "input": "14\n1000 1\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1 1", "output": "13001" }, { "input": "2\n2 4\n2 1", "output": "3" }, { "input": "3\n1 100\n100 3\n200 1", "output": "200" }, { "input": "7\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1 1", "output": "6001" }, { "input": "2\n5 5\n15 5", "output": "15" }, { "input": "2\n2 2\n2 4", "output": "6" }, { "input": "2\n1 1\n10 1", "output": "10" }, { "input": "2\n10 1\n100 1", "output": "100" }, { "input": "15\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1 1", "output": "14001" }, { "input": "3\n2 1\n1 3\n4 7", "output": "11" }, { "input": "2\n5 5\n100 5", "output": "100" }, { "input": "2\n1 10\n2 30", "output": "2" }, { "input": "4\n2 2\n2 2\n2 2\n2 2", "output": "8" }, { "input": "1\n10 10", "output": "10" } ]

1,548,411,327

2,147,483,647

Python 3

OK

TESTS

51

124

0

n = int(input()) cur = 0 for i in range(n): s, d = map(int, input().split()) q = max(0, cur - s + d) // d cur = q * d + s print(cur)

Title: Borya's Diagnosis Time Limit: None seconds Memory Limit: None megabytes Problem Description: It seems that Borya is seriously sick. He is going visit *n* doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor. Doctors have a strange working schedule. The doctor *i* goes to work on the *s**i*-th day and works every *d**i* day. So, he works on days *s**i*,<=*s**i*<=+<=*d**i*,<=*s**i*<=+<=2*d**i*,<=.... The doctor's appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors? Input Specification: First line contains an integer *n* — number of doctors (1<=≤<=*n*<=≤<=1000). Next *n* lines contain two numbers *s**i* and *d**i* (1<=≤<=*s**i*,<=*d**i*<=≤<=1000). Output Specification: Output a single integer — the minimum day at which Borya can visit the last doctor. Demo Input: ['3\n2 2\n1 2\n2 2\n', '2\n10 1\n6 5\n'] Demo Output: ['4\n', '11\n'] Note: In the first sample case, Borya can visit all doctors on days 2, 3 and 4. In the second sample case, Borya can visit all doctors on days 10 and 11.

```python n = int(input()) cur = 0 for i in range(n): s, d = map(int, input().split()) q = max(0, cur - s + d) // d cur = q * d + s print(cur) ```

3

681

A

A Good Contest

PROGRAMMING

800

[ "implementation" ]

null

Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance. Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it. Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest . The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters. It is guaranteed that all handles are distinct.

Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.

[ "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n", "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n" ]

[ "YES", "NO" ]

In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest. In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.

500

[ { "input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES" }, { "input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "NO" }, { "input": "1\nDb -3373 3591", "output": "NO" }, { "input": "5\nQ2bz 960 2342\nhmX 2710 -1348\ngbAe -1969 -963\nE -160 196\npsi 2665 -3155", "output": "NO" }, { "input": "9\nmwAz9lQ 1786 -1631\nnYgYFXZQfY -1849 -1775\nKU4jF -1773 -3376\nopR 3752 2931\nGl -1481 -1002\nR -1111 3778\n0i9B21DC 3650 289\nQ8L2dS0 358 -3305\ng -2662 3968", "output": "NO" }, { "input": "5\nzMSBcOUf -2883 -2238\nYN -3314 -1480\nfHpuccQn06 -1433 -589\naM1NVEPQi 399 3462\n_L 2516 -3290", "output": "NO" }, { "input": "1\na 2400 2401", "output": "YES" }, { "input": "1\nfucker 4000 4000", "output": "NO" }, { "input": "1\nJora 2400 2401", "output": "YES" }, { "input": "1\nACA 2400 2420", "output": "YES" }, { "input": "1\nAca 2400 2420", "output": "YES" }, { "input": "1\nSub_d 2401 2402", "output": "YES" }, { "input": "2\nHack 2400 2401\nDum 1243 555", "output": "YES" }, { "input": "1\nXXX 2400 2500", "output": "YES" }, { "input": "1\nfucker 2400 2401", "output": "YES" }, { "input": "1\nX 2400 2500", "output": "YES" }, { "input": "1\nvineet 2400 2401", "output": "YES" }, { "input": "1\nabc 2400 2500", "output": "YES" }, { "input": "1\naaaaa 2400 2401", "output": "YES" }, { "input": "1\nhoge 2400 2401", "output": "YES" }, { "input": "1\nInfinity 2400 2468", "output": "YES" }, { "input": "1\nBurunduk1 2400 2401", "output": "YES" }, { "input": "1\nFuck 2400 2401", "output": "YES" }, { "input": "1\nfuck 2400 2401", "output": "YES" }, { "input": "3\nApplejack 2400 2401\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "YES" }, { "input": "1\nalex 2400 2401", "output": "YES" }, { "input": "1\nA 2400 2401", "output": "YES" }, { "input": "1\na 2400 2455", "output": "YES" }, { "input": "1\nlol 2400 2401", "output": "YES" }, { "input": "2\nBurunduk1 2400 2537\nBudAlNik 2084 2214", "output": "YES" }, { "input": "1\naaaaaa 2400 2401", "output": "YES" }, { "input": "1\nBurunduk1 2400 2500", "output": "YES" }, { "input": "1\nds 2400 2410", "output": "YES" }, { "input": "1\nas 2400 2401", "output": "YES" }, { "input": "1\nabc 2400 2401", "output": "YES" }, { "input": "3\nBudAlNik 2084 2214\nsubscriber 2833 2749\nBurunduk1 2526 2537", "output": "YES" }, { "input": "1\ncaonima 2400 2401", "output": "YES" }, { "input": "1\narr 2400 2500", "output": "YES" }, { "input": "1\nx 2400 2401", "output": "YES" }, { "input": "1\narrr 2400 2500", "output": "YES" }, { "input": "1\nabc 2400 2405", "output": "YES" }, { "input": "3\nBurunduk1 2400 2420\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES" }, { "input": "1\nBurunduk1 2400 2537", "output": "YES" }, { "input": "1\nHELLO 2400 2401", "output": "YES" }, { "input": "1\neatmore 2400 2500", "output": "YES" }, { "input": "1\nb 2400 2401", "output": "YES" }, { "input": "3\nBurunduk1 2400 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES" }, { "input": "1\nApplejack 2400 2410", "output": "YES" }, { "input": "1\nabacaba 2400 2451", "output": "YES" }, { "input": "1\nrekt_n00b 2500 2600", "output": "YES" } ]

1,561,213,214

2,147,483,647

Python 3

OK

TESTS

60

109

0

n = int(input()) flag = 'NO' for i in range(n): text, f, s = input().split() if int(s) > int(f) >= 2400: flag = 'YES' print(flag)

Title: A Good Contest Time Limit: None seconds Memory Limit: None megabytes Problem Description: Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance. Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it. Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same? Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest . The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters. It is guaranteed that all handles are distinct. Output Specification: Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. Demo Input: ['3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n', '3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n'] Demo Output: ['YES', 'NO'] Note: In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest. In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.

```python n = int(input()) flag = 'NO' for i in range(n): text, f, s = input().split() if int(s) > int(f) >= 2400: flag = 'YES' print(flag) ```

3

629

B

Far Relative’s Problem

PROGRAMMING

1,100

[ "brute force" ]

null

Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible. Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party. Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends. Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive.

Print the maximum number of people that may come to Famil Door's party.

[ "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n", "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n" ]

[ "2\n", "4\n" ]

In the first sample, friends 3 and 4 can come on any day in range [117, 128]. In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140.

1,000

[ { "input": "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128", "output": "2" }, { "input": "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200", "output": "4" }, { "input": "1\nF 68 307", "output": "0" }, { "input": "40\nM 55 363\nF 117 252\nM 157 282\nF 322 345\nM 330 363\nF 154 231\nF 216 352\nF 357 365\nM 279 292\nF 353 359\nF 82 183\nM 78 297\nM 231 314\nM 107 264\nF 34 318\nM 44 244\nF 42 339\nM 253 307\nM 128 192\nF 119 328\nM 135 249\nF 303 358\nF 348 352\nF 8 364\nF 126 303\nM 226 346\nF 110 300\nF 47 303\nF 201 311\nF 287 288\nM 270 352\nM 227 351\nF 8 111\nF 39 229\nM 163 315\nF 269 335\nF 147 351\nF 96 143\nM 97 99\nM 177 295", "output": "22" }, { "input": "2\nF 1 1\nM 1 1", "output": "2" }, { "input": "4\nM 1 2\nM 2 3\nF 3 4\nF 4 5", "output": "2" }, { "input": "1\nF 1 2", "output": "0" }, { "input": "18\nF 3 4\nF 3 4\nF 3 4\nF 3 4\nF 5 6\nF 5 6\nM 3 4\nM 3 4\nM 5 6\nM 5 6\nM 5 6\nM 5 6\nF 7 8\nF 7 8\nF 7 8\nM 7 8\nM 7 8\nM 7 8", "output": "6" }, { "input": "2\nM 1 1\nF 1 1", "output": "2" }, { "input": "3\nM 1 1\nF 1 1\nM 1 1", "output": "2" } ]

1,489,481,431

2,147,483,647

Python 3

OK

TESTS

76

546

5,017,600

n = int(input()) gender = [] fromdate = [] todate = [] for k in range(n): a, b, c = input().split() gender.append(a) fromdate.append(int(b)) todate.append(int(c)) minimum = min(fromdate) maximum = max(todate) #print("min = {} max = {}".format(minimum,maximum)) ans = 0 for i in range(minimum, maximum + 1): male = 0 women = 0 for j in range(n): if fromdate[j] <= i and todate[j] >= i: if gender[j] == 'M': male += 1 else: women += 1 count = min(male, women) * 2 #print(count) if count > ans: ans = count print(ans)

Title: Far Relative’s Problem Time Limit: None seconds Memory Limit: None megabytes Problem Description: Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible. Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party. Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends. Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive. Output Specification: Print the maximum number of people that may come to Famil Door's party. Demo Input: ['4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n', '6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n'] Demo Output: ['2\n', '4\n'] Note: In the first sample, friends 3 and 4 can come on any day in range [117, 128]. In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140.

```python n = int(input()) gender = [] fromdate = [] todate = [] for k in range(n): a, b, c = input().split() gender.append(a) fromdate.append(int(b)) todate.append(int(c)) minimum = min(fromdate) maximum = max(todate) #print("min = {} max = {}".format(minimum,maximum)) ans = 0 for i in range(minimum, maximum + 1): male = 0 women = 0 for j in range(n): if fromdate[j] <= i and todate[j] >= i: if gender[j] == 'M': male += 1 else: women += 1 count = min(male, women) * 2 #print(count) if count > ans: ans = count print(ans) ```

3

769

B

News About Credit

PROGRAMMING

1,200

[ "*special", "greedy", "two pointers" ]

null

Polycarp studies at the university in the group which consists of *n* students (including himself). All they are registrated in the social net "TheContacnt!". Not all students are equally sociable. About each student you know the value *a**i* — the maximum number of messages which the *i*-th student is agree to send per day. The student can't send messages to himself. In early morning Polycarp knew important news that the programming credit will be tomorrow. For this reason it is necessary to urgently inform all groupmates about this news using private messages. Your task is to make a plan of using private messages, so that: - the student *i* sends no more than *a**i* messages (for all *i* from 1 to *n*); - all students knew the news about the credit (initially only Polycarp knew it); - the student can inform the other student only if he knows it himself. Let's consider that all students are numerated by distinct numbers from 1 to *n*, and Polycarp always has the number 1. In that task you shouldn't minimize the number of messages, the moment of time, when all knew about credit or some other parameters. Find any way how to use private messages which satisfies requirements above.

The first line contains the positive integer *n* (2<=≤<=*n*<=≤<=100) — the number of students. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* equals to the maximum number of messages which can the *i*-th student agree to send. Consider that Polycarp always has the number 1.

Print -1 to the first line if it is impossible to inform all students about credit. Otherwise, in the first line print the integer *k* — the number of messages which will be sent. In each of the next *k* lines print two distinct integers *f* and *t*, meaning that the student number *f* sent the message with news to the student number *t*. All messages should be printed in chronological order. It means that the student, who is sending the message, must already know this news. It is assumed that students can receive repeated messages with news of the credit. If there are several answers, it is acceptable to print any of them.

[ "4\n1 2 1 0\n", "6\n2 0 1 3 2 0\n", "3\n0 2 2\n" ]

[ "3\n1 2\n2 4\n2 3\n", "6\n1 3\n3 4\n1 2\n4 5\n5 6\n4 6\n", "-1\n" ]

In the first test Polycarp (the student number 1) can send the message to the student number 2, who after that can send the message to students number 3 and 4. Thus, all students knew about the credit.

1,000

[ { "input": "4\n1 2 1 0", "output": "3\n1 2\n2 3\n2 4" }, { "input": "6\n2 0 1 3 2 0", "output": "5\n1 4\n1 5\n4 3\n4 2\n4 6" }, { "input": "3\n0 2 2", "output": "-1" }, { "input": "2\n0 0", "output": "-1" }, { "input": "2\n1 0", "output": "1\n1 2" }, { "input": "2\n0 1", "output": "-1" }, { "input": "2\n1 1", "output": "1\n1 2" }, { "input": "3\n1 1 0", "output": "2\n1 2\n2 3" }, { "input": "3\n0 1 1", "output": "-1" }, { "input": "3\n1 0 0", "output": "-1" }, { "input": "3\n2 0 0", "output": "2\n1 2\n1 3" }, { "input": "3\n1 0 1", "output": "2\n1 3\n3 2" }, { "input": "3\n1 1 1", "output": "2\n1 2\n2 3" }, { "input": "40\n3 3 2 1 0 0 0 4 5 4 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 2 3 2 0 1 0 0 2 0 3 0 1 0", "output": "-1" }, { "input": "100\n1 0 0 2 0 2 0 0 2 0 0 2 0 0 2 2 2 1 1 2 1 2 2 2 1 2 0 1 0 1 0 2 2 2 0 1 2 0 0 2 0 2 0 1 1 0 1 0 2 0 0 2 1 2 1 2 2 2 2 1 0 2 0 0 1 0 2 0 0 2 0 1 0 2 1 1 2 2 2 2 0 0 2 0 2 1 0 0 0 1 0 2 2 2 0 1 0 1 1 0", "output": "99\n1 4\n4 6\n4 9\n6 12\n6 15\n9 16\n9 17\n12 20\n12 22\n15 23\n15 24\n16 26\n16 32\n17 33\n17 34\n20 37\n20 40\n22 42\n22 49\n23 52\n23 54\n24 56\n24 57\n26 58\n26 59\n32 62\n32 67\n33 70\n33 74\n34 77\n34 78\n37 79\n37 80\n40 83\n40 85\n42 92\n42 93\n49 94\n49 18\n52 19\n52 21\n54 25\n54 28\n56 30\n56 36\n57 44\n57 45\n58 47\n58 53\n59 55\n59 60\n62 65\n62 72\n67 75\n67 76\n70 86\n70 90\n74 96\n74 98\n77 99\n77 2\n78 3\n78 5\n79 7\n79 8\n80 10\n80 11\n83 13\n83 14\n85 27\n85 29\n92 31\n92 35\n93 38\n93 3..." }, { "input": "4\n2 0 0 0", "output": "-1" }, { "input": "4\n2 0 0 1", "output": "3\n1 4\n1 2\n4 3" }, { "input": "4\n2 0 1 0", "output": "3\n1 3\n1 2\n3 4" }, { "input": "4\n2 1 0 0", "output": "3\n1 2\n1 3\n2 4" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "-1" }, { "input": "100\n99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "99\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87\n..." }, { "input": "100\n98 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "-1" }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "99\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n76 77\n..." }, { "input": "100\n1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "99\n1 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 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0 0 0 4 1 2 0 0 0 1 0 3 2 0 0 2 0 3 0 1 4 2 1 0 0 1 1 0 1 0 0 4 0 1 0 1 0 1 1 0 0 0 0 1 2 0 0 0 3 3 0 2", "output": "99\n1 52\n1 69\n1 80\n52 10\n52 16\n52 19\n52 24\n69 48\n69 60\n69 66\n69 97\n80 98\n80 2\n80 9\n80 11\n10 26\n10 27\n10 28\n16 37\n16 40\n16 54\n19 61\n19 64\n19 70\n24 93\n24 100\n24 3\n48 5\n48 12\n48 17\n60 18\n60 21\n60 22\n66 23\n66 30\n66 31\n97 32\n97 35\n97 38\n98 39\n98 41\n98 45\n2 46\n2 53\n9 58\n9 68\n11 71\n11 74\n26 75\n26 77\n27 82\n27 84\n28 86\n28 87\n37 92\n37 4\n40 6\n40 7\n54 8\n54 13\n61 14\n61 15\n64 20\n64 25\n70 29\n70 33\n93 34\n93 36\n100 42\n100 43\n3 44\n5 47\n12 49\n17 50\n18 ..." }, { "input": "100\n66 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0", "output": "99\n1 97\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n97 67\n97 68\n97 69\n97 70\n97 71\n97 72\n97 73\n97 74\n97 75\n97 76\n97 77\n97 78\n97 79\n97 80\n97 81\n97 82\n97 83\n..." }, { "input": "20\n0 0 3 0 0 0 3 4 2 0 2 0 0 0 0 1 0 1 0 1", "output": "-1" }, { "input": "60\n3 0 0 1 0 0 0 0 3 1 3 4 0 0 0 3 0 0 0 2 0 3 4 1 3 3 0 2 0 4 1 5 3 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 1 1 0 1 0 3 0 0", "output": "-1" }, { "input": "80\n4 0 0 0 0 0 0 3 0 3 0 0 0 4 3 0 1 0 2 0 0 0 5 0 5 0 0 0 0 4 0 3 0 0 0 1 0 0 2 0 5 2 0 0 4 4 0 3 0 0 0 0 0 0 0 2 5 0 2 0 0 0 0 0 0 0 0 0 0 3 0 0 3 5 0 0 0 0 0 0", "output": "-1" }, { "input": "100\n2 0 0 2 0 0 0 0 0 2 0 0 0 5 0 0 0 0 0 0 0 1 0 7 0 0 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 6 0 7 4 0 0 0 0 5 0 0 0 0 0 0 7 4 0 0 0 0 0 0 7 7 0 0 0 0 0 2 0 0 0 0 0 0 0 0 4 7 7 0 0 0", "output": "-1" }, { "input": "100\n1 0 0 0 1 2 3 2 1 0 1 2 3 1 3 1 0 0 1 1 0 0 2 1 2 1 3 3 1 0 0 1 0 2 2 0 3 0 1 1 1 2 0 2 0 1 0 2 0 1 2 2 0 0 0 0 1 0 0 0 1 1 4 0 2 0 1 0 2 0 2 2 2 1 1 0 0 2 0 3 1 0 0 1 1 0 1 0 2 3 2 0 1 2 0 0 0 0 0 1", "output": "99\n1 63\n63 7\n63 13\n63 15\n63 27\n7 28\n7 37\n7 80\n13 90\n13 6\n13 8\n15 12\n15 23\n15 25\n27 34\n27 35\n27 42\n28 44\n28 48\n28 51\n37 52\n37 65\n37 69\n80 71\n80 72\n80 73\n90 78\n90 89\n90 91\n6 94\n6 5\n8 9\n8 11\n12 14\n12 16\n23 19\n23 20\n25 24\n25 26\n34 29\n34 32\n35 39\n35 40\n42 41\n42 46\n44 50\n44 57\n48 61\n48 62\n51 67\n51 74\n52 75\n52 81\n65 84\n65 85\n69 87\n69 93\n71 100\n71 2\n72 3\n72 4\n73 10\n73 17\n78 18\n78 21\n89 22\n89 30\n91 31\n91 33\n94 36\n94 38\n5 43\n9 45\n11 47\n14 49\n..." }, { "input": "100\n5 0 0 1 1 0 0 0 1 0 0 0 0 0 5 0 2 1 0 1 0 6 0 0 0 3 0 0 0 0 0 0 0 0 1 0 3 0 0 0 0 0 0 4 0 0 3 1 1 0 0 4 0 0 0 0 0 0 3 2 3 3 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 2 1 0 5 1 0 8 0 1 0 0 10 4 0 0 0 6 4 0 0 0 0 1", "output": "99\n1 89\n1 84\n1 22\n1 94\n1 15\n89 81\n89 44\n89 52\n89 90\n89 95\n89 26\n89 37\n89 47\n89 59\n89 61\n84 62\n84 17\n84 60\n84 75\n84 78\n84 4\n84 5\n84 9\n22 18\n22 20\n22 35\n22 48\n22 49\n22 79\n94 82\n94 86\n94 100\n94 2\n94 3\n94 6\n15 7\n15 8\n15 10\n15 11\n15 12\n81 13\n81 14\n81 16\n81 19\n81 21\n44 23\n44 24\n44 25\n44 27\n52 28\n52 29\n52 30\n52 31\n90 32\n90 33\n90 34\n90 36\n95 38\n95 39\n95 40\n95 41\n26 42\n26 43\n26 45\n37 46\n37 50\n37 51\n47 53\n47 54\n47 55\n59 56\n59 57\n59 58\n61 63\n6..." }, { "input": "100\n47 0 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 4 0 1 0 2 0 0 1 0 0 0 0 0 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0", "output": "99\n1 47\n1 12\n1 66\n1 33\n1 93\n1 37\n1 81\n1 25\n1 35\n1 40\n1 46\n1 61\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 34\n1 36\n1 38\n1 39\n1 41\n1 42\n47 43\n47 44\n47 45\n47 48\n47 49\n47 50\n47 51\n47 52\n47 53\n47 54\n47 55\n47 56\n47 57\n47 58\n47 59\n47 60\n47 62\n47 63\n12 64\n12 65\n12 67\n12 68\n12 69\n12 70\n12 71\n12 72\n12 73\n66 74\n66 75\n66 76\n66 77\n66 78\n66 79\n66..." }, { "input": "100\n1 0 2 1 1 1 0 0 0 3 2 1 1 1 0 1 0 1 1 1 0 1 1 1 0 1 1 1 1 2 1 0 2 1 1 1 0 0 1 2 1 3 1 1 0 0 2 1 0 1 1 1 2 1 2 0 3 1 2 0 1 1 2 2 1 1 1 1 1 2 0 0 2 1 1 0 1 2 1 1 1 1 1 0 1 1 1 0 3 0 0 2 2 0 0 0 2 1 2 0", "output": "99\n1 10\n10 42\n10 57\n10 89\n42 3\n42 11\n42 30\n57 33\n57 40\n57 47\n89 53\n89 55\n89 59\n3 63\n3 64\n11 70\n11 73\n30 78\n30 92\n33 93\n33 97\n40 99\n40 4\n47 5\n47 6\n53 12\n53 13\n55 14\n55 16\n59 18\n59 19\n63 20\n63 22\n64 23\n64 24\n70 26\n70 27\n73 28\n73 29\n78 31\n78 34\n92 35\n92 36\n93 39\n93 41\n97 43\n97 44\n99 48\n99 50\n4 51\n5 52\n6 54\n12 58\n13 61\n14 62\n16 65\n18 66\n19 67\n20 68\n22 69\n23 74\n24 75\n26 77\n27 79\n28 80\n29 81\n31 82\n34 83\n35 85\n36 86\n39 87\n41 98\n43 2\n44 7\n4..." }, { "input": "100\n83 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 11 0", "output": "99\n1 99\n1 83\n1 22\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n99 84\n99 85\n99 ..." }, { "input": "100\n1 1 0 1 0 1 1 1 2 1 0 1 1 0 1 2 1 1 1 1 2 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 2 2 2 2 1 1 0 1 2 1 1 0 1 1 0 1 0 1 1 0 0 1 1 1 1 1 1 2 0 2 1 2 1 1 0 0 1 1 1 1 0 2 4 2 1 1 1 0 1 2 1 1 1 0 2 1 2", "output": "99\n1 86\n86 9\n86 16\n86 21\n86 44\n9 45\n9 46\n16 47\n16 52\n21 71\n21 73\n44 75\n44 85\n45 87\n45 93\n46 98\n46 100\n47 2\n47 4\n52 6\n52 7\n71 8\n71 10\n73 12\n73 13\n75 15\n75 17\n85 18\n85 19\n87 20\n87 22\n93 23\n93 25\n98 27\n98 28\n100 29\n100 30\n2 31\n4 32\n6 33\n7 34\n8 35\n10 36\n12 37\n13 38\n15 39\n17 41\n18 42\n19 48\n20 49\n22 51\n23 53\n25 54\n27 56\n28 57\n29 59\n30 61\n31 62\n32 65\n33 66\n34 67\n35 68\n36 69\n37 70\n38 74\n39 76\n41 77\n42 80\n48 81\n49 82\n51 83\n53 88\n54 89\n56 90\n..." }, { "input": "100\n99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "99\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87\n..." }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "99\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76\n76 77\n..." }, { "input": "2\n0 100", "output": "-1" }, { "input": "2\n100 0", "output": "1\n1 2" }, { "input": "2\n100 100", "output": "1\n1 2" } ]

1,546,334,404

2,147,483,647

PyPy 3

OK

TESTS

60

140

0

n = int(input()) call = list(map(int, input().split())) cp = [] ans = [] for i in range(1, n): x = call[i] cp.append([x, i + 1]) cp.sort(reverse=True) l = [[call[0], 1]] j = 0 while len(cp) > 0 and len(l) > 0 and j < len(cp): s = l[0] x = s[0] p = s[1] for i in range(j, min(j + x, len(cp))): ans.append([str(p), str(cp[i][1])]) l.append(cp[i]) j = min(j + x, len(cp)) l.remove(l[0]) if j < len(cp): print(-1) else: print(len(ans)) for i in range(len(ans)): print(' '.join(ans[i]))

Title: News About Credit Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp studies at the university in the group which consists of *n* students (including himself). All they are registrated in the social net "TheContacnt!". Not all students are equally sociable. About each student you know the value *a**i* — the maximum number of messages which the *i*-th student is agree to send per day. The student can't send messages to himself. In early morning Polycarp knew important news that the programming credit will be tomorrow. For this reason it is necessary to urgently inform all groupmates about this news using private messages. Your task is to make a plan of using private messages, so that: - the student *i* sends no more than *a**i* messages (for all *i* from 1 to *n*); - all students knew the news about the credit (initially only Polycarp knew it); - the student can inform the other student only if he knows it himself. Let's consider that all students are numerated by distinct numbers from 1 to *n*, and Polycarp always has the number 1. In that task you shouldn't minimize the number of messages, the moment of time, when all knew about credit or some other parameters. Find any way how to use private messages which satisfies requirements above. Input Specification: The first line contains the positive integer *n* (2<=≤<=*n*<=≤<=100) — the number of students. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* equals to the maximum number of messages which can the *i*-th student agree to send. Consider that Polycarp always has the number 1. Output Specification: Print -1 to the first line if it is impossible to inform all students about credit. Otherwise, in the first line print the integer *k* — the number of messages which will be sent. In each of the next *k* lines print two distinct integers *f* and *t*, meaning that the student number *f* sent the message with news to the student number *t*. All messages should be printed in chronological order. It means that the student, who is sending the message, must already know this news. It is assumed that students can receive repeated messages with news of the credit. If there are several answers, it is acceptable to print any of them. Demo Input: ['4\n1 2 1 0\n', '6\n2 0 1 3 2 0\n', '3\n0 2 2\n'] Demo Output: ['3\n1 2\n2 4\n2 3\n', '6\n1 3\n3 4\n1 2\n4 5\n5 6\n4 6\n', '-1\n'] Note: In the first test Polycarp (the student number 1) can send the message to the student number 2, who after that can send the message to students number 3 and 4. Thus, all students knew about the credit.

```python n = int(input()) call = list(map(int, input().split())) cp = [] ans = [] for i in range(1, n): x = call[i] cp.append([x, i + 1]) cp.sort(reverse=True) l = [[call[0], 1]] j = 0 while len(cp) > 0 and len(l) > 0 and j < len(cp): s = l[0] x = s[0] p = s[1] for i in range(j, min(j + x, len(cp))): ans.append([str(p), str(cp[i][1])]) l.append(cp[i]) j = min(j + x, len(cp)) l.remove(l[0]) if j < len(cp): print(-1) else: print(len(ans)) for i in range(len(ans)): print(' '.join(ans[i])) ```

3

1,005

A

Tanya and Stairways

PROGRAMMING

800

[ "implementation" ]

null

Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$. You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya. The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.

[ "7\n1 2 3 1 2 3 4\n", "4\n1 1 1 1\n", "5\n1 2 3 4 5\n", "5\n1 2 1 2 1\n" ]

[ "2\n3 4 ", "4\n1 1 1 1 ", "1\n5 ", "3\n2 2 1 " ]

none

0

[ { "input": "7\n1 2 3 1 2 3 4", "output": "2\n3 4 " }, { "input": "4\n1 1 1 1", "output": "4\n1 1 1 1 " }, { "input": "5\n1 2 3 4 5", "output": "1\n5 " }, { "input": "5\n1 2 1 2 1", "output": "3\n2 2 1 " }, { "input": "1\n1", "output": "1\n1 " }, { "input": "48\n1 2 3 4 1 2 3 1 1 2 3 1 2 3 4 1 1 2 3 4 1 2 3 4 1 2 3 4 1 1 2 1 2 1 2 1 1 2 1 2 1 2 3 1 2 1 2 1", "output": "20\n4 3 1 3 4 1 4 4 4 1 2 2 2 1 2 2 3 2 2 1 " }, { "input": "2\n1 2", "output": "1\n2 " }, { "input": "3\n1 1 2", "output": "2\n1 2 " }, { "input": "4\n1 1 2 3", "output": "2\n1 3 " }, { "input": "8\n1 2 3 1 2 3 4 5", "output": "2\n3 5 " }, { "input": "5\n1 1 1 2 3", "output": "3\n1 1 3 " } ]

1,637,832,748

2,147,483,647

Python 3

OK

TESTS

16

62

0

n = int(input()) li = list(map(int, input().split())) count = 1 newli = [] for i in range(1,n): if li[i]<li[i-1]: count +=1 newli.append(li[i-1]) if li[i]==li[i-1]: count +=1 newli.append(li[i-1]) newli.append(li[-1]) print(count) print(" ".join([str(x) for x in newli]))

Title: Tanya and Stairways Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$. You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. Input Specification: The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya. The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order. The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. Output Specification: In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways. Demo Input: ['7\n1 2 3 1 2 3 4\n', '4\n1 1 1 1\n', '5\n1 2 3 4 5\n', '5\n1 2 1 2 1\n'] Demo Output: ['2\n3 4 ', '4\n1 1 1 1 ', '1\n5 ', '3\n2 2 1 '] Note: none

```python n = int(input()) li = list(map(int, input().split())) count = 1 newli = [] for i in range(1,n): if li[i]<li[i-1]: count +=1 newli.append(li[i-1]) if li[i]==li[i-1]: count +=1 newli.append(li[i-1]) newli.append(li[-1]) print(count) print(" ".join([str(x) for x in newli])) ```

3

255

A

Greg's Workout

PROGRAMMING

800

[ "implementation" ]

null

Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times. Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise. Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.

The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.

Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise. It is guaranteed that the input is such that the answer to the problem is unambiguous.

[ "2\n2 8\n", "3\n5 1 10\n", "7\n3 3 2 7 9 6 8\n" ]

[ "biceps\n", "back\n", "chest\n" ]

In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises. In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises. In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.

500

[ { "input": "2\n2 8", "output": "biceps" }, { "input": "3\n5 1 10", "output": "back" }, { "input": "7\n3 3 2 7 9 6 8", "output": "chest" }, { "input": "4\n5 6 6 2", "output": "chest" }, { "input": "5\n8 2 2 6 3", "output": "chest" }, { "input": "6\n8 7 2 5 3 4", "output": "chest" }, { "input": "8\n7 2 9 10 3 8 10 6", "output": "chest" }, { "input": "9\n5 4 2 3 4 4 5 2 2", "output": "chest" }, { "input": "10\n4 9 8 5 3 8 8 10 4 2", "output": "biceps" }, { "input": "11\n10 9 7 6 1 3 9 7 1 3 5", "output": "chest" }, { "input": "12\n24 22 6 16 5 21 1 7 2 19 24 5", "output": "chest" }, { "input": "13\n24 10 5 7 16 17 2 7 9 20 15 2 24", "output": "chest" }, { "input": "14\n13 14 19 8 5 17 9 16 15 9 5 6 3 7", "output": "back" }, { "input": "15\n24 12 22 21 25 23 21 5 3 24 23 13 12 16 12", "output": "chest" }, { "input": "16\n12 6 18 6 25 7 3 1 1 17 25 17 6 8 17 8", "output": "biceps" }, { "input": "17\n13 8 13 4 9 21 10 10 9 22 14 23 22 7 6 14 19", "output": "chest" }, { "input": "18\n1 17 13 6 11 10 25 13 24 9 21 17 3 1 17 12 25 21", "output": "back" }, { "input": "19\n22 22 24 25 19 10 7 10 4 25 19 14 1 14 3 18 4 19 24", "output": "chest" }, { "input": "20\n9 8 22 11 18 14 15 10 17 11 2 1 25 20 7 24 4 25 9 20", "output": "chest" }, { "input": "1\n10", "output": "chest" }, { "input": "2\n15 3", "output": "chest" }, { "input": "3\n21 11 19", "output": "chest" }, { "input": "4\n19 24 13 15", "output": "chest" }, { "input": "5\n4 24 1 9 19", "output": "biceps" }, { "input": "6\n6 22 24 7 15 24", "output": "back" }, { "input": "7\n10 8 23 23 14 18 14", "output": "chest" }, { "input": "8\n5 16 8 9 17 16 14 7", "output": "biceps" }, { "input": "9\n12 3 10 23 6 4 22 13 12", "output": "chest" }, { "input": "10\n1 9 20 18 20 17 7 24 23 2", "output": "back" }, { "input": "11\n22 25 8 2 18 15 1 13 1 11 4", "output": "biceps" }, { "input": "12\n20 12 14 2 15 6 24 3 11 8 11 14", "output": "chest" }, { "input": "13\n2 18 8 8 8 20 5 22 15 2 5 19 18", "output": "back" }, { "input": "14\n1 6 10 25 17 13 21 11 19 4 15 24 5 22", "output": "biceps" }, { "input": "15\n13 5 25 13 17 25 19 21 23 17 12 6 14 8 6", "output": "back" }, { "input": "16\n10 15 2 17 22 12 14 14 6 11 4 13 9 8 21 14", "output": "chest" }, { "input": "17\n7 22 9 22 8 7 20 22 23 5 12 11 1 24 17 20 10", "output": "biceps" }, { "input": "18\n18 15 4 25 5 11 21 25 12 14 25 23 19 19 13 6 9 17", "output": "chest" }, { "input": "19\n3 1 3 15 15 25 10 25 23 10 9 21 13 23 19 3 24 21 14", "output": "back" }, { "input": "20\n19 18 11 3 6 14 3 3 25 3 1 19 25 24 23 12 7 4 8 6", "output": "back" }, { "input": "1\n19", "output": "chest" }, { "input": "2\n1 7", "output": "biceps" }, { "input": "3\n18 18 23", "output": "back" }, { "input": "4\n12 15 1 13", "output": "chest" }, { "input": "5\n11 14 25 21 21", "output": "biceps" }, { "input": "6\n11 9 12 11 22 18", "output": "biceps" }, { "input": "7\n11 1 16 20 21 25 20", "output": "chest" }, { "input": "8\n1 2 20 9 3 22 17 4", "output": "back" }, { "input": "9\n19 2 10 19 15 20 3 1 13", "output": "back" }, { "input": "10\n11 2 11 8 21 16 2 3 19 9", "output": "back" }, { "input": "20\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 24", "output": "chest" }, { "input": "12\n4 24 21 3 13 24 22 13 12 21 1 15", "output": "back" }, { "input": "13\n14 14 16 2 13 5 1 14 9 4 16 8 3", "output": "biceps" }, { "input": "14\n1 9 15 4 11 8 25 3 9 14 13 2 1 11", "output": "biceps" }, { "input": "15\n4 19 10 6 16 12 5 11 7 23 1 24 11 7 17", "output": "back" }, { "input": "16\n2 8 2 8 13 22 20 12 22 23 18 13 18 22 11 17", "output": "chest" }, { "input": "17\n24 5 5 16 10 8 22 6 4 13 10 10 5 23 8 20 8", "output": "chest" }, { "input": "18\n14 8 9 12 11 18 24 1 14 24 18 5 12 17 1 10 1 22", "output": "chest" }, { "input": "19\n21 2 10 6 9 1 24 5 2 19 10 13 10 7 19 2 6 13 24", "output": "chest" }, { "input": "20\n7 1 14 17 6 6 18 13 12 3 25 4 3 19 22 24 16 14 1 23", "output": "biceps" }, { "input": "1\n19", "output": "chest" }, { "input": "20\n2 1 2 2 1 2 2 1 2 1 1 1 1 1 1 1 1 1 1 22", "output": "biceps" } ]

1,649,615,359

2,147,483,647

Python 3

OK

TESTS

61

92

0

n=int(input('')) l=list(map(int,input().split())) c,bc,bk=[],[],[] for i in range(0,n,3): c.append(l[i]) for j in range(1,n,3): bc.append(l[j]) for k in range(2,n,3): bk.append(l[k]) t=[sum(c),sum(bc),sum(bk)] m=t.index(max(t)) if m==0: print('chest') elif m==1: print('biceps') else: print('back')

Title: Greg's Workout Time Limit: None seconds Memory Limit: None megabytes Problem Description: Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times. Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise. Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises. Output Specification: Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise. It is guaranteed that the input is such that the answer to the problem is unambiguous. Demo Input: ['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n'] Demo Output: ['biceps\n', 'back\n', 'chest\n'] Note: In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises. In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises. In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.

```python n=int(input('')) l=list(map(int,input().split())) c,bc,bk=[],[],[] for i in range(0,n,3): c.append(l[i]) for j in range(1,n,3): bc.append(l[j]) for k in range(2,n,3): bk.append(l[k]) t=[sum(c),sum(bc),sum(bk)] m=t.index(max(t)) if m==0: print('chest') elif m==1: print('biceps') else: print('back') ```

3

448

A

Rewards

PROGRAMMING

800

[ "implementation" ]

null

Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals. Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules: - any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals. Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.

The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100). The numbers in the lines are separated by single spaces.

Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).

[ "1 1 1\n1 1 1\n4\n", "1 1 3\n2 3 4\n2\n", "1 0 0\n1 0 0\n1\n" ]

[ "YES\n", "YES\n", "NO\n" ]

none

500

[ { "input": "1 1 1\n1 1 1\n4", "output": "YES" }, { "input": "1 1 3\n2 3 4\n2", "output": "YES" }, { "input": "1 0 0\n1 0 0\n1", "output": "NO" }, { "input": "0 0 0\n0 0 0\n1", "output": "YES" }, { "input": "100 100 100\n100 100 100\n100", "output": "YES" }, { "input": "100 100 100\n100 100 100\n1", "output": "NO" }, { "input": "1 10 100\n100 10 1\n20", "output": "NO" }, { "input": "1 1 1\n0 0 0\n1", "output": "YES" }, { "input": "0 0 0\n1 1 1\n1", "output": "YES" }, { "input": "5 5 5\n0 0 0\n2", "output": "NO" }, { "input": "0 0 0\n10 10 10\n2", "output": "NO" }, { "input": "21 61 39\n63 58 69\n44", "output": "YES" }, { "input": "18 95 4\n7 1 75\n46", "output": "YES" }, { "input": "64 27 81\n72 35 23\n48", "output": "YES" }, { "input": "6 6 6\n11 11 11\n7", "output": "NO" }, { "input": "1 2 3\n2 4 6\n3", "output": "NO" }, { "input": "1 2 3\n2 4 6\n4", "output": "YES" }, { "input": "99 99 99\n99 99 99\n89", "output": "NO" }, { "input": "5 0 0\n15 0 0\n2", "output": "NO" }, { "input": "10 10 10\n0 0 0\n1", "output": "NO" }, { "input": "1 1 1\n1 1 1\n15", "output": "YES" }, { "input": "2 3 5\n2 3 5\n2", "output": "NO" }, { "input": "2 2 2\n3 3 5\n3", "output": "NO" }, { "input": "1 2 2\n2 4 4\n1", "output": "NO" }, { "input": "1 2 3\n1 5 5\n2", "output": "NO" } ]

1,655,994,472

2,147,483,647

PyPy 3-64

OK

TESTS

25

186

12,800,000

import sys import math import collections import heapq import decimal input=sys.stdin.readline a=[int(i) for i in input().split()] b=[int(i) for i in input().split()] n=int(input()) if(math.ceil(sum(a)/5)+math.ceil(sum(b)/10)<=n): print("YES") else: print("NO")

Title: Rewards Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals. Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules: - any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals. Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled. Input Specification: The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100). The numbers in the lines are separated by single spaces. Output Specification: Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes). Demo Input: ['1 1 1\n1 1 1\n4\n', '1 1 3\n2 3 4\n2\n', '1 0 0\n1 0 0\n1\n'] Demo Output: ['YES\n', 'YES\n', 'NO\n'] Note: none

```python import sys import math import collections import heapq import decimal input=sys.stdin.readline a=[int(i) for i in input().split()] b=[int(i) for i in input().split()] n=int(input()) if(math.ceil(sum(a)/5)+math.ceil(sum(b)/10)<=n): print("YES") else: print("NO") ```

3

363

B

Fence

PROGRAMMING

1,100

[ "brute force", "dp" ]

null

There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights. Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible. Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).

The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.

Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.

[ "7 3\n1 2 6 1 1 7 1\n" ]

[ "3\n" ]

In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

1,000

[ { "input": "7 3\n1 2 6 1 1 7 1", "output": "3" }, { "input": "1 1\n100", "output": "1" }, { "input": "2 1\n10 20", "output": "1" }, { "input": "10 5\n1 2 3 1 2 2 3 1 4 5", "output": "1" }, { "input": "10 2\n3 1 4 1 4 6 2 1 4 6", "output": "7" }, { "input": "2 2\n20 10", "output": "1" }, { "input": "2 1\n20 1", "output": "2" }, { "input": "3 1\n1 2 3", "output": "1" }, { "input": "3 1\n2 1 3", "output": "2" }, { "input": "3 1\n3 2 1", "output": "3" }, { "input": "3 2\n1 2 3", "output": "1" }, { "input": "3 2\n3 2 1", "output": "2" }, { "input": "3 3\n1 2 3", "output": "1" }, { "input": "4 2\n9 8 11 7", "output": "1" }, { "input": "4 2\n10 1 2 3", "output": "2" }, { "input": "6 3\n56 56 56 2 1 2", "output": "4" }, { "input": "8 3\n1 1 1 1 2 60 90 1", "output": "1" }, { "input": "4 1\n1 5 2 2", "output": "1" }, { "input": "4 2\n4 6 7 4", "output": "1" }, { "input": "10 4\n1 1 1 4 4 4 4 4 4 3", "output": "1" }, { "input": "6 3\n1 2 1 3 1 1", "output": "1" }, { "input": "5 2\n100 100 100 1 1", "output": "4" } ]

1,685,903,866

2,147,483,647

Python 3

OK

TESTS

51

109

15,769,600

n, k = list(map(int, input().split())) arr = list(map(int, input().split())) from math import * summ = 0 min_sum = inf for right in range(k): summ += arr[right] min_sum = min(min_sum, summ) idx = 0 for i in range(k, n): summ = summ - arr[i - k] + arr[i] if summ < min_sum: min_sum = summ idx = i - k + 1 print(idx + 1)

Title: Fence Time Limit: None seconds Memory Limit: None megabytes Problem Description: There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights. Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible. Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic). Input Specification: The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence. Output Specification: Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them. Demo Input: ['7 3\n1 2 6 1 1 7 1\n'] Demo Output: ['3\n'] Note: In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

```python n, k = list(map(int, input().split())) arr = list(map(int, input().split())) from math import * summ = 0 min_sum = inf for right in range(k): summ += arr[right] min_sum = min(min_sum, summ) idx = 0 for i in range(k, n): summ = summ - arr[i - k] + arr[i] if summ < min_sum: min_sum = summ idx = i - k + 1 print(idx + 1) ```

3

707

A

Brain's Photos

PROGRAMMING

800

[ "implementation" ]

null

Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead. As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such). Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour! As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white. Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors: - 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black) The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.

The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively. Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.

Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.

[ "2 2\nC M\nY Y\n", "3 2\nW W\nW W\nB B\n", "1 1\nW\n" ]

[ "#Color", "#Black&White", "#Black&White" ]

none

500

[ { "input": "2 2\nC M\nY Y", "output": "#Color" }, { "input": "3 2\nW W\nW W\nB B", "output": "#Black&White" }, { "input": "1 1\nW", "output": "#Black&White" }, { "input": "2 3\nW W W\nB G Y", "output": "#Color" }, { "input": "1 1\nW", "output": "#Black&White" }, { "input": "5 5\nW G B Y M\nG B Y M C\nB Y M C W\nY M C W G\nM C W G B", "output": "#Color" }, { "input": "1 6\nC M Y W G B", "output": "#Color" }, { "input": "1 3\nW G B", "output": "#Black&White" }, { "input": "1 1\nW", "output": "#Black&White" }, { "input": "5 5\nW G B W G\nG B W G B\nB W G B W\nW G B W G\nG B W G B", "output": "#Black&White" }, { "input": "2 3\nW W W\nB G C", "output": "#Color" }, { "input": "2 3\nW W W\nB G M", "output": "#Color" }, { "input": "3 3\nC B W\nB Y M\nB B W", "output": "#Color" }, { "input": "1 3\nW C W", "output": "#Color" }, { "input": "3 3\nB W B\nB C W\nB W W", "output": "#Color" }, { "input": "1 2\nW Y", "output": "#Color" }, { "input": "1 1\nG", "output": "#Black&White" }, { "input": "1 3\nB G W", "output": "#Black&White" }, { "input": "2 2\nW W\nB C", "output": "#Color" }, { "input": "1 1\nM", "output": "#Color" }, { "input": "1 2\nW C", "output": "#Color" }, { "input": "2 3\nW W M\nW W M", "output": "#Color" }, { "input": "3 2\nW W\nW W\nB C", "output": "#Color" }, { "input": "2 3\nW W C\nW W W", "output": "#Color" }, { "input": "1 3\nG G G", "output": "#Black&White" }, { "input": "1 1\nC", "output": "#Color" }, { "input": "1 2\nC W", "output": "#Color" }, { "input": "1 3\nW W C", "output": "#Color" }, { "input": "2 2\nW B\nB G", "output": "#Black&White" }, { "input": "2 2\nB B\nY Y", "output": "#Color" }, { "input": "2 2\nG G\nC C", "output": "#Color" }, { "input": "1 2\nB C", "output": "#Color" }, { "input": "1 4\nG G G C", "output": "#Color" }, { "input": "1 1\nB", "output": "#Black&White" }, { "input": "2 1\nY\nB", "output": "#Color" } ]

1,692,145,421

2,147,483,647

Python 3

OK

TESTS

50

46

0

A = [] def Color (A) : for i in A: for j in i: if j!='B' and j!='W' and j!='G' : return '#Color' return '#Black&White' a = list(map(int,input().split())) for i in range(a[0]): b = list(map(str, input().split())) A.append(b) print(Color(A))

Title: Brain's Photos Time Limit: None seconds Memory Limit: None megabytes Problem Description: Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead. As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such). Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour! As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white. Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors: - 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black) The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. Input Specification: The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively. Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. Output Specification: Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. Demo Input: ['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n'] Demo Output: ['#Color', '#Black&White', '#Black&White'] Note: none

```python A = [] def Color (A) : for i in A: for j in i: if j!='B' and j!='W' and j!='G' : return '#Color' return '#Black&White' a = list(map(int,input().split())) for i in range(a[0]): b = list(map(str, input().split())) A.append(b) print(Color(A)) ```

3

707

C

Pythagorean Triples

PROGRAMMING

1,500

[ "math", "number theory" ]

null

Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples. For example, triples (3,<=4,<=5), (5,<=12,<=13) and (6,<=8,<=10) are Pythagorean triples. Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse. Katya had no problems with completing this task. Will you do the same?

The only line of the input contains single integer *n* (1<=≤<=*n*<=≤<=109) — the length of some side of a right triangle.

Print two integers *m* and *k* (1<=≤<=*m*,<=*k*<=≤<=1018), such that *n*, *m* and *k* form a Pythagorean triple, in the only line. In case if there is no any Pythagorean triple containing integer *n*, print <=-<=1 in the only line. If there are many answers, print any of them.

[ "3\n", "6\n", "1\n", "17\n", "67\n" ]

[ "4 5", "8 10", "-1", "144 145", "2244 2245" ]

Illustration for the first sample.

1,500

[ { "input": "3", "output": "4 5" }, { "input": "6", "output": "8 10" }, { "input": "1", "output": "-1" }, { "input": "17", "output": "144 145" }, { "input": "67", "output": "2244 2245" }, { "input": "10", "output": "24 26" }, { "input": "14", "output": "48 50" }, { "input": "22", "output": "120 122" }, { "input": "23", "output": "264 265" }, { "input": "246", "output": "15128 15130" }, { "input": "902", "output": "203400 203402" }, { "input": "1000000000", "output": "1250000000 750000000" }, { "input": "1998", "output": "998000 998002" }, { "input": "2222222", "output": "1234567654320 1234567654322" }, { "input": "2222226", "output": "1234572098768 1234572098770" }, { "input": "1111110", "output": "308641358024 308641358026" }, { "input": "9999998", "output": "24999990000000 24999990000002" }, { "input": "1024", "output": "1280 768" }, { "input": "8388608", "output": "10485760 6291456" }, { "input": "4", "output": "5 3" }, { "input": "8", "output": "10 6" }, { "input": "16", "output": "20 12" }, { "input": "492", "output": "615 369" }, { "input": "493824", "output": "617280 370368" }, { "input": "493804", "output": "617255 370353" }, { "input": "493800", "output": "617250 370350" }, { "input": "2048", "output": "2560 1536" }, { "input": "8388612", "output": "10485765 6291459" }, { "input": "44", "output": "55 33" }, { "input": "444", "output": "555 333" }, { "input": "4444", "output": "5555 3333" }, { "input": "44444", "output": "55555 33333" }, { "input": "444444", "output": "555555 333333" }, { "input": "4444444", "output": "5555555 3333333" }, { "input": "100000000", "output": "125000000 75000000" }, { "input": "2", "output": "-1" }, { "input": "3", "output": "4 5" }, { "input": "5", "output": "12 13" }, { "input": "7", "output": "24 25" }, { "input": "9", "output": "40 41" }, { "input": "11", "output": "60 61" }, { "input": "13", "output": "84 85" }, { "input": "15", "output": "112 113" }, { "input": "19", "output": "180 181" }, { "input": "111", "output": "6160 6161" }, { "input": "113", "output": "6384 6385" }, { "input": "115", "output": "6612 6613" }, { "input": "117", "output": "6844 6845" }, { "input": "119", "output": "7080 7081" }, { "input": "111111", "output": "6172827160 6172827161" }, { "input": "111113", "output": "6173049384 6173049385" }, { "input": "111115", "output": "6173271612 6173271613" }, { "input": "111117", "output": "6173493844 6173493845" }, { "input": "111119", "output": "6173716080 6173716081" }, { "input": "9999993", "output": "49999930000024 49999930000025" }, { "input": "9999979", "output": "49999790000220 49999790000221" }, { "input": "9999990", "output": "24999950000024 24999950000026" }, { "input": "9999991", "output": "49999910000040 49999910000041" }, { "input": "9999992", "output": "12499990 7499994" }, { "input": "9999973", "output": "49999730000364 49999730000365" }, { "input": "9999994", "output": "24999970000008 24999970000010" }, { "input": "9999995", "output": "49999950000012 49999950000013" }, { "input": "9999996", "output": "12499995 7499997" }, { "input": "9999997", "output": "49999970000004 49999970000005" }, { "input": "9999978", "output": "24999890000120 24999890000122" }, { "input": "99999993", "output": "4999999300000024 4999999300000025" }, { "input": "99999979", "output": "4999997900000220 4999997900000221" }, { "input": "99999990", "output": "2499999500000024 2499999500000026" }, { "input": "99999991", "output": "4999999100000040 4999999100000041" }, { "input": "99999992", "output": "124999990 74999994" }, { "input": "99999973", "output": "4999997300000364 4999997300000365" }, { "input": "99999994", "output": "2499999700000008 2499999700000010" }, { "input": "99999995", "output": "4999999500000012 4999999500000013" }, { "input": "99999996", "output": "124999995 74999997" }, { "input": "99999997", "output": "4999999700000004 4999999700000005" }, { "input": "99999978", "output": "2499998900000120 2499998900000122" }, { "input": "987654323", "output": "487730530870294164 487730530870294165" }, { "input": "2", "output": "-1" }, { "input": "4", "output": "5 3" }, { "input": "8", "output": "10 6" }, { "input": "64", "output": "80 48" }, { "input": "999999999", "output": "499999999000000000 499999999000000001" }, { "input": "16", "output": "20 12" }, { "input": "999999937", "output": "499999937000001984 499999937000001985" }, { "input": "999999998", "output": "249999999000000000 249999999000000002" }, { "input": "433494437", "output": "93958713454973484 93958713454973485" }, { "input": "484916147", "output": "117571834810662804 117571834810662805" }, { "input": "999999929", "output": "499999929000002520 499999929000002521" }, { "input": "982451653", "output": "482605625241216204 482605625241216205" }, { "input": "2048", "output": "2560 1536" } ]

1,681,307,754

2,147,483,647

Python 3

OK

TESTS

90

46

0

n = int(input()) if n < 3: print(-1) elif n % 2 == 1: k = 1 a = (n**2//k - k)//2 b = (n**2//k + k)//2 print(a, b) else: k = 2 a = (n ** 2 // k - k) // 2 b = (n ** 2 // k + k) // 2 print(a, b)

Title: Pythagorean Triples Time Limit: None seconds Memory Limit: None megabytes Problem Description: Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples. For example, triples (3,<=4,<=5), (5,<=12,<=13) and (6,<=8,<=10) are Pythagorean triples. Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse. Katya had no problems with completing this task. Will you do the same? Input Specification: The only line of the input contains single integer *n* (1<=≤<=*n*<=≤<=109) — the length of some side of a right triangle. Output Specification: Print two integers *m* and *k* (1<=≤<=*m*,<=*k*<=≤<=1018), such that *n*, *m* and *k* form a Pythagorean triple, in the only line. In case if there is no any Pythagorean triple containing integer *n*, print <=-<=1 in the only line. If there are many answers, print any of them. Demo Input: ['3\n', '6\n', '1\n', '17\n', '67\n'] Demo Output: ['4 5', '8 10', '-1', '144 145', '2244 2245'] Note: Illustration for the first sample.

```python n = int(input()) if n < 3: print(-1) elif n % 2 == 1: k = 1 a = (n**2//k - k)//2 b = (n**2//k + k)//2 print(a, b) else: k = 2 a = (n ** 2 // k - k) // 2 b = (n ** 2 // k + k) // 2 print(a, b) ```

3

52

A

123-sequence

PROGRAMMING

900

[ "implementation" ]

A. 123-sequence

2

256

There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).

Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.

[ "9\n1 3 2 2 2 1 1 2 3\n" ]

[ "5\n" ]

In the example all the numbers equal to 1 and 3 should be replaced by 2.

500

[ { "input": "9\n1 3 2 2 2 1 1 2 3", "output": "5" }, { "input": "6\n3 3 2 2 1 3", "output": "3" }, { "input": "12\n3 1 3 1 2 1 3 2 2 1 2 1", "output": "7" }, { "input": "15\n3 2 1 1 1 1 3 2 2 3 3 1 2 3 2", "output": "10" }, { "input": "2\n2 1", "output": "1" }, { "input": "2\n3 2", "output": "1" }, { "input": "2\n3 1", "output": "1" }, { "input": "18\n2 3 2 1 2 3 2 1 2 3 3 3 1 2 3 3 3 2", "output": "10" }, { "input": "30\n2 1 3 2 3 2 2 2 2 3 2 2 3 2 1 1 3 1 3 2 1 2 3 1 1 3 3 1 3 1", "output": "19" }, { "input": "1\n1", "output": "0" }, { "input": "1\n2", "output": "0" }, { "input": "1\n3", "output": "0" } ]

1,600,524,465

2,147,483,647

Python 3

OK

TESTS

23

904

14,643,200

n=int(input()) count=0 one=0 two=0 three=0 x=list(map(int,input().split())) for i in x: if i==1: one=one+1 elif i==2: two=two+1 elif i==3: three=three+1 max_=max(one,two ,three) answer=n-max_ print(answer)

Title: 123-sequence Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3). Output Specification: Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal. Demo Input: ['9\n1 3 2 2 2 1 1 2 3\n'] Demo Output: ['5\n'] Note: In the example all the numbers equal to 1 and 3 should be replaced by 2.

```python n=int(input()) count=0 one=0 two=0 three=0 x=list(map(int,input().split())) for i in x: if i==1: one=one+1 elif i==2: two=two+1 elif i==3: three=three+1 max_=max(one,two ,three) answer=n-max_ print(answer) ```

3.746725

1,009

C

Annoying Present

PROGRAMMING

1,700

[ "greedy", "math" ]

null

Alice got an array of length $n$ as a birthday present once again! This is the third year in a row! And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice. Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1 \le i \le n$) and for every $j \in [1, n]$ adds $x + d \cdot dist(i, j)$ to the value of the $j$-th cell. $dist(i, j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i, j) = |i - j|$, where $|x|$ is an absolute value of $x$). For example, if Alice currently has an array $[2, 1, 2, 2]$ and Bob chooses position $3$ for $x = -1$ and $d = 2$ then the array will become $[2 - 1 + 2 \cdot 2,~1 - 1 + 2 \cdot 1,~2 - 1 + 2 \cdot 0,~2 - 1 + 2 \cdot 1]$ = $[5, 2, 1, 3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$). Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric. What is the maximum arithmetic mean value Bob can achieve?

The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of elements of the array and the number of changes. Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-10^3 \le x_i, d_i \le 10^3$) — the parameters for the $i$-th change.

Print the maximal average arithmetic mean of the elements Bob can achieve. Your answer is considered correct if its absolute or relative error doesn't exceed $10^{-6}$.

[ "2 3\n-1 3\n0 0\n-1 -4\n", "3 2\n0 2\n5 0\n" ]

[ "-2.500000000000000\n", "7.000000000000000\n" ]

none

0

[ { "input": "2 3\n-1 3\n0 0\n-1 -4", "output": "-2.500000000000000" }, { "input": "3 2\n0 2\n5 0", "output": "7.000000000000000" }, { "input": "8 8\n-21 -60\n-96 -10\n-4 -19\n-27 -4\n57 -15\n-95 62\n-42 1\n-17 64", "output": "-16.500000000000000" }, { "input": "1 1\n0 0", "output": "0.000000000000000" }, { "input": "100000 1\n1000 1000", "output": "50000500.000000000000000" }, { "input": "11 1\n0 -10", "output": "-27.272727272727273" }, { "input": "3 1\n1 -1", "output": "0.333333333333333" }, { "input": "1 2\n-1 -1\n-2 -2", "output": "-3.000000000000000" }, { "input": "1 2\n0 -1\n0 1", "output": "0.000000000000000" }, { "input": "1 1\n1 -2", "output": "1.000000000000000" }, { "input": "3 1\n2 -1", "output": "1.333333333333333" }, { "input": "3 1\n0 -1", "output": "-0.666666666666667" }, { "input": "1 1\n-1000 -1000", "output": "-1000.000000000000000" }, { "input": "1 1\n0 -5", "output": "0.000000000000000" }, { "input": "15 3\n2 0\n2 -5\n-2 5", "output": "18.333333333333332" }, { "input": "9 1\n0 -5", "output": "-11.111111111111111" }, { "input": "7 1\n0 -1", "output": "-1.714285714285714" }, { "input": "3 1\n-2 -2", "output": "-3.333333333333333" }, { "input": "3 1\n5 -5", "output": "1.666666666666667" }, { "input": "1 1\n-1 -1", "output": "-1.000000000000000" }, { "input": "7 1\n-1 -5", "output": "-9.571428571428571" }, { "input": "3 2\n-2 -2\n-2 -2", "output": "-6.666666666666667" }, { "input": "5 1\n0 -4", "output": "-4.800000000000000" }, { "input": "5 1\n-1 -5", "output": "-7.000000000000000" }, { "input": "5 1\n0 -2", "output": "-2.400000000000000" }, { "input": "3 5\n1 -1000\n1 -1000\n1 -1000\n1 -1000\n1 -1000", "output": "-3328.333333333333485" }, { "input": "1 1\n0 -1", "output": "0.000000000000000" }, { "input": "1 2\n0 -3\n0 -3", "output": "0.000000000000000" }, { "input": "7 1\n2 -3", "output": "-3.142857142857143" }, { "input": "3 2\n-1 -1\n-1 -1", "output": "-3.333333333333333" }, { "input": "5 1\n-1 -162", "output": "-195.400000000000006" }, { "input": "5 10\n-506 -243\n727 -141\n-548 -306\n740 880\n-744 -116\n-84 182\n-859 -108\n64 86\n135 446\n69 -184", "output": "864.399999999999977" }, { "input": "5 1\n0 -1", "output": "-1.200000000000000" }, { "input": "5 12\n634 895\n143 730\n901 245\n386 486\n395 -111\n-469 -104\n-681 -623\n-900 843\n889 -883\n476 -304\n777 986\n206 -491", "output": "8107.800000000000182" }, { "input": "3 3\n4 2\n5 0\n6 -1", "output": "16.333333333333332" }, { "input": "1 3\n4 2\n5 0\n6 -1", "output": "15.000000000000000" }, { "input": "85 10\n-223 435\n-771 455\n72 -940\n490 -178\n400 -117\n169 -527\n836 610\n849 944\n572 -237\n-428 -428", "output": "53047.388235294114565" }, { "input": "69 10\n-8 4\n-3 3\n7 5\n5 -9\n8 1\n7 -5\n-8 -8\n9 3\n1 1\n0 6", "output": "420.579710144927560" }, { "input": "1 10\n1 1\n1 0\n1 0\n1 0\n-1 0\n0 1\n1 0\n0 0\n2 1\n9 2", "output": "15.000000000000000" }, { "input": "5 4\n0 1\n0 2\n0 3\n0 -9", "output": "1.200000000000000" } ]

1,682,173,812

2,147,483,647

PyPy 3-64

OK

TESTS

54

124

3,891,200

import sys input = sys.stdin.readline def f(u): return u*(u+1)//2 n, m = map(int, input().split()) x1, x2 = n//2, (n-1)//2 x, y, c, d = f(x1) + f(x2), f(n-1), 0, 0 for i in range(m): a, b = map(int, input().split()) c += a d += max(b*x, b*y) print((c*n+d)/n)

Title: Annoying Present Time Limit: None seconds Memory Limit: None megabytes Problem Description: Alice got an array of length $n$ as a birthday present once again! This is the third year in a row! And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice. Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1 \le i \le n$) and for every $j \in [1, n]$ adds $x + d \cdot dist(i, j)$ to the value of the $j$-th cell. $dist(i, j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i, j) = |i - j|$, where $|x|$ is an absolute value of $x$). For example, if Alice currently has an array $[2, 1, 2, 2]$ and Bob chooses position $3$ for $x = -1$ and $d = 2$ then the array will become $[2 - 1 + 2 \cdot 2,~1 - 1 + 2 \cdot 1,~2 - 1 + 2 \cdot 0,~2 - 1 + 2 \cdot 1]$ = $[5, 2, 1, 3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$). Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric. What is the maximum arithmetic mean value Bob can achieve? Input Specification: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of elements of the array and the number of changes. Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-10^3 \le x_i, d_i \le 10^3$) — the parameters for the $i$-th change. Output Specification: Print the maximal average arithmetic mean of the elements Bob can achieve. Your answer is considered correct if its absolute or relative error doesn't exceed $10^{-6}$. Demo Input: ['2 3\n-1 3\n0 0\n-1 -4\n', '3 2\n0 2\n5 0\n'] Demo Output: ['-2.500000000000000\n', '7.000000000000000\n'] Note: none

```python import sys input = sys.stdin.readline def f(u): return u*(u+1)//2 n, m = map(int, input().split()) x1, x2 = n//2, (n-1)//2 x, y, c, d = f(x1) + f(x2), f(n-1), 0, 0 for i in range(m): a, b = map(int, input().split()) c += a d += max(b*x, b*y) print((c*n+d)/n) ```

3

92

A

Chips

PROGRAMMING

800

[ "implementation", "math" ]

A. Chips

2

256

There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*. The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.

The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.

Print the number of chips the presenter ended up with.

[ "4 11\n", "17 107\n", "3 8\n" ]

[ "0\n", "2\n", "1\n" ]

In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes. In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.

500

[ { "input": "4 11", "output": "0" }, { "input": "17 107", "output": "2" }, { "input": "3 8", "output": "1" }, { "input": "46 7262", "output": "35" }, { "input": "32 6864", "output": "0" }, { "input": "36 6218", "output": "14" }, { "input": "25 9712", "output": "11" }, { "input": "9 7601", "output": "5" }, { "input": "1 9058", "output": "0" }, { "input": "29 7772", "output": "26" }, { "input": "45 9465", "output": "14" }, { "input": "46 866", "output": "5" }, { "input": "29 1241", "output": "20" }, { "input": "17 4248", "output": "12" }, { "input": "20 8082", "output": "11" }, { "input": "50 9555", "output": "0" }, { "input": "4 7455", "output": "2" }, { "input": "36 880", "output": "4" }, { "input": "24 7440", "output": "9" }, { "input": "44 7888", "output": "12" }, { "input": "1 1", "output": "0" }, { "input": "50 10000", "output": "40" }, { "input": "1 10000", "output": "0" }, { "input": "50 1", "output": "0" }, { "input": "50 50", "output": "5" } ]

1,507,366,718

2,147,483,647

Python 3

OK

TESTS

25

124

0

n,m = map(int,input().split()) s = int(n*(n+1)/2) m = m%s for i in range(1,n): if(m<i): break m= m-i print(m)

Title: Chips Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*. The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. Output Specification: Print the number of chips the presenter ended up with. Demo Input: ['4 11\n', '17 107\n', '3 8\n'] Demo Output: ['0\n', '2\n', '1\n'] Note: In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes. In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.

```python n,m = map(int,input().split()) s = int(n*(n+1)/2) m = m%s for i in range(1,n): if(m<i): break m= m-i print(m) ```

3.969

774

D

Lie or Truth

PROGRAMMING

1,500

[ "*special", "constructive algorithms", "implementation", "sortings" ]

null

Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*. While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*. Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way). Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother.

The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order. It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes.

Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes).

[ "5 2 4\n3 4 2 3 1\n3 2 3 4 1\n", "3 1 2\n1 2 3\n3 1 2\n", "4 2 4\n1 1 1 1\n1 1 1 1\n" ]

[ "TRUTH\n", "LIE\n", "TRUTH\n" ]

In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]). In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother. In the third example for any values *l* and *r* there is a situation when Stepan said the truth.

0

[ { "input": "5 2 4\n3 4 2 3 1\n3 2 3 4 1", "output": "TRUTH" }, { "input": "3 1 2\n1 2 3\n3 1 2", "output": "LIE" }, { "input": "4 2 4\n1 1 1 1\n1 1 1 1", "output": "TRUTH" }, { "input": "5 1 3\n2 2 2 1 2\n2 2 2 1 2", "output": "TRUTH" }, { "input": "7 1 4\n2 5 5 5 4 3 4\n2 5 5 5 4 3 4", "output": "TRUTH" }, { "input": "10 1 10\n6 7 6 1 10 10 9 5 3 9\n7 10 9 6 1 5 9 3 10 6", "output": "TRUTH" }, { "input": "1 1 1\n1\n1", "output": "TRUTH" }, { "input": "4 3 4\n1 2 3 4\n2 1 3 4", "output": "LIE" }, { "input": "7 2 4\n1 2 3 4 5 7 6\n1 2 3 4 5 6 7", "output": "LIE" }, { "input": "5 1 2\n1 2 3 4 5\n1 2 3 5 4", "output": "LIE" }, { "input": "8 3 6\n5 3 1 1 1 1 3 5\n3 3 1 1 1 1 5 5", "output": "LIE" }, { "input": "4 2 2\n2 1 2 2\n1 2 2 2", "output": "LIE" } ]

1,491,407,788

1,288

Python 3

OK

TESTS

52

140

15,155,200

import sys def main(): yes = True array = sys.stdin.readline().strip().split(' ') array1 = sys.stdin.readline().strip().split(' ') array2 = sys.stdin.readline().strip().split(' ') for i in range(len(array1)): if i < int(array[1])-1 or i >= int(array[2]): if array1[i] != array2[i]: yes = False break if yes: print("TRUTH") else: print("LIE") return main()

Title: Lie or Truth Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*. While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*. Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way). Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother. Input Specification: The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan. The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order. The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order. It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes. Output Specification: Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes). Demo Input: ['5 2 4\n3 4 2 3 1\n3 2 3 4 1\n', '3 1 2\n1 2 3\n3 1 2\n', '4 2 4\n1 1 1 1\n1 1 1 1\n'] Demo Output: ['TRUTH\n', 'LIE\n', 'TRUTH\n'] Note: In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]). In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother. In the third example for any values *l* and *r* there is a situation when Stepan said the truth.

```python import sys def main(): yes = True array = sys.stdin.readline().strip().split(' ') array1 = sys.stdin.readline().strip().split(' ') array2 = sys.stdin.readline().strip().split(' ') for i in range(len(array1)): if i < int(array[1])-1 or i >= int(array[2]): if array1[i] != array2[i]: yes = False break if yes: print("TRUTH") else: print("LIE") return main() ```

3

776

B

Sherlock and his girlfriend

PROGRAMMING

1,200

[ "constructive algorithms", "number theory" ]

null

Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry. He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1. Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used. Help Sherlock complete this trivial task.

The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.

The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints. The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price. If there are multiple ways to color the pieces using *k* colors, you can output any of them.

[ "3\n", "4\n" ]

[ "2\n1 1 2 ", "2\n2 1 1 2\n" ]

In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively. In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.

1,000

[ { "input": "3", "output": "2\n1 1 2 " }, { "input": "4", "output": "2\n1 1 2 1 " }, { "input": "17", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 " }, { "input": "25", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 " }, { "input": "85", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 " }, { "input": "105", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 " }, { "input": "123", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 " }, { "input": "452", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "641", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "293", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "733", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "1", "output": "1\n1 " }, { "input": "10", "output": "2\n1 1 2 1 2 1 2 2 2 1 " }, { "input": "287", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "3202", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "728", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "3509", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "5137", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 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2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "2", "output": "1\n1 1 " }, { "input": "1", "output": "1\n1 " }, { "input": "2", "output": "1\n1 1 " }, { "input": "100000", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." }, { "input": "99971", "output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..." } ]

1,521,092,554

2,147,483,647

Python 3

OK

TESTS

53

623

6,758,400

n=int(input()) a=[0]*(n+2) for i in range(2,n+2): for j in range(i,n+2,i): a[j]+=1 if n>2:print(2) else:print(1) for i in range(2,n+2): if a[i]==1: print(1,end=' ') else: print(2,end=' ')

Title: Sherlock and his girlfriend Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry. He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1. Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used. Help Sherlock complete this trivial task. Input Specification: The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces. Output Specification: The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints. The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price. If there are multiple ways to color the pieces using *k* colors, you can output any of them. Demo Input: ['3\n', '4\n'] Demo Output: ['2\n1 1 2 ', '2\n2 1 1 2\n'] Note: In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively. In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.

```python n=int(input()) a=[0]*(n+2) for i in range(2,n+2): for j in range(i,n+2,i): a[j]+=1 if n>2:print(2) else:print(1) for i in range(2,n+2): if a[i]==1: print(1,end=' ') else: print(2,end=' ') ```

3

133

A

HQ9+

PROGRAMMING

900

[ "implementation" ]

null

HQ9+ is a joke programming language which has only four one-character instructions: - "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator. Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored. You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.

The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.

Output "YES", if executing the program will produce any output, and "NO" otherwise.

[ "Hi!\n", "Codeforces\n" ]

[ "YES\n", "NO\n" ]

In the first case the program contains only one instruction — "H", which prints "Hello, World!". In the second case none of the program characters are language instructions.

500

[ { "input": "Hi!", "output": "YES" }, { "input": "Codeforces", "output": "NO" }, { "input": "a+b=c", "output": "NO" }, { "input": "hq-lowercase", "output": "NO" }, { "input": "Q", "output": "YES" }, { "input": "9", "output": "YES" }, { "input": "H", "output": "YES" }, { "input": "+", "output": "NO" }, { "input": "~", "output": "NO" }, { "input": "dEHsbM'gS[\\brZ_dpjXw8f?L[4E\"s4Zc9*(,j:>p$}m7HD[_9nOWQ\\uvq2mHWR", "output": "YES" }, { "input": "tt6l=RHOfStm.;Qd$-}zDes*E,.F7qn5-b%HC", "output": "YES" }, { "input": "@F%K2=%RyL/", "output": "NO" }, { "input": "juq)k(FT.^G=G\\zcqnO\"uJIE1_]KFH9S=1c\"mJ;F9F)%>&.WOdp09+k`Yc6}\"6xw,Aos:M\\_^^:xBb[CcsHm?J", "output": "YES" }, { "input": "6G_\"Fq#<AWyHG=Rci1t%#Jc#x<Fpg'N@t%F=``YO7\\Zd;6PkMe<#91YgzTC)", "output": "YES" }, { "input": "Fvg_~wC>SO4lF}*c`Q;mII9E{4.QodbqN]C", "output": "YES" }, { "input": "p-UXsbd&f", "output": "NO" }, { "input": "<]D7NMA)yZe=`?RbP5lsa.l_Mg^V:\"-0x+$3c,q&L%18Ku<HcA\\s!^OQblk^x{35S'>yz8cKgVHWZ]kV0>_", "output": "YES" }, { "input": "f.20)8b+.R}Gy!DbHU3v(.(=Q^`z[_BaQ}eO=C1IK;b2GkD\\{\\Bf\"!#qh]", "output": "YES" }, { "input": "}do5RU<(w<q[\"-NR)IAH_HyiD{", "output": "YES" }, { "input": "Iy^.,Aw*,5+f;l@Q;jLK'G5H-r1Pfmx?ei~`CjMmUe{K:lS9cu4ay8rqRh-W?Gqv!e-j*U)!Mzn{E8B6%~aSZ~iQ_QwlC9_cX(o8", "output": "YES" }, { "input": "sKLje,:q>-D,;NvQ3,qN3-N&tPx0nL/,>Ca|z\"k2S{NF7btLa3_TyXG4XZ:`(t&\"'^M|@qObZxv", "output": "YES" }, { "input": "%z:c@1ZsQ@\\6U/NQ+M9R>,$bwG`U1+C\\18^:S},;kw!&4r|z`", "output": "YES" }, { "input": "OKBB5z7ud81[Tn@P\"nDUd,>@", "output": "NO" }, { "input": "y{0;neX]w0IenPvPx0iXp+X|IzLZZaRzBJ>q~LhMhD$x-^GDwl;,a'<bAqH8QrFwbK@oi?I'W.bZ]MlIQ/x(0YzbTH^l.)]0Bv", "output": "YES" }, { "input": "EL|xIP5_+Caon1hPpQ0[8+r@LX4;b?gMy>;/WH)pf@Ur*TiXu*e}b-*%acUA~A?>MDz#!\\Uh", "output": "YES" }, { "input": "UbkW=UVb>;z6)p@Phr;^Dn.|5O{_i||:Rv|KJ_ay~V(S&Jp", "output": "NO" }, { "input": "!3YPv@2JQ44@)R2O_4`GO", "output": "YES" }, { "input": "Kba/Q,SL~FMd)3hOWU'Jum{9\"$Ld4:GW}D]%tr@G{hpG:PV5-c'VIZ~m/6|3I?_4*1luKnOp`%p|0H{[|Y1A~4-ZdX,Rw2[\\", "output": "YES" }, { "input": "NRN*=v>;oU7[acMIJn*n^bWm!cm3#E7Efr>{g-8bl\"DN4~_=f?[T;~Fq#&)aXq%</GcTJD^e$@Extm[e\"C)q_L", "output": "NO" }, { "input": "y#<fv{_=$MP!{D%I\\1OqjaqKh[pqE$KvYL<9@*V'j8uH0/gQdA'G;&y4Cv6&", "output": "YES" }, { "input": "+SE_Pg<?7Fh,z&uITQut2a-mk8X8La`c2A}", "output": "YES" }, { "input": "Uh3>ER](J", "output": "NO" }, { "input": "!:!{~=9*\\P;Z6F?HC5GadFz)>k*=u|+\"Cm]ICTmB!`L{&oS/z6b~#Snbp/^\\Q>XWU-vY+/dP.7S=-#&whS@,", "output": "YES" }, { "input": "KimtYBZp+ISeO(uH;UldoE6eAcp|9u?SzGZd6j-e}[}u#e[Cx8.qgY]$2!", "output": "YES" }, { "input": "[:[SN-{r>[l+OggH3v3g{EPC*@YBATT@", "output": "YES" }, { "input": "'jdL(vX", "output": "NO" }, { "input": "Q;R+aay]cL?Zh*uG\"YcmO*@Dts*Gjp}D~M7Z96+<4?9I3aH~0qNdO(RmyRy=ci,s8qD_kwj;QHFzD|5,5", "output": "YES" }, { "input": "{Q@#<LU_v^qdh%gGxz*pu)Y\"]k-l-N30WAxvp2IE3:jD0Wi4H/xWPH&s", "output": "YES" }, { "input": "~@Gb(S&N$mBuBUMAky-z^{5VwLNTzYg|ZUZncL@ahS?K*As<$iNUARM3r43J'jJB)$ujfPAq\"G<S9flGyakZg!2Z.-NJ|2{F>]", "output": "YES" }, { "input": "Jp5Aa>aP6fZ!\\6%A}<S}j{O4`C6y$8|i3IW,WHy&\"ioE&7zP\"'xHAY;:x%@SnS]Mr{R|})gU", "output": "YES" }, { "input": "ZA#:U)$RI^sE\\vuAt]x\"2zipI!}YEu2<j$:H0_9/~eB?#->", "output": "YES" }, { "input": "&ppw0._:\\p-PuWM@l}%%=", "output": "NO" }, { "input": "P(^pix\"=oiEZu8?@d@J(I`Xp5TN^T3\\Z7P5\"ZrvZ{2Fwz3g-8`U!)(1$a<g+9Q|COhDoH;HwFY02Pa|ZGp$/WZBR=>6Jg!yr", "output": "YES" }, { "input": "`WfODc\\?#ax~1xu@[ao+o_rN|L7%v,p,nDv>3+6cy.]q3)+A6b!q*Hc+#.t4f~vhUa~$^q", "output": "YES" }, { "input": ",)TH9N}'6t2+0Yg?S#6/{_.,!)9d}h'wG|sY&'Ul4D0l0", "output": "YES" }, { "input": "VXB&r9Z)IlKOJ:??KDA", "output": "YES" }, { "input": "\")1cL>{o\\dcYJzu?CefyN^bGRviOH&P7rJS3PT4:0V3F)%\\}L=AJouYsj_>j2|7^1NWu*%NbOP>ngv-ls<;b-4Sd3Na0R", "output": "YES" }, { "input": "2Y}\\A)>row{~c[g>:'.|ZC8%UTQ/jcdhK%6O)QRC.kd@%y}LJYk=V{G5pQK/yKJ%{G3C", "output": "YES" }, { "input": "O.&=qt(`z(", "output": "NO" }, { "input": "_^r6fyIc/~~;>l%9?aVEi7-{=,[<aMiB'-scSg$$|\"jAzY0N>QkHHGBZj2c\"=fhRlWd5;5K|GgU?7h]!;wl@", "output": "YES" }, { "input": "+/`sAd&eB29E=Nu87${.u6GY@$^a$,}s^!p!F}B-z8<<wORb<S7;HM1a,gp", "output": "YES" }, { "input": "U_ilyOGMT+QiW/M8/D(1=6a7)_FA,h4`8", "output": "YES" }, { "input": "!0WKT:$O", "output": "NO" }, { "input": "1EE*I%EQz6$~pPu7|(r7nyPQt4uGU@]~H'4uII?b1_Wn)K?ZRHrr0z&Kr;}aO3<mN=3:{}QgPxI|Ncm4#)", "output": "YES" }, { "input": "[u3\"$+!:/.<Dp1M7tH}:zxjt],^kv}qP;y12\"`^'/u*h%AFmPJ>e1#Yly", "output": "YES" }, { "input": "'F!_]tB<A&UO+p?7liE>(x&RFgG2~\\(", "output": "NO" }, { "input": "Qv)X8", "output": "YES" }, { "input": "aGv7,J@&g1(}E3g6[LuDZwZl2<v7IwQA%\"R(?ouBD>_=y\"3Kf%^>vON<a^T\\G^ootgE@whWmZo=[ex|F", "output": "YES" }, { "input": "e{}2vQ+/r@p0}cLKNe4MCk", "output": "YES" }, { "input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnkH", "output": "YES" }, { "input": "Qzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnky", "output": "YES" }, { "input": "mzbmweyydiadtlcouegmdbyfwurpwb9uvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnky", "output": "YES" }, { "input": "1H1", "output": "YES" }, { "input": "+Q", "output": "YES" }, { "input": "1ab", "output": "NO" }, { "input": "!", "output": "NO" }, { "input": "0+", "output": "NO" }, { "input": "+H", "output": "YES" }, { "input": "cH", "output": "YES" }, { "input": "+8", "output": "NO" }, { "input": "8+", "output": "NO" }, { "input": "++++++++++++++++++++++++++", "output": "NO" }, { "input": "(+)", "output": "NO" }, { "input": "H+", "output": "YES" }, { "input": "a!", "output": "NO" }, { "input": "++++++++++++++", "output": "NO" }, { "input": "+++++++++++++++++++++++++++++++++++++++++++++++++++++++++", "output": "NO" }, { "input": "8", "output": "NO" }, { "input": "3", "output": "NO" }, { "input": "HQ9", "output": "YES" }, { "input": "+++H", "output": "YES" }, { "input": "++++++++++", "output": "NO" }, { "input": "HHHHHHH", "output": "YES" }, { "input": "abacabaH", "output": "YES" }, { "input": "+G", "output": "NO" } ]

1,689,393,338

2,147,483,647

Python 3

OK

TESTS

85

92

0

print(["NO", "YES"][bool(set("HQ9") & set(input()))])

Title: HQ9+ Time Limit: None seconds Memory Limit: None megabytes Problem Description: HQ9+ is a joke programming language which has only four one-character instructions: - "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator. Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored. You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output. Input Specification: The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive. Output Specification: Output "YES", if executing the program will produce any output, and "NO" otherwise. Demo Input: ['Hi!\n', 'Codeforces\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first case the program contains only one instruction — "H", which prints "Hello, World!". In the second case none of the program characters are language instructions.

```python print(["NO", "YES"][bool(set("HQ9") & set(input()))]) ```

3

678

A

Johny Likes Numbers

PROGRAMMING

800

[ "implementation", "math" ]

null

Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.

The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).

Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.

[ "5 3\n", "25 13\n", "26 13\n" ]

[ "6\n", "26\n", "39\n" ]

none

0

[ { "input": "5 3", "output": "6" }, { "input": "25 13", "output": "26" }, { "input": "26 13", "output": "39" }, { "input": "1 1", "output": "2" }, { "input": "8 8", "output": "16" }, { "input": "14 15", "output": "15" }, { "input": "197 894", "output": "894" }, { "input": "6058 8581", "output": "8581" }, { "input": "97259 41764", "output": "125292" }, { "input": "453145 333625", "output": "667250" }, { "input": "2233224 4394826", "output": "4394826" }, { "input": "76770926 13350712", "output": "80104272" }, { "input": "687355301 142098087", "output": "710490435" }, { "input": "1000000000 999999999", "output": "1999999998" }, { "input": "1000000000 1000000000", "output": "2000000000" }, { "input": "999999999 1000000000", "output": "1000000000" }, { "input": "1000000000 1", "output": "1000000001" }, { "input": "1000000000 2", "output": "1000000002" }, { "input": "999999999 1", "output": "1000000000" }, { "input": "100000000 1", "output": "100000001" }, { "input": "999999999 500000000", "output": "1000000000" }, { "input": "999999990 10", "output": "1000000000" }, { "input": "1000000000 999999997", "output": "1999999994" }, { "input": "999999999 2", "output": "1000000000" }, { "input": "999999984 1", "output": "999999985" }, { "input": "999999983 1", "output": "999999984" }, { "input": "666666666 1", "output": "666666667" }, { "input": "1000000000 990000000", "output": "1980000000" }, { "input": "41 48", "output": "48" }, { "input": "123456 2", "output": "123458" }, { "input": "111 111", "output": "222" }, { "input": "878787 1", "output": "878788" }, { "input": "121 1", "output": "122" }, { "input": "114514 114514", "output": "229028" }, { "input": "500000001 1000000000", "output": "1000000000" }, { "input": "999999997 1", "output": "999999998" }, { "input": "100000000 10", "output": "100000010" } ]

1,663,689,304

2,147,483,647

Python 3

OK

TESTS

37

46

0

n, k = input().split() n = int(n) k = int(k) print(n-(n%k)+k)

Title: Johny Likes Numbers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*. Input Specification: The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109). Output Specification: Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*. Demo Input: ['5 3\n', '25 13\n', '26 13\n'] Demo Output: ['6\n', '26\n', '39\n'] Note: none

```python n, k = input().split() n = int(n) k = int(k) print(n-(n%k)+k) ```

3

22

A

Second Order Statistics

PROGRAMMING

800

[ "brute force" ]

A. Second Order Statistics

2

256

Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.

The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.

If the given sequence has the second order statistics, output this order statistics, otherwise output NO.

[ "4\n1 2 2 -4\n", "5\n1 2 3 1 1\n" ]

[ "1\n", "2\n" ]

none

0

[ { "input": "4\n1 2 2 -4", "output": "1" }, { "input": "5\n1 2 3 1 1", "output": "2" }, { "input": "1\n28", "output": "NO" }, { "input": "2\n-28 12", "output": "12" }, { "input": "3\n-83 40 -80", "output": "-80" }, { "input": "8\n93 77 -92 26 21 -48 53 91", "output": "-48" }, { "input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49", "output": "-86" }, { "input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94", "output": "-98" }, { "input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46", "output": "-91" }, { "input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89", "output": "-98" }, { "input": "2\n-1 -1", "output": "NO" }, { "input": "3\n-2 -2 -2", "output": "NO" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "NO" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100", "output": "100" }, { "input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47", "output": "40" }, { "input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90", "output": "-64" }, { "input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93", "output": "-93" }, { "input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83", "output": "-86" }, { "input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43", "output": "-95" }, { "input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92", "output": "-92" }, { "input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12", "output": "-98" }, { "input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100", "output": "-98" }, { "input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92", "output": "-98" }, { "input": "3\n1 2 3", "output": "2" }, { "input": "3\n1 3 2", "output": "2" }, { "input": "3\n2 1 3", "output": "2" }, { "input": "3\n2 3 1", "output": "2" }, { "input": "3\n3 1 2", "output": "2" }, { "input": "3\n3 2 1", "output": "2" }, { "input": "9\n99 99 99 99 100 100 100 100 100", "output": "100" }, { "input": "5\n-100 -100 -100 -100 -100", "output": "NO" } ]

1,671,630,512

2,147,483,647

Python 3

OK

TESTS

31

92

0

n1=int(input()) lis=list(map(int,input().split())) lis=list(set(lis)) lis.sort() if len(lis)>1: print(lis[1]) else: print("NO")

Title: Second Order Statistics Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. Input Specification: The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. Output Specification: If the given sequence has the second order statistics, output this order statistics, otherwise output NO. Demo Input: ['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n'] Demo Output: ['1\n', '2\n'] Note: none

```python n1=int(input()) lis=list(map(int,input().split())) lis=list(set(lis)) lis.sort() if len(lis)>1: print(lis[1]) else: print("NO") ```

3.977

630

A

Again Twenty Five!

PROGRAMMING

800

[ "number theory" ]

null

The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions." Could you pass the interview in the machine vision company in IT City?

The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.

Output the last two digits of 5*n* without spaces between them.

[ "2\n" ]

[ "25" ]

none

0

[ { "input": "2", "output": "25" }, { "input": "7", "output": "25" }, { "input": "1000000000000000000", "output": "25" }, { "input": "2000000000000000000", "output": "25" }, { "input": "987654321012345678", "output": "25" } ]

1,694,440,621

2,147,483,647

Python 3

OK

TESTS

5

31

0

n= int(input()) if n==0: m='0' p=1 print(m,end='') print(p) elif n==1: m=0 p=5 print(m,end='') print(p) else: m=25 print(m)

Title: Again Twenty Five! Time Limit: None seconds Memory Limit: None megabytes Problem Description: The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions." Could you pass the interview in the machine vision company in IT City? Input Specification: The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5. Output Specification: Output the last two digits of 5*n* without spaces between them. Demo Input: ['2\n'] Demo Output: ['25'] Note: none

```python n= int(input()) if n==0: m='0' p=1 print(m,end='') print(p) elif n==1: m=0 p=5 print(m,end='') print(p) else: m=25 print(m) ```

3

628

B

New Skateboard

PROGRAMMING

1,300

[ "dp" ]

null

Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4. You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero. A substring of a string is a nonempty sequence of consecutive characters. For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04. As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9.

Print integer *a* — the number of substrings of the string *s* that are divisible by 4. Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

[ "124\n", "04\n", "5810438174\n" ]

[ "4\n", "3\n", "9\n" ]

none

0

[ { "input": "124", "output": "4" }, { "input": "04", "output": "3" }, { "input": "5810438174", "output": "9" }, { "input": "1", "output": "0" }, { "input": "039", "output": "1" }, { "input": "97247", "output": "6" }, { "input": "5810438174", "output": "9" }, { "input": "12883340691714056185860211260984431382156326935244", "output": "424" }, { "input": "2144315253572020279108092911160072328496568665545836825277616363478721946398140227406814602154768031", "output": "1528" }, { "input": "80124649014054971081213608137817466046254652492627741860478258558206397113198232823859870363821007188476405951611069347299689170240023979048198711745011542774268179055311013054073075176122755643483380248999657649211459997766221072399103579977409770898200358240970169892326442892826731631357561876251276209119521202062222947560634301788787748428236988789594458520867663257476744168528121470923031438015546006185059454402637036376247785881323277542968298682307854655591317046086531554595892680980142608", "output": "30826" }, { "input": "123456", "output": "7" }, { "input": "4", "output": "1" }, { "input": "123", "output": "1" } ]

1,679,977,212

2,147,483,647

Python 3

OK

TESTS

22

265

614,400

a=input() l=r=0 for i in range(len(a)): r=l+(a[i]in'048')+i*(i and int(a[i-1:i+1])%4==0) l=r print(r)

Title: New Skateboard Time Limit: None seconds Memory Limit: None megabytes Problem Description: Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4. You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero. A substring of a string is a nonempty sequence of consecutive characters. For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04. As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Input Specification: The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9. Output Specification: Print integer *a* — the number of substrings of the string *s* that are divisible by 4. Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. Demo Input: ['124\n', '04\n', '5810438174\n'] Demo Output: ['4\n', '3\n', '9\n'] Note: none

```python a=input() l=r=0 for i in range(len(a)): r=l+(a[i]in'048')+i*(i and int(a[i-1:i+1])%4==0) l=r print(r) ```

3

651

A

Joysticks

PROGRAMMING

1,100

[ "dp", "greedy", "implementation", "math" ]

null

Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger). Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops. Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.

The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.

Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.

[ "3 5\n", "4 4\n" ]

[ "6\n", "5\n" ]

In the first sample game lasts for 6 minute by using the following algorithm: - at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%. After that the first joystick is completely discharged and the game is stopped.

500

[ { "input": "3 5", "output": "6" }, { "input": "4 4", "output": "5" }, { "input": "100 100", "output": "197" }, { "input": "1 100", "output": "98" }, { "input": "100 1", "output": "98" }, { "input": "1 4", "output": "2" }, { "input": "1 1", "output": "0" }, { "input": "8 8", "output": "13" }, { "input": "7 2", "output": "7" }, { "input": "24 15", "output": "36" }, { "input": "19 30", "output": "47" }, { "input": "15 31", "output": "44" }, { "input": "14 15", "output": "27" }, { "input": "58 33", "output": "89" }, { "input": "15 25", "output": "38" }, { "input": "59 45", "output": "102" }, { "input": "3 73", "output": "74" }, { "input": "48 1", "output": "47" }, { "input": "100 25", "output": "122" }, { "input": "40 49", "output": "86" }, { "input": "85 73", "output": "155" }, { "input": "29 1", "output": "28" }, { "input": "74 25", "output": "97" }, { "input": "24 57", "output": "78" }, { "input": "23 12", "output": "33" }, { "input": "2 99", "output": "99" }, { "input": "98 2", "output": "97" }, { "input": "2 97", "output": "97" }, { "input": "30 54", "output": "81" }, { "input": "32 53", "output": "82" }, { "input": "32 54", "output": "84" }, { "input": "1 2", "output": "1" }, { "input": "2 1", "output": "1" }, { "input": "2 2", "output": "1" }, { "input": "1 3", "output": "2" }, { "input": "3 1", "output": "2" }, { "input": "1 4", "output": "2" }, { "input": "2 3", "output": "3" }, { "input": "3 2", "output": "3" } ]

1,623,950,979

2,147,483,647

Python 3

OK

TESTS

39

62

0

a,b=map(int,input().split()) m=0 if a==1 and b==1: print("0") else: while a>0 and b>0: if a==b: a+=1 b-=2 m+=1 elif a==1: a+=1 b-=2 m+=1 elif b==1: b+=1 a-=2 m+=1 elif a>b: b+=1 a-=2 m+=1 else: a+=1 b-=2 m+=1 print(m)

Title: Joysticks Time Limit: None seconds Memory Limit: None megabytes Problem Description: Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger). Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops. Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent. Input Specification: The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively. Output Specification: Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged. Demo Input: ['3 5\n', '4 4\n'] Demo Output: ['6\n', '5\n'] Note: In the first sample game lasts for 6 minute by using the following algorithm: - at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%. After that the first joystick is completely discharged and the game is stopped.

```python a,b=map(int,input().split()) m=0 if a==1 and b==1: print("0") else: while a>0 and b>0: if a==b: a+=1 b-=2 m+=1 elif a==1: a+=1 b-=2 m+=1 elif b==1: b+=1 a-=2 m+=1 elif a>b: b+=1 a-=2 m+=1 else: a+=1 b-=2 m+=1 print(m) ```

3

505

A

Mr. Kitayuta's Gift

PROGRAMMING

1,100

[ "brute force", "implementation", "strings" ]

null

Mr. Kitayuta has kindly given you a string *s* consisting of lowercase English letters. You are asked to insert exactly one lowercase English letter into *s* to make it a palindrome. A palindrome is a string that reads the same forward and backward. For example, "noon", "testset" and "a" are all palindromes, while "test" and "kitayuta" are not. You can choose any lowercase English letter, and insert it to any position of *s*, possibly to the beginning or the end of *s*. You have to insert a letter even if the given string is already a palindrome. If it is possible to insert one lowercase English letter into *s* so that the resulting string will be a palindrome, print the string after the insertion. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one palindrome that can be obtained, you are allowed to print any of them.

The only line of the input contains a string *s* (1<=≤<=|*s*|<=≤<=10). Each character in *s* is a lowercase English letter.

If it is possible to turn *s* into a palindrome by inserting one lowercase English letter, print the resulting string in a single line. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one solution, any of them will be accepted.

[ "revive\n", "ee\n", "kitayuta\n" ]

[ "reviver\n", "eye", "NA\n" ]

For the first sample, insert 'r' to the end of "revive" to obtain a palindrome "reviver". For the second sample, there is more than one solution. For example, "eve" will also be accepted. For the third sample, it is not possible to turn "kitayuta" into a palindrome by just inserting one letter.

500

[ { "input": "revive", "output": "reviver" }, { "input": "ee", "output": "eee" }, { "input": "kitayuta", "output": "NA" }, { "input": "evima", "output": "NA" }, { "input": "a", "output": "aa" }, { "input": "yutampo", "output": "NA" }, { "input": "fft", "output": "tfft" }, { "input": "shuseki", "output": "NA" }, { "input": "lsdijfjisl", "output": "lsdijfjidsl" }, { "input": "ewcdcwerp", "output": "NA" }, { "input": "noon", "output": "nooon" }, { "input": "add", "output": "adda" }, { "input": "iq", "output": "qiq" }, { "input": "hogloid", "output": "NA" }, { "input": "yosupo", "output": "NA" }, { "input": "stat", "output": "stats" }, { "input": "level", "output": "levvel" }, { "input": "iwiwi", "output": "iwiiwi" }, { "input": "racecar", "output": "raceecar" }, { "input": "dog", "output": "NA" }, { "input": "squirrel", "output": "NA" }, { "input": "slime", "output": "NA" }, { "input": "palindrome", "output": "NA" }, { "input": "nqwcvcwqn", "output": "nqwcvvcwqn" }, { "input": "lsdijfjisl", "output": "lsdijfjidsl" }, { "input": "jewiiwefj", "output": "jfewiiwefj" }, { "input": "vnwmvvmonv", "output": "NA" }, { "input": "zennyuu", "output": "NA" }, { "input": "mrtoudai", "output": "NA" }, { "input": "z", "output": "zz" }, { "input": "babccba", "output": "babccbab" }, { "input": "battaba", "output": "abattaba" }, { "input": "bbb", "output": "bbbb" }, { "input": "zaa", "output": "zaaz" }, { "input": "abbabab", "output": "NA" }, { "input": "racecar", "output": "raceecar" }, { "input": "zzz", "output": "zzzz" }, { "input": "eviver", "output": "reviver" }, { "input": "aaaaa", "output": "aaaaaa" }, { "input": "eee", "output": "eeee" } ]

1,422,577,299

2,147,483,647

Python 3

OK

TESTS

40

62

0

def ispal(s): i = 0 j = len(s) - 1 while i < j: if s[i] != s[j]: return False i += 1 j -= 1 return True alpha = 'abcdefghijklmnopqrstuvwxyz' found = 0 s = input() for i in range(0,len(s)+1): for j in range(len(alpha)): temp = s[:i] + alpha[j] + s[i:] if ispal(temp): found = 1 print(temp) break if found: break if not found: print("NA")

Title: Mr. Kitayuta's Gift Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mr. Kitayuta has kindly given you a string *s* consisting of lowercase English letters. You are asked to insert exactly one lowercase English letter into *s* to make it a palindrome. A palindrome is a string that reads the same forward and backward. For example, "noon", "testset" and "a" are all palindromes, while "test" and "kitayuta" are not. You can choose any lowercase English letter, and insert it to any position of *s*, possibly to the beginning or the end of *s*. You have to insert a letter even if the given string is already a palindrome. If it is possible to insert one lowercase English letter into *s* so that the resulting string will be a palindrome, print the string after the insertion. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one palindrome that can be obtained, you are allowed to print any of them. Input Specification: The only line of the input contains a string *s* (1<=≤<=|*s*|<=≤<=10). Each character in *s* is a lowercase English letter. Output Specification: If it is possible to turn *s* into a palindrome by inserting one lowercase English letter, print the resulting string in a single line. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one solution, any of them will be accepted. Demo Input: ['revive\n', 'ee\n', 'kitayuta\n'] Demo Output: ['reviver\n', 'eye', 'NA\n'] Note: For the first sample, insert 'r' to the end of "revive" to obtain a palindrome "reviver". For the second sample, there is more than one solution. For example, "eve" will also be accepted. For the third sample, it is not possible to turn "kitayuta" into a palindrome by just inserting one letter.

```python def ispal(s): i = 0 j = len(s) - 1 while i < j: if s[i] != s[j]: return False i += 1 j -= 1 return True alpha = 'abcdefghijklmnopqrstuvwxyz' found = 0 s = input() for i in range(0,len(s)+1): for j in range(len(alpha)): temp = s[:i] + alpha[j] + s[i:] if ispal(temp): found = 1 print(temp) break if found: break if not found: print("NA") ```

3

607

A

Chain Reaction

PROGRAMMING

1,600

[ "binary search", "dp" ]

null

There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated. Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.

The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons. The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*.

Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.

[ "4\n1 9\n3 1\n6 1\n7 4\n", "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n" ]

[ "1\n", "3\n" ]

For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2. For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.

500

[ { "input": "4\n1 9\n3 1\n6 1\n7 4", "output": "1" }, { "input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1", "output": "3" }, { "input": "1\n0 1", "output": "0" }, { "input": "1\n0 1000000", "output": "0" }, { "input": "1\n1000000 1000000", "output": "0" }, { "input": "7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 6\n7 7", "output": "4" }, { "input": "5\n1 1\n3 1\n5 1\n7 10\n8 10", "output": "2" }, { "input": "11\n110 90\n100 70\n90 10\n80 10\n70 1\n60 1\n50 10\n40 1\n30 1\n10 1\n20 1", "output": "4" } ]

1,681,265,733

2,147,483,647

Python 3

OK

TESTS

41

561

8,499,200

from sys import stdin from bisect import bisect_left def main(): n = int(input()) beacons = [tuple(map(int, input().split())) for _ in range(n)] beacons.sort() # dp[i] stores the number of beacons that can be saved when considering the beacons from the rightmost one to the i-th one dp = [0] * n for i in range(n): last_undestroyed = bisect_left(beacons, (beacons[i][0] - beacons[i][1],)) - 1 dp[i] = 1 if last_undestroyed == -1 else dp[last_undestroyed] + 1 # The minimum number of beacons destroyed ans = n - max(dp) print(ans) main()

Title: Chain Reaction Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* beacons located at distinct positions on a number line. The *i*-th beacon has position *a**i* and power level *b**i*. When the *i*-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance *b**i* inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated. Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed. Input Specification: The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the initial number of beacons. The *i*-th of next *n* lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=1<=000<=000, 1<=≤<=*b**i*<=≤<=1<=000<=000) — the position and power level of the *i*-th beacon respectively. No two beacons will have the same position, so *a**i*<=≠<=*a**j* if *i*<=≠<=*j*. Output Specification: Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added. Demo Input: ['4\n1 9\n3 1\n6 1\n7 4\n', '7\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n'] Demo Output: ['1\n', '3\n'] Note: For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2. For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337 with power level 42.

```python from sys import stdin from bisect import bisect_left def main(): n = int(input()) beacons = [tuple(map(int, input().split())) for _ in range(n)] beacons.sort() # dp[i] stores the number of beacons that can be saved when considering the beacons from the rightmost one to the i-th one dp = [0] * n for i in range(n): last_undestroyed = bisect_left(beacons, (beacons[i][0] - beacons[i][1],)) - 1 dp[i] = 1 if last_undestroyed == -1 else dp[last_undestroyed] + 1 # The minimum number of beacons destroyed ans = n - max(dp) print(ans) main() ```

3

94

A

Restoring Password

PROGRAMMING

900

[ "implementation", "strings" ]

A. Restoring Password

2

256

Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification.

The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.

Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.

[ "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n", "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n" ]

[ "12345678\n", "30234919\n" ]

none

500

[ { "input": "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110", "output": "12345678" }, { "input": "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000", "output": "30234919" }, { "input": "00010101101110110101100110101100010101100010101111000101011010011010110010000011\n0101010110\n0001001101\n1001101011\n0000100011\n0010101111\n1110110101\n0001010110\n0110111000\n0000111110\n0010000011", "output": "65264629" }, { "input": "10100100010010010011011001101000100100110110011010011001101011000100110110011010\n1111110011\n1001000111\n1001000100\n1100010011\n0110011010\n0010000001\n1110101110\n0010000110\n0010010011\n1010010001", "output": "98484434" }, { "input": "00101100011111010001001000000110110000000110010011001111111010110010001011000000\n0010000001\n0110010011\n0010000010\n1011001000\n0011111110\n0110001000\n1111010001\n1011000000\n0000100110\n0010110001", "output": "96071437" }, { "input": "10001110111110000001000010001010001110110000100010100010111101101101010000100010\n0000010110\n1101010111\n1000101111\n0001011110\n0011110101\n0101100100\n0110110101\n0000100010\n1000111011\n1110000001", "output": "89787267" }, { "input": "10010100011001010001010101001101010100110100111011001010111100011001000010100000\n0011100000\n1001100100\n0001100100\n0010100000\n0101010011\n0010101110\n0010101111\n0100111011\n1001010001\n1111111110", "output": "88447623" }, { "input": "01101100111000000101011011001110000001011111111000111111100001011010001001011001\n1000000101\n0101101000\n0101110101\n1101011110\n0000101100\n1111111000\n0001001101\n0110111011\n0110110011\n1001011001", "output": "80805519" }, { "input": "11100011000100010110010011101010101010011110001100011010111110011000011010110111\n1110001100\n0110101111\n0100111010\n0101000000\n1001100001\n1010101001\n0000100010\n1010110111\n1100011100\n0100010110", "output": "09250147" }, { "input": "10000110110000010100000010001000111101110110101011110111000100001101000000100010\n0000010100\n0000110001\n0110101011\n1101110001\n1000011011\n0000110100\n0011110111\n1000110010\n0000100010\n0000011011", "output": "40862358" }, { "input": "01000000010000000110100101000110110000100100000001101100001000011111111001010001\n1011000010\n1111101010\n0111110011\n0000000110\n0000001001\n0001111111\n0110010010\n0100000001\n1011001000\n1001010001", "output": "73907059" }, { "input": "01111000111110011001110101110011110000111110010001101100110110100111101011001101\n1110010001\n1001100000\n1100001000\n1010011110\n1011001101\n0111100011\n1101011100\n1110011001\n1111000011\n0010000101", "output": "57680434" }, { "input": "01001100101000100010001011110001000101001001100010010000001001001100101001011111\n1001011111\n1110010111\n0111101011\n1000100010\n0011100101\n0100000010\n0010111100\n0100010100\n1001100010\n0100110010", "output": "93678590" }, { "input": "01110111110000111011101010110110101011010100110111000011101101110101011101001000\n0110000101\n1010101101\n1101010111\n1101011100\n0100110111\n0111011111\n1100011001\n0111010101\n0000111011\n1101001000", "output": "58114879" }, { "input": "11101001111100110101110011010100110011011110100111010110110011000111000011001101\n1100011100\n1100110101\n1011101000\n0011011110\n0011001101\n0100010001\n1110100111\n1010101100\n1110110100\n0101101100", "output": "61146904" }, { "input": "10101010001011010001001001011000100101100001011011101010101110101010001010101000\n0010110101\n1010011010\n1010101000\n1011010001\n1010101011\n0010010110\n0110100010\n1010100101\n0001011011\n0110100001", "output": "23558422" }, { "input": "11110101001100010000110100001110101011011111010100110001000001001010001001101111\n0101101100\n1001101111\n1010101101\n0100101000\n1111110000\n0101010010\n1100010000\n1111010100\n1101000011\n1011111111", "output": "76827631" }, { "input": "10001100110000110111100011001101111110110011110101000011011100001101110000110111\n0011110101\n0101100011\n1000110011\n1011011001\n0111111011\n0101111011\n0000110111\n0100001110\n1000000111\n0110110111", "output": "26240666" }, { "input": "10000100010000111101100100111101111011101000001001100001000110000010010000111101\n1001001111\n0000111101\n1000010001\n0110011101\n0110101000\n1011111001\n0111101110\n1000001001\n1101011111\n0001010100", "output": "21067271" }, { "input": "01101111000110111100011011110001101111001010001100101000110001010101100100000010\n1010001100\n0011010011\n0101010110\n1111001100\n1100011000\n0100101100\n1001100101\n0110111100\n0011001101\n0100000010", "output": "77770029" }, { "input": "10100111011010001011111000000111100000010101000011000010111101010000111010011101\n1010011101\n1010111111\n0110100110\n1111000100\n1110000001\n0000101111\n0011111000\n1000110001\n0101000011\n1010001011", "output": "09448580" }, { "input": "10000111111000011111001010101010010011111001001111000010010100100011000010001100\n1101101110\n1001001111\n0000100101\n1100111010\n0010101010\n1110000110\n1100111101\n0010001100\n1110000001\n1000011111", "output": "99411277" }, { "input": "10110110111011001111101100111100111111011011011011001111110110010011100010000111\n0111010011\n0111101100\n1001101010\n0101000101\n0010000111\n0011111101\n1011001111\n1101111000\n1011011011\n1001001110", "output": "86658594" }, { "input": "01001001100101100011110110111100000110001111001000100000110111110010000000011000\n0100100110\n1000001011\n1000111110\n0000011000\n0101100011\n1101101111\n1111001000\n1011011001\n1000001101\n0010101000", "output": "04536863" }, { "input": "10010100011101000011100100001100101111000010111100000010010000001001001101011101\n1001000011\n1101000011\n1001010001\n1101011101\n1000010110\n0011111101\n0010111100\n0000100100\n1010001000\n0101000110", "output": "21066773" }, { "input": "01111111110101111111011111111111010010000001100000101000100100111001011010001001\n0111111111\n0101111111\n0100101101\n0001100000\n0011000101\n0011100101\n1101001000\n0010111110\n1010001001\n1111000111", "output": "01063858" }, { "input": "00100011111001001010001111000011101000001110100000000100101011101000001001001010\n0010001111\n1001001010\n1010011001\n0011100111\n1000111000\n0011110000\n0000100010\n0001001010\n1111110111\n1110100000", "output": "01599791" }, { "input": "11011101000100110100110011010101100011111010011010010011010010010010100110101111\n0100110100\n1001001010\n0001111101\n1101011010\n1101110100\n1100110101\n0110101111\n0110001111\n0001101000\n1010011010", "output": "40579016" }, { "input": "10000010111101110110011000111110000011100110001111100100000111000011011000001011\n0111010100\n1010110110\n1000001110\n1110000100\n0110001111\n1101110110\n1100001101\n1000001011\n0000000101\n1001000001", "output": "75424967" }, { "input": "11101100101110111110111011111010001111111111000001001001000010001111111110110010\n0101100001\n1111010011\n1110111110\n0100110100\n1110011111\n1000111111\n0010010000\n1110110010\n0011000010\n1111000001", "output": "72259657" }, { "input": "01011110100101111010011000001001100000101001110011010111101011010000110110010101\n0100111100\n0101110011\n0101111010\n0110000010\n0101001111\n1101000011\n0110010101\n0111011010\n0001101110\n1001110011", "output": "22339256" }, { "input": "01100000100101111000100001100010000110000010100100100001100000110011101001110000\n0101111000\n1001110000\n0001000101\n0110110111\n0010100100\n1000011000\n1101110110\n0110000010\n0001011010\n0011001110", "output": "70554591" }, { "input": "11110011011000001001111100110101001000010100100000110011001110011111100100100001\n1010011000\n1111001101\n0100100001\n1111010011\n0100100000\n1001111110\n1010100111\n1000100111\n1000001001\n1100110011", "output": "18124952" }, { "input": "10001001011000100101010110011101011001110010000001010110000101000100101111101010\n0101100001\n1100001100\n1111101010\n1000100101\n0010000001\n0100010010\n0010110110\n0101100111\n0000001110\n1101001110", "output": "33774052" }, { "input": "00110010000111001001001100100010010111101011011110001011111100000101000100000001\n0100000001\n1011011110\n0010111111\n0111100111\n0100111001\n0000010100\n1001011110\n0111001001\n0100010011\n0011001000", "output": "97961250" }, { "input": "01101100001000110101101100101111101110010011010111100011010100010001101000110101\n1001101001\n1000110101\n0110110000\n0111100100\n0011010111\n1110111001\n0001000110\n0000000100\n0001101001\n1011001011", "output": "21954161" }, { "input": "10101110000011010110101011100000101101000110100000101101101101110101000011110010\n0110100000\n1011011011\n0011110010\n0001110110\n0010110100\n1100010010\n0001101011\n1010111000\n0011010110\n0111010100", "output": "78740192" }, { "input": "11000101011100100111010000010001000001001100101100000011000000001100000101011010\n1100010101\n1111101011\n0101011010\n0100000100\n1000110111\n1100100111\n1100101100\n0111001000\n0000110000\n0110011111", "output": "05336882" }, { "input": "11110100010000101110010110001000001011100101100010110011011011111110001100110110\n0101100010\n0100010001\n0000101110\n1100110110\n0101000101\n0011001011\n1111010001\n1000110010\n1111111000\n1010011111", "output": "62020383" }, { "input": "00011001111110000011101011010001010111100110100101000110011111011001100000001100\n0111001101\n0101011110\n0001100111\n1101011111\n1110000011\n0000001100\n0111010001\n1101100110\n1010110100\n0110100101", "output": "24819275" }, { "input": "10111110010011111001001111100101010111010011111001001110101000111110011001111101\n0011111001\n0101011101\n0100001010\n0001110010\n1001111101\n0011101010\n1111001001\n1100100001\n1001101000\n1011111001", "output": "90010504" }, { "input": "01111101111100101010001001011110111001110111110111011111011110110111111011011111\n1111110111\n0010000101\n0110000100\n0111111011\n1011100111\n1100101010\n1011011111\n1100010001\n0111110111\n0010010111", "output": "85948866" }, { "input": "01111100000111110000110010111001111100001001101010110010111010001000101001101010\n0100010101\n1011110101\n1010100100\n1010000001\n1001101010\n0101100110\n1000100010\n0111110000\n1100101110\n0110010110", "output": "77874864" }, { "input": "11100011010000000010011110010111001011111001000111000000001000000000100111100101\n0000000010\n1110001101\n0011010101\n0111100101\n1001000111\n1101001111\n0111010110\n1100101111\n0110000000\n1101101011", "output": "10374003" }, { "input": "01111011100111101110011001000110001111101000111110100100100001011111001011100010\n0110010100\n1100010001\n0111101110\n1001001000\n1010011011\n1000111110\n0010110101\n1011100010\n0101111100\n0110010001", "output": "22955387" }, { "input": "11011010001100000011000100110011010101000110011110110000001100111100001000011111\n0000100010\n1000011111\n1101101000\n0110011110\n0011110000\n1100000011\n0010001100\n0101101000\n0001001100\n1101010100", "output": "25893541" }, { "input": "01011001011111010010101111011001000011001100011101101111011011010011101011110110\n0100001100\n0101100101\n1111111011\n1111010010\n1111101100\n1100011101\n1011000011\n1101001110\n1011110110\n0110001010", "output": "13805878" }, { "input": "11110011011000111111001100111110001111111100000010111100110100110011111111001101\n1111001101\n1001101010\n1100110010\n0011001111\n0001011110\n1000110011\n1000111111\n0110001010\n1001011101\n1100000010", "output": "06369030" }, { "input": "01110011110010000011011001011000001000010110010110011001100001100110001100101000\n0000100001\n0110011000\n1010000010\n1110011101\n0111001111\n1100101000\n0010000011\n0110010000\n1100100101\n0110010110", "output": "46909115" }, { "input": "00001011001111110111111111011111111101110101110100010111010010100101100001010110\n1111110111\n0001010110\n0111011011\n0111000001\n1010010110\n0101110100\n0001000101\n0000111000\n0110100001\n0000101100", "output": "90005541" } ]

1,697,709,407

2,147,483,647

PyPy 3-64

OK

TESTS

50

124

0

n=input() m=[] seq_size = 10 seq_num = 8 a='' for i in range(seq_num): m.append(n[i*seq_size:(i+1)*seq_size]) mas=[] for i in range(10): l=input() mas.append(l) for i in m: a+=str(mas.index(i)) print(a)

Title: Restoring Password Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password". Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address. Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits. Help Igor K. restore his ISQ account by the encrypted password and encryption specification. Input Specification: The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9. Output Specification: Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists. Demo Input: ['01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n', '10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n'] Demo Output: ['12345678\n', '30234919\n'] Note: none

```python n=input() m=[] seq_size = 10 seq_num = 8 a='' for i in range(seq_num): m.append(n[i*seq_size:(i+1)*seq_size]) mas=[] for i in range(10): l=input() mas.append(l) for i in m: a+=str(mas.index(i)) print(a) ```

3.969

499

B

Lecture

PROGRAMMING

1,000

[ "implementation", "strings" ]

null

You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.

The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.

Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.

[ "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n", "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n" ]

[ "codeforces round letter round\n", "hbnyiyc joll joll un joll\n" ]

none

500

[ { "input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest", "output": "codeforces round letter round" }, { "input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll", "output": "hbnyiyc joll joll un joll" }, { "input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b", "output": "f jpggvr b hnxsd b" }, { "input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb", "output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs" }, { "input": "1 1\namit am\namit", "output": "am" }, { "input": "1 1\na c\na", "output": "a" } ]

1,698,690,299

2,147,483,647

Python 3

OK

TESTS

37

358

819,200

IN = lambda x:list(map(x, input().split())) n, m = IN(x = int) l1 = [] l2 = [] for x in range(m): a,b = IN(x = str) l1 += [a]; l2 += [b] for y in input().split(): if y in l1: if len(l2[l1.index(y)]) < len(y): print(l2[l1.index(y)], end = " ") else: print(y, end = " ") else: if len(l1[l2.index(y)]) <= len(y): print(l1[l2.index(y)], end = " ") else: print(y, end = " ")

Title: Lecture Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes. You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning. You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language. You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes. Input Specification: The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages. The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once. The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}. All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters. Output Specification: Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input. Demo Input: ['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n'] Demo Output: ['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n'] Note: none

```python IN = lambda x:list(map(x, input().split())) n, m = IN(x = int) l1 = [] l2 = [] for x in range(m): a,b = IN(x = str) l1 += [a]; l2 += [b] for y in input().split(): if y in l1: if len(l2[l1.index(y)]) < len(y): print(l2[l1.index(y)], end = " ") else: print(y, end = " ") else: if len(l1[l2.index(y)]) <= len(y): print(l1[l2.index(y)], end = " ") else: print(y, end = " ") ```

3

465

A

inc ARG

PROGRAMMING

900

[ "implementation" ]

null

Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit. Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded. Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell. The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.

Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.

[ "4\n1100\n", "4\n1111\n" ]

[ "3\n", "4\n" ]

In the first sample the cell ends up with value 0010, in the second sample — with 0000.

500

[ { "input": "4\n1100", "output": "3" }, { "input": "4\n1111", "output": "4" }, { "input": "1\n0", "output": "1" }, { "input": "1\n1", "output": "1" }, { "input": "2\n00", "output": "1" }, { "input": "2\n01", "output": "1" }, { "input": "2\n10", "output": "2" }, { "input": "2\n11", "output": "2" }, { "input": "10\n0000000000", "output": "1" }, { "input": "20\n11111111110110001100", "output": "11" }, { "input": "50\n01011110100010000001010000100001001101001101101011", "output": "1" }, { "input": "60\n111111111101111111111111111111111111111111111111111111111111", "output": "11" }, { "input": "60\n111111111111111111111111111111111111111111111111111111111111", "output": "60" }, { "input": "66\n111111010010011001110011000111000100011110011001111110011111111101", "output": "7" }, { "input": "90\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "1" }, { "input": "91\n1011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "2" }, { "input": "92\n11111111101011111011111111111111111111111011101111111111011111111111111101111111101011111101", "output": "10" }, { "input": "100\n0001011110100011001100100010111001000001111101101001001001001011110100101101010000000110100101110010", "output": "1" }, { "input": "100\n0111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "1" }, { "input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "1" }, { "input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "output": "1" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "100" }, { "input": "100\n1011001110001000011111110011000100001110010110111101110110011011011000010100110001111100000010110010", "output": "2" }, { "input": "100\n1101111011001111111111110011110111101110111111111111111111111111111111011111111111110111111111111111", "output": "3" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111", "output": "100" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110", "output": "100" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111101", "output": "99" }, { "input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111100000", "output": "96" }, { "input": "100\n1111111111111111111111111111111110011000110110001010101100100100001000010010010010000100110000010101", "output": "34" } ]

1,688,650,266

2,147,483,647

PyPy 3-64

OK

TESTS

29

61

0

import sys input = sys.stdin.readline n = int(input()) data = input().rstrip()[::-1] tmp = bin(int(data, 2) + 1)[2:] tmp = (n - len(tmp)) * '0' + tmp cnt = 0 for i in range(n): if data[-(i + 1)] != tmp[-(i + 1)]: cnt += 1 print(cnt)

Title: inc ARG Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit. Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded. Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation? Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell. The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit. Output Specification: Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell. Demo Input: ['4\n1100\n', '4\n1111\n'] Demo Output: ['3\n', '4\n'] Note: In the first sample the cell ends up with value 0010, in the second sample — with 0000.

```python import sys input = sys.stdin.readline n = int(input()) data = input().rstrip()[::-1] tmp = bin(int(data, 2) + 1)[2:] tmp = (n - len(tmp)) * '0' + tmp cnt = 0 for i in range(n): if data[-(i + 1)] != tmp[-(i + 1)]: cnt += 1 print(cnt) ```

3

987

A

Infinity Gauntlet

PROGRAMMING

800

[ "implementation" ]

null

You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems: - the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color. Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.

In the first line of input there is one integer $n$ ($0 \le n \le 6$) — the number of Gems in Infinity Gauntlet. In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.

In the first line output one integer $m$ ($0 \le m \le 6$) — the number of absent Gems. Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.

[ "4\nred\npurple\nyellow\norange\n", "0\n" ]

[ "2\nSpace\nTime\n", "6\nTime\nMind\nSoul\nPower\nReality\nSpace\n" ]

In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space. In the second sample Thanos doesn't have any Gems, so he needs all six.

500

[ { "input": "4\nred\npurple\nyellow\norange", "output": "2\nSpace\nTime" }, { "input": "0", "output": "6\nMind\nSpace\nPower\nTime\nReality\nSoul" }, { "input": "6\npurple\nblue\nyellow\nred\ngreen\norange", "output": "0" }, { "input": "1\npurple", "output": "5\nTime\nReality\nSoul\nSpace\nMind" }, { "input": "3\nblue\norange\npurple", "output": "3\nTime\nReality\nMind" }, { "input": "2\nyellow\nred", "output": "4\nPower\nSoul\nSpace\nTime" }, { "input": "1\ngreen", "output": "5\nReality\nSpace\nPower\nSoul\nMind" }, { "input": "2\npurple\ngreen", "output": "4\nReality\nMind\nSpace\nSoul" }, { "input": "1\nblue", "output": "5\nPower\nReality\nSoul\nTime\nMind" }, { "input": "2\npurple\nblue", "output": "4\nMind\nSoul\nTime\nReality" }, { "input": "2\ngreen\nblue", "output": "4\nReality\nMind\nPower\nSoul" }, { "input": "3\npurple\ngreen\nblue", "output": "3\nMind\nReality\nSoul" }, { "input": "1\norange", "output": "5\nReality\nTime\nPower\nSpace\nMind" }, { "input": "2\npurple\norange", "output": "4\nReality\nMind\nTime\nSpace" }, { "input": "2\norange\ngreen", "output": "4\nSpace\nMind\nReality\nPower" }, { "input": "3\norange\npurple\ngreen", "output": "3\nReality\nSpace\nMind" }, { "input": "2\norange\nblue", "output": "4\nTime\nMind\nReality\nPower" }, { "input": "3\nblue\ngreen\norange", "output": "3\nPower\nMind\nReality" }, { "input": "4\nblue\norange\ngreen\npurple", "output": "2\nMind\nReality" }, { "input": "1\nred", "output": "5\nTime\nSoul\nMind\nPower\nSpace" }, { "input": "2\nred\npurple", "output": "4\nMind\nSpace\nTime\nSoul" }, { "input": "2\nred\ngreen", "output": "4\nMind\nSpace\nPower\nSoul" }, { "input": "3\nred\npurple\ngreen", "output": "3\nSoul\nSpace\nMind" }, { "input": "2\nblue\nred", "output": "4\nMind\nTime\nPower\nSoul" }, { "input": "3\nred\nblue\npurple", "output": "3\nTime\nMind\nSoul" }, { "input": "3\nred\nblue\ngreen", "output": "3\nSoul\nPower\nMind" }, { "input": "4\npurple\nblue\ngreen\nred", "output": "2\nMind\nSoul" }, { "input": "2\norange\nred", "output": "4\nPower\nMind\nTime\nSpace" }, { "input": "3\nred\norange\npurple", "output": "3\nMind\nSpace\nTime" }, { "input": "3\nred\norange\ngreen", "output": "3\nMind\nSpace\nPower" }, { "input": "4\nred\norange\ngreen\npurple", "output": "2\nSpace\nMind" }, { "input": "3\nblue\norange\nred", "output": "3\nPower\nMind\nTime" }, { "input": "4\norange\nblue\npurple\nred", "output": "2\nTime\nMind" }, { "input": "4\ngreen\norange\nred\nblue", "output": "2\nMind\nPower" }, { "input": "5\npurple\norange\nblue\nred\ngreen", "output": "1\nMind" }, { "input": "1\nyellow", "output": "5\nPower\nSoul\nReality\nSpace\nTime" }, { "input": "2\npurple\nyellow", "output": "4\nTime\nReality\nSpace\nSoul" }, { "input": "2\ngreen\nyellow", "output": "4\nSpace\nReality\nPower\nSoul" }, { "input": "3\npurple\nyellow\ngreen", "output": "3\nSoul\nReality\nSpace" }, { "input": "2\nblue\nyellow", "output": "4\nTime\nReality\nPower\nSoul" }, { "input": "3\nyellow\nblue\npurple", "output": "3\nSoul\nReality\nTime" }, { "input": "3\ngreen\nyellow\nblue", "output": "3\nSoul\nReality\nPower" }, { "input": "4\nyellow\nblue\ngreen\npurple", "output": "2\nReality\nSoul" }, { "input": "2\nyellow\norange", "output": "4\nTime\nSpace\nReality\nPower" }, { "input": "3\nyellow\npurple\norange", "output": "3\nSpace\nReality\nTime" }, { "input": "3\norange\nyellow\ngreen", "output": "3\nSpace\nReality\nPower" }, { "input": "4\ngreen\nyellow\norange\npurple", "output": "2\nSpace\nReality" }, { "input": "3\nyellow\nblue\norange", "output": "3\nTime\nReality\nPower" }, { "input": "4\norange\npurple\nblue\nyellow", "output": "2\nReality\nTime" }, { "input": "4\nblue\norange\nyellow\ngreen", "output": "2\nReality\nPower" }, { "input": "5\ngreen\nyellow\norange\nblue\npurple", "output": "1\nReality" }, { "input": "3\nyellow\npurple\nred", "output": "3\nTime\nSoul\nSpace" }, { "input": "3\nred\ngreen\nyellow", "output": "3\nPower\nSoul\nSpace" }, { "input": "4\nred\npurple\ngreen\nyellow", "output": "2\nSpace\nSoul" }, { "input": "3\nred\nyellow\nblue", "output": "3\nPower\nSoul\nTime" }, { "input": "4\nblue\nyellow\nred\npurple", "output": "2\nTime\nSoul" }, { "input": "4\nblue\nyellow\nred\ngreen", "output": "2\nSoul\nPower" }, { "input": "5\nred\nyellow\ngreen\nblue\npurple", "output": "1\nSoul" }, { "input": "3\nred\nyellow\norange", "output": "3\nPower\nSpace\nTime" }, { "input": "4\norange\ngreen\nyellow\nred", "output": "2\nPower\nSpace" }, { "input": "5\norange\nred\ngreen\nyellow\npurple", "output": "1\nSpace" }, { "input": "4\nyellow\nred\norange\nblue", "output": "2\nTime\nPower" }, { "input": "5\npurple\nblue\norange\nyellow\nred", "output": "1\nTime" }, { "input": "5\norange\nblue\nyellow\nred\ngreen", "output": "1\nPower" } ]

1,581,040,004

2,147,483,647

Python 3

OK

TESTS

64

124

0

a=['purple', 'green', 'blue', 'orange', 'red', 'yellow'] b=['Power', 'Time', 'Space', 'Soul', 'Reality', 'Mind'] n=int(input()) c=[input() for _ in [0]*n] d=[b[x] for x,y in enumerate(a) if y not in c] print(len(d),*d,sep='\n')

Title: Infinity Gauntlet Time Limit: None seconds Memory Limit: None megabytes Problem Description: You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems: - the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color. Using colors of Gems you saw in the Gauntlet determine the names of absent Gems. Input Specification: In the first line of input there is one integer $n$ ($0 \le n \le 6$) — the number of Gems in Infinity Gauntlet. In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters. Output Specification: In the first line output one integer $m$ ($0 \le m \le 6$) — the number of absent Gems. Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase. Demo Input: ['4\nred\npurple\nyellow\norange\n', '0\n'] Demo Output: ['2\nSpace\nTime\n', '6\nTime\nMind\nSoul\nPower\nReality\nSpace\n'] Note: In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space. In the second sample Thanos doesn't have any Gems, so he needs all six.

```python a=['purple', 'green', 'blue', 'orange', 'red', 'yellow'] b=['Power', 'Time', 'Space', 'Soul', 'Reality', 'Mind'] n=int(input()) c=[input() for _ in [0]*n] d=[b[x] for x,y in enumerate(a) if y not in c] print(len(d),*d,sep='\n') ```

3

507

C

Guess Your Way Out!

PROGRAMMING

1,700

[ "implementation", "math", "trees" ]

null

Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height *h*. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node. Let's index all the leaf nodes from the left to the right from 1 to 2*h*. The exit is located at some node *n* where 1<=≤<=*n*<=≤<=2*h*, the player doesn't know where the exit is so he has to guess his way out! Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules: - Character 'L' means "go to the left child of the current node"; - Character 'R' means "go to the right child of the current node"; - If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node; - If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command; - If he reached a leaf node that is not the exit, he returns to the parent of the current node; - If he reaches an exit, the game is finished. Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?

Input consists of two integers *h*,<=*n* (1<=≤<=*h*<=≤<=50, 1<=≤<=*n*<=≤<=2*h*).

Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.

[ "1 2\n", "2 3\n", "3 6\n", "10 1024\n" ]

[ "2", "5", "10", "2046" ]

A perfect binary tree of height *h* is a binary tree consisting of *h* + 1 levels. Level 0 consists of a single node called root, level *h* consists of 2*h* nodes called leaves. Each node that is not a leaf has exactly two children, left and right one. Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit. <img class="tex-graphics" src="https://espresso.codeforces.com/e9d0715dc8cd9b4f6ac7a0fb137563f857660adc.png" style="max-width: 100.0%;max-height: 100.0%;"/>

1,500

[ { "input": "1 2", "output": "2" }, { "input": "2 3", "output": "5" }, { "input": "3 6", "output": "10" }, { "input": "10 1024", "output": "2046" }, { "input": "10 577", "output": "1345" }, { "input": "11 550", "output": "408" }, { "input": "19 12783", "output": "503251" }, { "input": "28 72803174", "output": "50649698" }, { "input": "39 457181784666", "output": "830699159852" }, { "input": "12 955", "output": "2871" }, { "input": "13 154", "output": "7770" }, { "input": "14 2334", "output": "9440" }, { "input": "15 15512", "output": "14926" }, { "input": "16 21395", "output": "2899" }, { "input": "17 80239", "output": "177237" }, { "input": "18 153276", "output": "328766" }, { "input": "20 589266", "output": "1505684" }, { "input": "21 1687606", "output": "3522472" }, { "input": "24 14428281", "output": "26969983" }, { "input": "29 113463931", "output": "347736449" }, { "input": "1 1", "output": "1" }, { "input": "3 8", "output": "14" }, { "input": "31 1819651953", "output": "3412135549" }, { "input": "33 2599588275", "output": "1357401405" }, { "input": "38 262402936512", "output": "519008349260" }, { "input": "4 13", "output": "27" }, { "input": "40 615535158153", "output": "1572205271927" }, { "input": "42 1042128038474", "output": "3195908899134" }, { "input": "45 17519319833295", "output": "17381304930499" }, { "input": "46 34999315964173", "output": "34646522010881" }, { "input": "49 295606900104348", "output": "820858833984106" }, { "input": "50 905353992267944", "output": "1871650493613618" }, { "input": "3 5", "output": "11" }, { "input": "4 14", "output": "26" }, { "input": "6 40", "output": "88" }, { "input": "7 31", "output": "95" }, { "input": "8 19", "output": "205" }, { "input": "10 359", "output": "91" }, { "input": "11 349", "output": "1057" }, { "input": "13 4796", "output": "10298" }, { "input": "20 742273", "output": "1182599" }, { "input": "22 3343393", "output": "7009189" }, { "input": "24 3543583", "output": "10865127" }, { "input": "25 678676", "output": "31527640" }, { "input": "27 109473899", "output": "209022797" }, { "input": "29 19827102", "output": "478963048" }, { "input": "8 204", "output": "422" }, { "input": "30 414940886", "output": "372407442" }, { "input": "32 3786259360", "output": "8003335020" }, { "input": "35 31233562499", "output": "60951693197" }, { "input": "38 99361414961", "output": "28342263489" }, { "input": "9 43", "output": "391" }, { "input": "40 874338951117", "output": "1800799608767" }, { "input": "33 2696188969", "output": "505562011" }, { "input": "35 12080044014", "output": "2415167450" }, { "input": "12 2715", "output": "4185" }, { "input": "29 524109003", "output": "1052258991" }, { "input": "50 1", "output": "1125899906842623" }, { "input": "50 1125899906842624", "output": "2251799813685246" }, { "input": "50 562949953421312", "output": "562949953421312" }, { "input": "50 844424930131968", "output": "1407374883553280" }, { "input": "50 375299968947542", "output": "50" } ]

1,606,941,942

2,147,483,647

PyPy 3

OK

TESTS

62

140

0

import sys import math MAXNUM = math.inf MINNUM = -1 * math.inf ASCIILOWER = 97 ASCIIUPPER = 65 def getInt(): return int(sys.stdin.readline().rstrip()) def getInts(): return map(int, sys.stdin.readline().rstrip().split(" ")) def getString(): return sys.stdin.readline().rstrip() def printOutput(ans): sys.stdout.write(str(ans) + "\n") def solve(h, n): heightNodes = [1 for _ in range(51)] for height in range(1, 51): heightNodes[height] = 2 ** height + heightNodes[height - 1] curPos = 2 ** (h - 1) depth = h add = 2 ** (h - 1) direction = -1 total = 0 # visited root while True: if depth == 0: break if (curPos >= n and direction == -1) or ( curPos < n and direction == 1 ): # this means n is to the left total += 1 depth -= 1 add //= 2 curPos += direction * (add) else: total += heightNodes[depth - 1] direction *= -1 return total def readinput(): h, n = getInts() printOutput(solve(h, n)) readinput()

Title: Guess Your Way Out! Time Limit: None seconds Memory Limit: None megabytes Problem Description: Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height *h*. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node. Let's index all the leaf nodes from the left to the right from 1 to 2*h*. The exit is located at some node *n* where 1<=≤<=*n*<=≤<=2*h*, the player doesn't know where the exit is so he has to guess his way out! Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules: - Character 'L' means "go to the left child of the current node"; - Character 'R' means "go to the right child of the current node"; - If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node; - If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command; - If he reached a leaf node that is not the exit, he returns to the parent of the current node; - If he reaches an exit, the game is finished. Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit? Input Specification: Input consists of two integers *h*,<=*n* (1<=≤<=*h*<=≤<=50, 1<=≤<=*n*<=≤<=2*h*). Output Specification: Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm. Demo Input: ['1 2\n', '2 3\n', '3 6\n', '10 1024\n'] Demo Output: ['2', '5', '10', '2046'] Note: A perfect binary tree of height *h* is a binary tree consisting of *h* + 1 levels. Level 0 consists of a single node called root, level *h* consists of 2*h* nodes called leaves. Each node that is not a leaf has exactly two children, left and right one. Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit. <img class="tex-graphics" src="https://espresso.codeforces.com/e9d0715dc8cd9b4f6ac7a0fb137563f857660adc.png" style="max-width: 100.0%;max-height: 100.0%;"/>

```python import sys import math MAXNUM = math.inf MINNUM = -1 * math.inf ASCIILOWER = 97 ASCIIUPPER = 65 def getInt(): return int(sys.stdin.readline().rstrip()) def getInts(): return map(int, sys.stdin.readline().rstrip().split(" ")) def getString(): return sys.stdin.readline().rstrip() def printOutput(ans): sys.stdout.write(str(ans) + "\n") def solve(h, n): heightNodes = [1 for _ in range(51)] for height in range(1, 51): heightNodes[height] = 2 ** height + heightNodes[height - 1] curPos = 2 ** (h - 1) depth = h add = 2 ** (h - 1) direction = -1 total = 0 # visited root while True: if depth == 0: break if (curPos >= n and direction == -1) or ( curPos < n and direction == 1 ): # this means n is to the left total += 1 depth -= 1 add //= 2 curPos += direction * (add) else: total += heightNodes[depth - 1] direction *= -1 return total def readinput(): h, n = getInts() printOutput(solve(h, n)) readinput() ```

3

584

A

Olesya and Rodion

PROGRAMMING

1,000

[ "math" ]

null

Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them. Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.

The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.

Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.

[ "3 2\n" ]

[ "712" ]

none

500

[ { "input": "3 2", "output": "222" }, { "input": "2 2", "output": "22" }, { "input": "4 3", "output": "3333" }, { "input": "5 3", "output": "33333" }, { "input": "10 7", "output": "7777777777" }, { "input": "2 9", "output": "99" }, { "input": "18 8", "output": "888888888888888888" }, { "input": "1 5", "output": "5" }, { "input": "1 10", "output": "-1" }, { "input": "100 5", "output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555" }, { "input": "10 2", "output": "2222222222" }, { "input": "18 10", "output": "111111111111111110" }, { "input": "1 9", "output": "9" }, { "input": "7 6", "output": "6666666" }, { "input": "4 4", "output": "4444" }, { "input": "14 7", "output": "77777777777777" }, { "input": "3 8", "output": "888" }, { "input": "1 3", "output": "3" }, { "input": "2 8", "output": "88" }, { "input": "3 8", "output": "888" }, { "input": "4 3", "output": "3333" }, { "input": "5 9", "output": "99999" }, { "input": "4 8", "output": "8888" }, { "input": "3 4", "output": "444" }, { "input": "9 4", "output": "444444444" }, { "input": "8 10", "output": "11111110" }, { "input": "1 6", "output": "6" }, { "input": "20 3", "output": "33333333333333333333" }, { "input": "15 10", "output": "111111111111110" }, { "input": "31 4", "output": "4444444444444444444444444444444" }, { "input": "18 9", "output": "999999999999999999" }, { "input": "72 4", "output": "444444444444444444444444444444444444444444444444444444444444444444444444" }, { "input": "76 8", "output": "8888888888888888888888888888888888888888888888888888888888888888888888888888" }, { "input": "12 5", "output": "555555555555" }, { "input": "54 5", "output": "555555555555555555555555555555555555555555555555555555" }, { "input": "96 10", "output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "15 9", "output": "999999999999999" }, { "input": "100 2", "output": "2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222" }, { "input": "99 3", "output": "333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333" }, { "input": "98 4", "output": "44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444" }, { "input": "97 5", "output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555" }, { "input": "100 6", "output": "6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666" }, { "input": "99 7", "output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "98 8", "output": "88888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888" }, { "input": "97 9", "output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999" }, { "input": "100 10", "output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "3 10", "output": "110" }, { "input": "2 4", "output": "44" }, { "input": "2 10", "output": "10" }, { "input": "10 10", "output": "1111111110" }, { "input": "4 10", "output": "1110" }, { "input": "25 10", "output": "1111111111111111111111110" }, { "input": "50 10", "output": "11111111111111111111111111111111111111111111111110" }, { "input": "5 10", "output": "11110" }, { "input": "99 10", "output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "100 10", "output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110" }, { "input": "1 2", "output": "2" }, { "input": "20 10", "output": "11111111111111111110" }, { "input": "99 7", "output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" }, { "input": "89 5", "output": "55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555" }, { "input": "6 10", "output": "111110" }, { "input": "3 8", "output": "888" }, { "input": "1 9", "output": "9" }, { "input": "1 4", "output": "4" }, { "input": "100 7", "output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777" } ]

1,696,287,260

2,147,483,647

PyPy 3-64

OK

TESTS

65

62

0

n,k = map(int,input().split()) if n==1 and k==10: print(-1) elif k==10: print('1'+'0'*(n-1)) else: print(str(k)*n)

Title: Olesya and Rodion Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them. Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. Input Specification: The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. Output Specification: Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. Demo Input: ['3 2\n'] Demo Output: ['712'] Note: none

```python n,k = map(int,input().split()) if n==1 and k==10: print(-1) elif k==10: print('1'+'0'*(n-1)) else: print(str(k)*n) ```

3

793

A

Oleg and shares

PROGRAMMING

900

[ "implementation", "math" ]

null

Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?

The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices.

Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible.

[ "3 3\n12 9 15\n", "2 2\n10 9\n", "4 1\n1 1000000000 1000000000 1000000000\n" ]

[ "3", "-1", "2999999997" ]

Consider the first example. Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds. There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3. In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal. In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time.

500

[ { "input": "3 3\n12 9 15", "output": "3" }, { "input": "2 2\n10 9", "output": "-1" }, { "input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997" }, { "input": "1 11\n123", "output": "0" }, { "input": "20 6\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14", "output": "151" }, { "input": "20 5\n59 54 19 88 55 100 54 3 6 13 99 38 36 71 59 6 64 85 45 54", "output": "-1" }, { "input": "100 10\n340 70 440 330 130 120 340 210 440 110 410 120 180 40 50 230 70 110 310 360 480 70 230 120 230 310 470 60 210 60 210 480 290 250 450 440 150 40 500 230 280 250 30 50 310 50 230 360 420 260 330 80 50 160 70 470 140 180 380 190 250 30 220 410 80 310 280 50 20 430 440 180 310 190 190 330 90 190 320 390 170 460 230 30 80 500 470 370 80 500 400 120 220 150 70 120 70 320 260 260", "output": "2157" }, { "input": "100 18\n489 42 300 366 473 105 220 448 70 488 201 396 168 281 67 235 324 291 313 387 407 223 39 144 224 233 72 318 229 377 62 171 448 119 354 282 147 447 260 384 172 199 67 326 311 431 337 142 281 202 404 468 38 120 90 437 33 420 249 372 367 253 255 411 309 333 103 176 162 120 203 41 352 478 216 498 224 31 261 493 277 99 375 370 394 229 71 488 246 194 233 13 66 111 366 456 277 360 116 354", "output": "-1" }, { "input": "4 2\n1 2 3 4", "output": "-1" }, { "input": "3 4\n3 5 5", "output": "-1" }, { "input": "3 2\n88888884 88888886 88888888", "output": "3" }, { "input": "2 1\n1000000000 1000000000", "output": "0" }, { "input": "4 2\n1000000000 100000000 100000000 100000000", "output": "450000000" }, { "input": "2 2\n1000000000 1000000000", "output": "0" }, { "input": "3 3\n3 2 1", "output": "-1" }, { "input": "3 4\n3 5 3", "output": "-1" }, { "input": "3 2\n1 2 2", "output": "-1" }, { "input": "4 2\n2 3 3 2", "output": "-1" }, { "input": "3 2\n1 2 4", "output": "-1" }, { "input": "3 2\n3 4 4", "output": "-1" }, { "input": "3 3\n4 7 10", "output": "3" }, { "input": "4 3\n2 2 5 1", "output": "-1" }, { "input": "3 3\n1 3 5", "output": "-1" }, { "input": "2 5\n5 9", "output": "-1" }, { "input": "2 3\n5 7", "output": "-1" }, { "input": "3 137\n1000000000 1000000000 1000000000", "output": "0" }, { "input": "5 1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "3 5\n1 2 5", "output": "-1" }, { "input": "3 3\n1000000000 1000000000 999999997", "output": "2" }, { "input": "2 4\n5 6", "output": "-1" }, { "input": "4 1\n1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "2 3\n5 8", "output": "1" }, { "input": "2 6\n8 16", "output": "-1" }, { "input": "5 3\n15 14 9 12 18", "output": "-1" }, { "input": "3 3\n1 2 3", "output": "-1" }, { "input": "3 3\n3 4 5", "output": "-1" }, { "input": "2 5\n8 17", "output": "-1" }, { "input": "2 1\n1 2", "output": "1" }, { "input": "1 1\n1000000000", "output": "0" }, { "input": "3 3\n5 3 4", "output": "-1" }, { "input": "3 6\n10 14 12", "output": "-1" }, { "input": "2 2\n3 5", "output": "1" }, { "input": "3 5\n1 3 4", "output": "-1" }, { "input": "4 3\n1 6 6 6", "output": "-1" }, { "input": "2 3\n1 8", "output": "-1" }, { "input": "3 5\n6 11 17", "output": "-1" }, { "input": "2 2\n1 4", "output": "-1" }, { "input": "2 4\n6 8", "output": "-1" }, { "input": "2 1\n2 3", "output": "1" }, { "input": "4 4\n1 5 8 14", "output": "-1" }, { "input": "3 3\n1 5 3", "output": "-1" }, { "input": "4 3\n1 2 2 5", "output": "-1" }, { "input": "3 2\n1 4 6", "output": "-1" }, { "input": "2 3\n6 9", "output": "1" }, { "input": "3 3\n2 3 4", "output": "-1" }, { "input": "3 2\n9 10 10", "output": "-1" }, { "input": "2 2\n9 12", "output": "-1" }, { "input": "2 2\n100000003 100000005", "output": "1" }, { "input": "2 3\n2 4", "output": "-1" }, { "input": "3 2\n2 3 5", "output": "-1" }, { "input": "3 3\n1 3 4", "output": "-1" }, { "input": "10 2\n2 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "4499999991" }, { "input": "3 5\n2 4 5", "output": "-1" }, { "input": "2 3\n7 10", "output": "1" }, { "input": "3 10\n10 13 17", "output": "-1" }, { "input": "2 3\n1 6", "output": "-1" }, { "input": "1 7\n1000000000", "output": "0" }, { "input": "2 4\n3 7", "output": "1" }, { "input": "2 3\n2 5", "output": "1" }, { "input": "20 1\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000", "output": "0" }, { "input": "3 3\n7 8 8", "output": "-1" }, { "input": "4 10\n1 11 100 11", "output": "-1" } ]

1,593,018,892

2,147,483,647

PyPy 3

OK

TESTS

88

234

34,099,200

n , k = map(int,input().split()) l = list(map(int,input().split())) st = set(l) cnt = 0 if len(st) == 1 : print(0) exit() else: x = min(l) for i in l : c = i - x if c % k == 0 : cnt += c // k else: print(-1) exit() print(cnt)

Title: Oleg and shares Time Limit: None seconds Memory Limit: None megabytes Problem Description: Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question? Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices. Output Specification: Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. Demo Input: ['3 3\n12 9 15\n', '2 2\n10 9\n', '4 1\n1 1000000000 1000000000 1000000000\n'] Demo Output: ['3', '-1', '2999999997'] Note: Consider the first example. Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds. There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3. In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal. In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time.

```python n , k = map(int,input().split()) l = list(map(int,input().split())) st = set(l) cnt = 0 if len(st) == 1 : print(0) exit() else: x = min(l) for i in l : c = i - x if c % k == 0 : cnt += c // k else: print(-1) exit() print(cnt) ```

3

155

A

I_love_\%username\%

PROGRAMMING

800

[ "brute force" ]

null

Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him. One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously). Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.

The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated. The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.

Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.

[ "5\n100 50 200 150 200\n", "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n" ]

[ "2\n", "4\n" ]

In the first sample the performances number 2 and 3 are amazing. In the second sample the performances number 2, 4, 9 and 10 are amazing.

500

[ { "input": "5\n100 50 200 150 200", "output": "2" }, { "input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242", "output": "4" }, { "input": "1\n6", "output": "0" }, { "input": "2\n2 1", "output": "1" }, { "input": "5\n100 36 53 7 81", "output": "2" }, { "input": "5\n7 36 53 81 100", "output": "4" }, { "input": "5\n100 81 53 36 7", "output": "4" }, { "input": "10\n8 6 3 4 9 10 7 7 1 3", "output": "5" }, { "input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862", "output": "6" }, { "input": "10\n1 3 3 4 6 7 7 8 9 10", "output": "7" }, { "input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137", "output": "9" }, { "input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198", "output": "5" }, { "input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101", "output": "5" }, { "input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498", "output": "7" }, { "input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679", "output": "9" }, { "input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992", "output": "57" }, { "input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83", "output": "4" }, { "input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901", "output": "57" }, { "input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100", "output": "99" }, { "input": "2\n0 10000", "output": "1" }, { "input": "2\n5 5", "output": "0" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n10 10", "output": "0" }, { "input": "1\n0", "output": "0" } ]

1,692,593,754

2,147,483,647

Python 3

OK

TESTS

37

92

0

n = int(input()) points = list(map(int, input().split())) amazing_performance = 0 highest_point = points[0] lowest_point = points[0] for i in range(1, n): if points[i] > highest_point: highest_point = points[i] amazing_performance += 1 elif points[i] < lowest_point: lowest_point = points[i] amazing_performance += 1 print(amazing_performance)

Title: I_love_\%username\% Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him. One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously). Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. Input Specification: The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated. The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. Output Specification: Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. Demo Input: ['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n'] Demo Output: ['2\n', '4\n'] Note: In the first sample the performances number 2 and 3 are amazing. In the second sample the performances number 2, 4, 9 and 10 are amazing.

```python n = int(input()) points = list(map(int, input().split())) amazing_performance = 0 highest_point = points[0] lowest_point = points[0] for i in range(1, n): if points[i] > highest_point: highest_point = points[i] amazing_performance += 1 elif points[i] < lowest_point: lowest_point = points[i] amazing_performance += 1 print(amazing_performance) ```

3

930

A

Peculiar apple-tree

PROGRAMMING

1,500

[ "dfs and similar", "graphs", "trees" ]

null

In Arcady's garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are *n* inflorescences, numbered from 1 to *n*. Inflorescence number 1 is situated near base of tree and any other inflorescence with number *i* (*i*<=><=1) is situated at the top of branch, which bottom is *p**i*-th inflorescence and *p**i*<=<<=*i*. Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in *a*-th inflorescence gets to *p**a*-th inflorescence. Apples that end up in first inflorescence are gathered by Arcady in exactly the same moment. Second peculiarity of this tree is that once two apples are in same inflorescence they annihilate. This happens with each pair of apples, e.g. if there are 5 apples in same inflorescence in same time, only one will not be annihilated and if there are 8 apples, all apples will be annihilated. Thus, there can be no more than one apple in each inflorescence in each moment of time. Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest.

First line of input contains single integer number *n* (2<=≤<=*n*<=≤<=100<=000) — number of inflorescences. Second line of input contains sequence of *n*<=-<=1 integer numbers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=<<=*i*), where *p**i* is number of inflorescence into which the apple from *i*-th inflorescence rolls down.

Single line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest.

[ "3\n1 1\n", "5\n1 2 2 2\n", "18\n1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4\n" ]

[ "1\n", "3\n", "4\n" ]

In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won't be able to collect them. In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it.

500

[ { "input": "3\n1 1", "output": "1" }, { "input": "5\n1 2 2 2", "output": "3" }, { "input": "18\n1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4", "output": "4" }, { "input": "2\n1", "output": "2" }, { "input": "3\n1 2", "output": "3" }, { "input": "20\n1 1 1 1 1 4 1 2 4 1 2 1 7 1 2 2 9 7 1", "output": "2" }, { "input": "20\n1 2 1 2 2 1 2 4 1 6 2 2 4 3 2 6 2 5 9", "output": "2" }, { "input": "20\n1 1 1 4 2 4 3 1 2 8 3 2 11 13 15 1 12 13 12", "output": "4" }, { "input": "20\n1 2 2 4 3 5 5 6 6 9 11 9 9 12 13 10 15 13 15", "output": "4" }, { "input": "20\n1 2 3 4 5 6 7 8 9 6 11 12 12 7 13 15 16 11 13", "output": "8" }, { "input": "10\n1 1 1 2 1 3 4 2 1", "output": "2" }, { "input": "30\n1 1 1 2 1 2 1 1 2 1 1 1 2 2 4 3 6 2 3 5 3 4 11 5 3 3 4 7 6", "output": "4" }, { "input": "40\n1 1 1 1 1 1 1 1 1 3 4 3 3 1 3 6 7 4 5 2 4 3 9 1 4 2 5 3 5 9 5 9 10 12 3 7 2 11 1", "output": "2" }, { "input": "50\n1 1 1 1 1 2 3 3 2 1 1 2 3 1 3 1 5 6 4 1 1 2 1 2 1 10 17 2 2 4 12 9 6 6 5 13 1 3 2 8 25 3 22 1 10 13 6 3 2", "output": "4" }, { "input": "10\n1 1 1 1 2 1 3 4 3", "output": "2" }, { "input": "30\n1 2 1 1 1 2 1 4 2 3 9 2 3 2 1 1 4 3 12 4 8 8 3 7 9 1 9 19 1", "output": "2" }, { "input": "40\n1 1 1 2 3 1 2 1 3 7 1 3 4 3 2 3 4 1 2 2 4 1 7 4 1 3 2 1 4 5 3 10 14 11 10 13 8 7 4", "output": "2" }, { "input": "50\n1 2 1 1 1 3 1 3 1 5 3 2 7 3 6 6 3 1 4 2 3 10 8 9 1 4 5 2 8 6 12 9 7 5 7 19 3 15 10 4 12 4 19 5 16 5 3 13 5", "output": "2" }, { "input": "10\n1 1 1 2 3 2 1 2 3", "output": "2" }, { "input": "30\n1 1 1 1 2 1 4 4 2 3 2 1 1 1 1 3 1 1 3 2 3 5 1 2 9 16 2 4 3", "output": "2" }, { "input": "40\n1 1 1 2 1 2 1 2 4 8 1 7 1 6 2 8 2 12 4 11 5 5 15 3 12 11 22 11 13 13 24 6 10 15 3 6 7 1 2", "output": "2" }, { "input": "50\n1 1 1 1 3 4 1 2 3 5 1 2 1 5 1 10 4 11 1 8 8 4 4 12 5 3 4 1 1 2 5 13 13 2 2 10 12 3 19 14 1 1 15 3 23 21 12 3 14", "output": "4" }, { "input": "10\n1 1 1 1 2 4 1 1 3", "output": "2" }, { "input": "30\n1 1 1 1 3 3 2 3 7 4 1 2 4 6 2 8 1 2 13 7 5 15 3 3 8 4 4 18 3", "output": "2" }, { "input": "40\n1 1 1 2 2 1 1 4 6 4 7 7 7 4 4 8 10 7 5 1 5 13 7 8 2 11 18 2 1 20 7 3 12 16 2 22 4 22 14", "output": "4" }, { "input": "50\n1 1 1 2 2 1 3 5 3 1 9 4 4 2 12 15 3 13 8 8 4 13 20 17 19 2 4 3 9 5 17 9 17 1 5 7 6 5 20 11 31 33 32 20 6 25 1 2 6", "output": "4" }, { "input": "10\n1 1 1 3 3 5 6 8 3", "output": "4" }, { "input": "30\n1 2 2 1 5 5 5 1 7 4 10 2 4 11 2 3 10 10 7 13 12 4 10 3 22 25 8 1 1", "output": "6" }, { "input": "40\n1 2 2 2 2 4 2 2 6 9 3 9 9 9 3 5 7 7 2 17 4 4 8 8 25 18 12 27 8 19 26 15 33 26 33 9 24 4 27", "output": "4" }, { "input": "50\n1 1 3 3 4 5 5 2 4 3 9 9 1 5 5 7 5 5 16 1 18 3 6 5 6 13 26 12 23 20 17 21 9 17 19 34 12 24 11 9 32 10 40 42 7 40 11 25 3", "output": "6" }, { "input": "10\n1 2 1 2 5 5 6 6 6", "output": "2" }, { "input": "30\n1 1 3 3 5 6 7 5 7 6 5 4 8 6 10 12 14 9 15 20 6 21 14 24 17 23 23 18 8", "output": "2" }, { "input": "40\n1 2 2 3 1 2 5 6 4 8 11 12 9 5 12 7 4 16 16 15 6 22 17 24 10 8 22 4 27 9 19 23 16 18 28 22 5 35 19", "output": "4" }, { "input": "50\n1 2 3 4 5 5 5 7 1 2 11 5 7 11 11 11 15 3 17 10 6 18 14 14 24 11 10 7 17 18 8 7 19 18 31 27 21 30 34 32 27 39 38 22 32 23 31 48 25", "output": "2" }, { "input": "10\n1 2 2 4 5 5 6 4 7", "output": "2" }, { "input": "30\n1 2 3 3 5 6 3 8 9 10 10 10 11 7 8 8 15 16 13 13 19 12 15 18 18 24 27 25 10", "output": "6" }, { "input": "40\n1 2 3 4 5 6 6 8 7 10 11 3 12 11 15 12 17 15 10 20 16 20 12 20 15 21 20 26 29 23 29 30 23 24 35 33 25 32 36", "output": "8" }, { "input": "50\n1 2 2 2 5 6 7 7 9 10 7 4 5 4 15 15 16 17 10 19 18 16 15 24 20 8 27 16 19 24 23 32 17 23 29 18 35 35 38 35 39 41 42 38 19 46 38 28 29", "output": "6" }, { "input": "10\n1 2 3 4 5 5 5 7 9", "output": "8" }, { "input": "30\n1 2 3 4 5 6 5 3 6 7 8 11 12 13 15 15 13 13 19 10 14 10 15 23 21 9 27 22 28", "output": "4" }, { "input": "40\n1 2 2 3 3 6 5 5 9 7 8 11 13 7 10 10 16 14 18 20 11 19 23 18 20 21 25 16 29 25 27 31 26 34 33 23 36 33 32", "output": "6" }, { "input": "50\n1 2 2 4 5 5 7 6 9 10 11 12 13 7 14 15 14 17 10 14 9 21 23 23 19 26 19 25 11 24 22 27 26 34 35 30 37 31 38 32 40 32 42 44 37 21 40 40 48", "output": "10" }, { "input": "10\n1 2 3 4 3 6 6 6 7", "output": "4" }, { "input": "30\n1 2 2 4 5 6 5 7 9 6 4 12 7 14 12 12 15 17 13 12 8 20 21 15 17 24 21 19 16", "output": "4" }, { "input": "40\n1 2 3 4 4 6 6 4 9 9 10 12 10 12 12 16 8 13 18 14 17 20 21 23 25 22 25 26 29 26 27 27 33 31 33 34 36 29 34", "output": "10" }, { "input": "50\n1 2 3 3 4 3 6 7 8 10 11 10 12 11 11 14 13 8 17 20 21 19 15 18 21 18 17 23 25 28 25 27 29 32 32 34 37 29 30 39 41 35 24 41 37 36 41 35 43", "output": "10" }, { "input": "99\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "1" }, { "input": "99\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98", "output": "99" }, { "input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "2" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99", "output": "100" } ]

1,644,592,407

2,147,483,647

Python 3

OK

TESTS

90

156

14,950,400

n=int(input()) a=[*map(int,input().split())] q={} for i in range(1,n+1): q[i]=0 for i in range(n-1): q[i+2]=q[a[i]]+1 z={} for i in q.values(): z[i]=z.get(i,0)+1 print(sum(z[i]%2 for i in z))

Title: Peculiar apple-tree Time Limit: None seconds Memory Limit: None megabytes Problem Description: In Arcady's garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are *n* inflorescences, numbered from 1 to *n*. Inflorescence number 1 is situated near base of tree and any other inflorescence with number *i* (*i*<=><=1) is situated at the top of branch, which bottom is *p**i*-th inflorescence and *p**i*<=<<=*i*. Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in *a*-th inflorescence gets to *p**a*-th inflorescence. Apples that end up in first inflorescence are gathered by Arcady in exactly the same moment. Second peculiarity of this tree is that once two apples are in same inflorescence they annihilate. This happens with each pair of apples, e.g. if there are 5 apples in same inflorescence in same time, only one will not be annihilated and if there are 8 apples, all apples will be annihilated. Thus, there can be no more than one apple in each inflorescence in each moment of time. Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest. Input Specification: First line of input contains single integer number *n* (2<=≤<=*n*<=≤<=100<=000) — number of inflorescences. Second line of input contains sequence of *n*<=-<=1 integer numbers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=<<=*i*), where *p**i* is number of inflorescence into which the apple from *i*-th inflorescence rolls down. Output Specification: Single line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest. Demo Input: ['3\n1 1\n', '5\n1 2 2 2\n', '18\n1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4\n'] Demo Output: ['1\n', '3\n', '4\n'] Note: In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won't be able to collect them. In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it.

```python n=int(input()) a=[*map(int,input().split())] q={} for i in range(1,n+1): q[i]=0 for i in range(n-1): q[i+2]=q[a[i]]+1 z={} for i in q.values(): z[i]=z.get(i,0)+1 print(sum(z[i]%2 for i in z)) ```

3

515

C

Drazil and Factorial

PROGRAMMING

1,400

[ "greedy", "math", "sortings" ]

null

Drazil is playing a math game with Varda. Let's define for positive integer *x* as a product of factorials of its digits. For example, . First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions: 1. *x* doesn't contain neither digit 0 nor digit 1. 2. = . Help friends find such number.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*. The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.

Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.

[ "4\n1234\n", "3\n555\n" ]

[ "33222\n", "555\n" ]

In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/>

1,000

[ { "input": "4\n1234", "output": "33222" }, { "input": "3\n555", "output": "555" }, { "input": "15\n012345781234578", "output": "7777553333222222222222" }, { "input": "1\n8", "output": "7222" }, { "input": "10\n1413472614", "output": "75333332222222" }, { "input": "8\n68931246", "output": "77553333332222222" }, { "input": "7\n4424368", "output": "75333332222222222" }, { "input": "6\n576825", "output": "7755532222" }, { "input": "5\n97715", "output": "7775332" }, { "input": "3\n915", "output": "75332" }, { "input": "2\n26", "output": "532" }, { "input": "1\n4", "output": "322" }, { "input": "15\n028745260720699", "output": "7777755533333332222222222" }, { "input": "13\n5761790121605", "output": "7775555333322" }, { "input": "10\n3312667105", "output": "755533332" }, { "input": "1\n7", "output": "7" }, { "input": "15\n989898989898989", "output": "777777777777777333333333333333322222222222222222222222222222" }, { "input": "15\n000000000000007", "output": "7" }, { "input": "15\n999999999999990", "output": "77777777777777333333333333333333333333333322222222222222" }, { "input": "1\n2", "output": "2" }, { "input": "1\n3", "output": "3" }, { "input": "1\n4", "output": "322" }, { "input": "1\n5", "output": "5" }, { "input": "1\n6", "output": "53" }, { "input": "1\n7", "output": "7" }, { "input": "1\n9", "output": "7332" }, { "input": "2\n09", "output": "7332" }, { "input": "13\n1337251172966", "output": "777555333333222" }, { "input": "15\n987654329876543", "output": "777777555533333333332222222222222" }, { "input": "9\n234567899", "output": "777755333333322222222" }, { "input": "2\n99", "output": "77333322" }, { "input": "2\n66", "output": "5533" }, { "input": "3\n999", "output": "777333333222" }, { "input": "5\n99999", "output": "77777333333333322222" }, { "input": "9\n123456789", "output": "77755333332222222" }, { "input": "9\n987654321", "output": "77755333332222222" }, { "input": "3\n666", "output": "555333" }, { "input": "2\n26", "output": "532" }, { "input": "6\n555777", "output": "777555" }, { "input": "10\n1234567899", "output": "777755333333322222222" }, { "input": "4\n6666", "output": "55553333" }, { "input": "4\n9754", "output": "775333222" }, { "input": "2\n95", "output": "75332" }, { "input": "14\n11122233344455", "output": "55333333222222222" }, { "input": "12\n836544897832", "output": "77777553333333222222222222222" } ]

1,625,917,931

2,147,483,647

PyPy 3

OK

TESTS

45

109

0

l=[""," ","2","3","322","5","53","7","7222","7332"] s="" n=int(input()) k=input() for i in range(n): s=s+l[int(k[i])] s=list(s) s.sort(reverse=True) print(*s,sep="")

Title: Drazil and Factorial Time Limit: None seconds Memory Limit: None megabytes Problem Description: Drazil is playing a math game with Varda. Let's define for positive integer *x* as a product of factorials of its digits. For example, . First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions: 1. *x* doesn't contain neither digit 0 nor digit 1. 2. = . Help friends find such number. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*. The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes. Output Specification: Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Demo Input: ['4\n1234\n', '3\n555\n'] Demo Output: ['33222\n', '555\n'] Note: In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/>

```python l=[""," ","2","3","322","5","53","7","7222","7332"] s="" n=int(input()) k=input() for i in range(n): s=s+l[int(k[i])] s=list(s) s.sort(reverse=True) print(*s,sep="") ```

3

667

A

Pouring Rain

PROGRAMMING

1,100

[ "geometry", "math" ]

null

A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation. Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom. You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously. Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds. Note one milliliter equals to one cubic centimeter.

The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where: - *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup.

If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104.

[ "1 2 3 100\n", "1 1 1 1\n" ]

[ "NO\n", "YES\n3.659792366325\n" ]

In the first example the water fills the cup faster than you can drink from it. In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds.

500

[ { "input": "1 2 3 100", "output": "NO" }, { "input": "1 1 1 1", "output": "YES\n3.659792366325" }, { "input": "48 7946 7992 72", "output": "NO" }, { "input": "72 6791 8546 46", "output": "NO" }, { "input": "100 5635 9099 23", "output": "NO" }, { "input": "20 287 3845 5", "output": "YES\n39.646277165210" }, { "input": "48 6428 9807 83", "output": "NO" }, { "input": "72 5272 4552 64", "output": "NO" }, { "input": "100 4117 5106 34", "output": "NO" }, { "input": "20 2961 9852 15", "output": "YES\n180.991437129723" }, { "input": "48 1805 3109 93", "output": "NO" }, { "input": "72 8534 7042 65", "output": "NO" }, { "input": "1 47 80 68", "output": "YES\n1.388102806810" }, { "input": "4 495 8813 1", "output": "YES\n0.706823517575" }, { "input": "5 2797 5925 9", "output": "YES\n9.553973511669" }, { "input": "1 8324 4362 23", "output": "YES\n1.505007106354" }, { "input": "6 1976 8455 3", "output": "YES\n6.674898722265" }, { "input": "7 2644 8080 5", "output": "YES\n12.900417790197" }, { "input": "3 4183 5491 98", "output": "YES\n6.162185601824" }, { "input": "2 8591 320 101", "output": "YES\n9999.259991757254" }, { "input": "10000 10000 10000 10000", "output": "NO" }, { "input": "2 5000 12 3", "output": "YES\n6099.653943875812" }, { "input": "10 1000 100 1", "output": "YES\n3659.792366325487" } ]

1,461,953,619

5,919

Python 3

OK

TESTS

23

62

4,608,000

import math (d,h,v,e) = tuple([int(p) for p in input().strip('\n').split(' ')]) increaseByRain = math.pi * (d/2) * (d/2) * e totalRate = v - increaseByRain if totalRate <= 0: print('NO') else: print('YES') currentVol = math.pi * (d/2) * (d/2) * h print(str(currentVol/totalRate))

Title: Pouring Rain Time Limit: None seconds Memory Limit: None megabytes Problem Description: A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation. Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom. You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously. Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 104 seconds. Note one milliliter equals to one cubic centimeter. Input Specification: The only line of the input contains four integer numbers *d*,<=*h*,<=*v*,<=*e* (1<=≤<=*d*,<=*h*,<=*v*,<=*e*<=≤<=104), where: - *d* — the diameter of your cylindrical cup, - *h* — the initial level of water in the cup, - *v* — the speed of drinking process from the cup in milliliters per second, - *e* — the growth of water because of rain if you do not drink from the cup. Output Specification: If it is impossible to make the cup empty, print "NO" (without quotes). Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=4. It is guaranteed that if the answer exists, it doesn't exceed 104. Demo Input: ['1 2 3 100\n', '1 1 1 1\n'] Demo Output: ['NO\n', 'YES\n3.659792366325\n'] Note: In the first example the water fills the cup faster than you can drink from it. In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds.

```python import math (d,h,v,e) = tuple([int(p) for p in input().strip('\n').split(' ')]) increaseByRain = math.pi * (d/2) * (d/2) * e totalRate = v - increaseByRain if totalRate <= 0: print('NO') else: print('YES') currentVol = math.pi * (d/2) * (d/2) * h print(str(currentVol/totalRate)) ```

3

946

A

Partition

PROGRAMMING

800

[ "greedy" ]

null

You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences. Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*?

The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*. The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*.

Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*.

[ "3\n1 -2 0\n", "6\n16 23 16 15 42 8\n" ]

[ "3\n", "120\n" ]

In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3. In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120.

0

[ { "input": "3\n1 -2 0", "output": "3" }, { "input": "6\n16 23 16 15 42 8", "output": "120" }, { "input": "1\n-1", "output": "1" }, { "input": "100\n-100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 -100", "output": "10000" }, { "input": "2\n-1 5", "output": "6" }, { "input": "3\n-2 0 1", "output": "3" }, { "input": "12\n-1 -2 -3 4 4 -6 -6 56 3 3 -3 3", "output": "94" }, { "input": "4\n1 -1 1 -1", "output": "4" }, { "input": "4\n100 -100 100 -100", "output": "400" }, { "input": "3\n-2 -5 10", "output": "17" }, { "input": "5\n1 -2 3 -4 5", "output": "15" }, { "input": "3\n-100 100 -100", "output": "300" }, { "input": "6\n1 -1 1 -1 1 -1", "output": "6" }, { "input": "6\n2 -2 2 -2 2 -2", "output": "12" }, { "input": "9\n12 93 -2 0 0 0 3 -3 -9", "output": "122" }, { "input": "6\n-1 2 4 -5 -3 55", "output": "70" }, { "input": "6\n-12 8 68 -53 1 -15", "output": "157" }, { "input": "2\n-2 1", "output": "3" }, { "input": "3\n100 -100 100", "output": "300" }, { "input": "5\n100 100 -1 -100 2", "output": "303" }, { "input": "6\n-5 -4 -3 -2 -1 0", "output": "15" }, { "input": "6\n4 4 4 -3 -3 2", "output": "20" }, { "input": "2\n-1 2", "output": "3" }, { "input": "1\n100", "output": "100" }, { "input": "5\n-1 -2 3 1 2", "output": "9" }, { "input": "5\n100 -100 100 -100 100", "output": "500" }, { "input": "5\n1 -1 1 -1 1", "output": "5" }, { "input": "4\n0 0 0 -1", "output": "1" }, { "input": "5\n100 -100 -1 2 100", "output": "303" }, { "input": "2\n75 0", "output": "75" }, { "input": "4\n55 56 -59 -58", "output": "228" }, { "input": "2\n9 71", "output": "80" }, { "input": "2\n9 70", "output": "79" }, { "input": "2\n9 69", "output": "78" }, { "input": "2\n100 -100", "output": "200" }, { "input": "4\n-9 4 -9 5", "output": "27" }, { "input": "42\n91 -27 -79 -56 80 -93 -23 10 80 94 61 -89 -64 81 34 99 31 -32 -69 92 79 -9 73 66 -8 64 99 99 58 -19 -40 21 1 -33 93 -23 -62 27 55 41 57 36", "output": "2348" }, { "input": "7\n-1 2 2 2 -1 2 -1", "output": "11" }, { "input": "6\n-12 8 17 -69 7 -88", "output": "201" }, { "input": "3\n1 -2 5", "output": "8" }, { "input": "6\n-2 3 -4 5 6 -1", "output": "21" }, { "input": "2\n-5 1", "output": "6" }, { "input": "4\n2 2 -2 4", "output": "10" }, { "input": "68\n21 47 -75 -25 64 83 83 -21 89 24 43 44 -35 34 -42 92 -96 -52 -66 64 14 -87 25 -61 -78 83 -96 -18 95 83 -93 -28 75 49 87 65 -93 -69 -2 95 -24 -36 -61 -71 88 -53 -93 -51 -81 -65 -53 -46 -56 6 65 58 19 100 57 61 -53 44 -58 48 -8 80 -88 72", "output": "3991" }, { "input": "5\n5 5 -10 -1 1", "output": "22" }, { "input": "3\n-1 2 3", "output": "6" }, { "input": "76\n57 -38 -48 -81 93 -32 96 55 -44 2 38 -46 42 64 71 -73 95 31 -39 -62 -1 75 -17 57 28 52 12 -11 82 -84 59 -86 73 -97 34 97 -57 -85 -6 39 -5 -54 95 24 -44 35 -18 9 91 7 -22 -61 -80 54 -40 74 -90 15 -97 66 -52 -49 -24 65 21 -93 -29 -24 -4 -1 76 -93 7 -55 -53 1", "output": "3787" }, { "input": "5\n-1 -2 1 2 3", "output": "9" }, { "input": "4\n2 2 -2 -2", "output": "8" }, { "input": "6\n100 -100 100 -100 100 -100", "output": "600" }, { "input": "100\n-59 -33 34 0 69 24 -22 58 62 -36 5 45 -19 -73 61 -9 95 42 -73 -64 91 -96 2 53 -8 82 -79 16 18 -5 -53 26 71 38 -31 12 -33 -1 -65 -6 3 -89 22 33 -27 -36 41 11 -47 -32 47 -56 -38 57 -63 -41 23 41 29 78 16 -65 90 -58 -12 6 -60 42 -36 -52 -54 -95 -10 29 70 50 -94 1 93 48 -71 -77 -16 54 56 -60 66 76 31 8 44 -61 -74 23 37 38 18 -18 29 41", "output": "4362" }, { "input": "2\n-1 1", "output": "2" }, { "input": "3\n1 -2 100", "output": "103" }, { "input": "5\n1 -2 3 1 2", "output": "9" }, { "input": "10\n100 -10 -100 10 10 10 10 10 10 10", "output": "280" }, { "input": "4\n2 0 -2 4", "output": "8" }, { "input": "4\n3 -3 1 -1", "output": "8" }, { "input": "3\n1 -1 1", "output": "3" }, { "input": "4\n2 5 -2 4", "output": "13" }, { "input": "2\n-2 2", "output": "4" }, { "input": "3\n1 -2 1", "output": "4" }, { "input": "5\n-1 -2 1 1 -1", "output": "6" }, { "input": "4\n-2 0 2 4", "output": "8" }, { "input": "8\n-42 7 87 -16 -5 65 -88 1", "output": "311" }, { "input": "3\n1 -3 4", "output": "8" }, { "input": "1\n1", "output": "1" }, { "input": "2\n0 1", "output": "1" }, { "input": "3\n-1 2 -1", "output": "4" }, { "input": "18\n-21 12 65 66 -24 62 82 35 -45 -47 28 37 5 -32 22 -14 -69 -95", "output": "761" }, { "input": "4\n-1 1 -1 1", "output": "4" }, { "input": "5\n-1 2 1 1 1", "output": "6" }, { "input": "3\n1 1 1", "output": "3" } ]

1,573,835,228

2,147,483,647

Python 3

OK

TESTS

72

109

0

n=int(input()) a=list(map(int,input().split())) sum=0 for i in a: sum+=abs(i) print(sum)

Title: Partition Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given a sequence *a* consisting of *n* integers. You may partition this sequence into two sequences *b* and *c* in such a way that every element belongs exactly to one of these sequences. Let *B* be the sum of elements belonging to *b*, and *C* be the sum of elements belonging to *c* (if some of these sequences is empty, then its sum is 0). What is the maximum possible value of *B*<=-<=*C*? Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in *a*. The second line contains *n* integers *a*1, *a*2, ..., *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100) — the elements of sequence *a*. Output Specification: Print the maximum possible value of *B*<=-<=*C*, where *B* is the sum of elements of sequence *b*, and *C* is the sum of elements of sequence *c*. Demo Input: ['3\n1 -2 0\n', '6\n16 23 16 15 42 8\n'] Demo Output: ['3\n', '120\n'] Note: In the first example we may choose *b* = {1, 0}, *c* = { - 2}. Then *B* = 1, *C* = - 2, *B* - *C* = 3. In the second example we choose *b* = {16, 23, 16, 15, 42, 8}, *c* = {} (an empty sequence). Then *B* = 120, *C* = 0, *B* - *C* = 120.

```python n=int(input()) a=list(map(int,input().split())) sum=0 for i in a: sum+=abs(i) print(sum) ```

3

954

A

Diagonal Walking

PROGRAMMING

800

[ "implementation" ]

null

Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible. In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left. Your problem is to print the minimum possible length of the sequence of moves after the replacements.

The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.

Print the minimum possible length of the sequence of moves after all replacements are done.

[ "5\nRUURU\n", "17\nUUURRRRRUUURURUUU\n" ]

[ "3\n", "13\n" ]

In the first test the shortened sequence of moves may be DUD (its length is 3). In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).

0

[ { "input": "5\nRUURU", "output": "3" }, { "input": "17\nUUURRRRRUUURURUUU", "output": "13" }, { "input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU", "output": "100" }, { "input": "100\nRRURRUUUURURRRURRRRURRRRRRURRUURRRUUURUURURRURUURUURRUURUURRURURUUUUURUUUUUURRUUURRRURRURRRUURRUUUUR", "output": "67" }, { "input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUURUUUUUUUUUUUUUUUUUUUUU", "output": "99" }, { "input": "3\nRUR", "output": "2" }, { "input": "1\nR", "output": "1" }, { "input": "5\nRURUU", "output": "3" }, { "input": "1\nU", "output": "1" }, { "input": "2\nUR", "output": "1" }, { "input": "23\nUUUUUUUUUUUUUUUUUUUUUUU", "output": "23" } ]

1,638,205,130

2,147,483,647

Python 3

OK

TESTS

11

46

0

# -*- coding: utf-8 -*- """diagonal walking .ipynb Automatically generated by Colaboratory. Original file is located at https://colab.research.google.com/drive/1pAmw3SAXBpNn-FeuwNJCdgpcYI-BxCEJ """ n=int(input()) a=input() count=n i=0 while i<n-1: if a[i]!=a[i+1]: count-=1 i+=2 else: i+=1 print(count)

Title: Diagonal Walking Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible. In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left. Your problem is to print the minimum possible length of the sequence of moves after the replacements. Input Specification: The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R. Output Specification: Print the minimum possible length of the sequence of moves after all replacements are done. Demo Input: ['5\nRUURU\n', '17\nUUURRRRRUUURURUUU\n'] Demo Output: ['3\n', '13\n'] Note: In the first test the shortened sequence of moves may be DUD (its length is 3). In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).

```python # -*- coding: utf-8 -*- """diagonal walking .ipynb Automatically generated by Colaboratory. Original file is located at https://colab.research.google.com/drive/1pAmw3SAXBpNn-FeuwNJCdgpcYI-BxCEJ """ n=int(input()) a=input() count=n i=0 while i<n-1: if a[i]!=a[i+1]: count-=1 i+=2 else: i+=1 print(count) ```

3

259

B

Little Elephant and Magic Square

PROGRAMMING

1,100

[ "brute force", "implementation" ]

null

Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes.

The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.

Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions.

[ "0 1 1\n1 0 1\n1 1 0\n", "0 3 6\n5 0 5\n4 7 0\n" ]

[ "1 1 1\n1 1 1\n1 1 1\n", "6 3 6\n5 5 5\n4 7 4\n" ]

none

1,000

[ { "input": "0 1 1\n1 0 1\n1 1 0", "output": "1 1 1\n1 1 1\n1 1 1" }, { "input": "0 3 6\n5 0 5\n4 7 0", "output": "6 3 6\n5 5 5\n4 7 4" }, { "input": "0 4 4\n4 0 4\n4 4 0", "output": "4 4 4\n4 4 4\n4 4 4" }, { "input": "0 54 48\n36 0 78\n66 60 0", "output": "69 54 48\n36 57 78\n66 60 45" }, { "input": "0 17 14\n15 0 15\n16 13 0", "output": "14 17 14\n15 15 15\n16 13 16" }, { "input": "0 97 56\n69 0 71\n84 43 0", "output": "57 97 56\n69 70 71\n84 43 83" }, { "input": "0 1099 1002\n1027 0 1049\n1074 977 0", "output": "1013 1099 1002\n1027 1038 1049\n1074 977 1063" }, { "input": "0 98721 99776\n99575 0 99123\n98922 99977 0", "output": "99550 98721 99776\n99575 99349 99123\n98922 99977 99148" }, { "input": "0 6361 2304\n1433 0 8103\n7232 3175 0", "output": "5639 6361 2304\n1433 4768 8103\n7232 3175 3897" }, { "input": "0 99626 99582\n99766 0 99258\n99442 99398 0", "output": "99328 99626 99582\n99766 99512 99258\n99442 99398 99696" }, { "input": "0 99978 99920\n99950 0 99918\n99948 99890 0", "output": "99904 99978 99920\n99950 99934 99918\n99948 99890 99964" }, { "input": "0 840 666\n612 0 948\n894 720 0", "output": "834 840 666\n612 780 948\n894 720 726" }, { "input": "0 28 10\n12 0 24\n26 8 0", "output": "16 28 10\n12 18 24\n26 8 20" }, { "input": "0 120 83\n98 0 90\n105 68 0", "output": "79 120 83\n98 94 90\n105 68 109" }, { "input": "0 86900 85807\n85836 0 86842\n86871 85778 0", "output": "86310 86900 85807\n85836 86339 86842\n86871 85778 86368" }, { "input": "0 74 78\n78 0 74\n74 78 0", "output": "76 74 78\n78 76 74\n74 78 76" }, { "input": "0 505 681\n605 0 657\n581 757 0", "output": "707 505 681\n605 631 657\n581 757 555" }, { "input": "0 662 918\n822 0 854\n758 1014 0", "output": "934 662 918\n822 838 854\n758 1014 742" }, { "input": "0 93 95\n93 0 97\n95 97 0", "output": "97 93 95\n93 95 97\n95 97 93" }, { "input": "0 709 712\n719 0 695\n702 705 0", "output": "700 709 712\n719 707 695\n702 705 714" }, { "input": "0 7 6\n9 0 1\n4 3 0", "output": "2 7 6\n9 5 1\n4 3 8" }, { "input": "0 9 2\n3 0 7\n8 1 0", "output": "4 9 2\n3 5 7\n8 1 6" }, { "input": "0 1 43\n13 0 61\n31 73 0", "output": "67 1 43\n13 37 61\n31 73 7" }, { "input": "0 100000 100000\n100000 0 100000\n100000 100000 0", "output": "100000 100000 100000\n100000 100000 100000\n100000 100000 100000" } ]

1,603,861,637

2,147,483,647

Python 3

OK

TESTS

24

218

0

l,q=[0]*3,0 for i in range(3):l[i]=list(map(int,input().split()));q+=sum(l[i]) for i in range(3):l[i][i]=q//2-sum(l[i]);print(*l[i])

Title: Little Elephant and Magic Square Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes. Input Specification: The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105. Output Specification: Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions. Demo Input: ['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n'] Demo Output: ['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n'] Note: none

```python l,q=[0]*3,0 for i in range(3):l[i]=list(map(int,input().split()));q+=sum(l[i]) for i in range(3):l[i][i]=q//2-sum(l[i]);print(*l[i]) ```

3

753

B

Interactive Bulls and Cows (Easy)

PROGRAMMING

1,600

[ "brute force", "constructive algorithms", "implementation" ]

null

This problem is a little bit unusual. Here you are to implement an interaction with a testing system. That means that you can make queries and get responses in the online mode. Please be sure to use the stream flushing operation after each query's output in order not to leave part of your output in some buffer. For example, in C++ you've got to use the fflush(stdout) function, in Java — call System.out.flush(), and in Pascal — flush(output). Bulls and Cows (also known as Cows and Bulls or Pigs and Bulls or Bulls and Cleots) is an old code-breaking paper and pencil game for two players, predating the similar commercially marketed board game Mastermind. On a sheet of paper, the first player thinks a secret string. This string consists only of digits and has the length 4. The digits in the string must be all different, no two or more equal digits are allowed. Then the second player tries to guess his opponent's string. For every guess the first player gives the number of matches. If the matching digits are on their right positions, they are "bulls", if on different positions, they are "cows". Thus a response is a pair of numbers — the number of "bulls" and the number of "cows". A try can contain equal digits. More formally, let's the secret string is *s* and the second player are trying to guess it with a string *x*. The number of "bulls" is a number of such positions *i* (1<=≤<=*i*<=≤<=4) where *s*[*i*]<==<=*x*[*i*]. The number of "cows" is a number of such digits *c* that *s* contains *c* in the position *i* (i.e. *s*[*i*]<==<=*c*), *x* contains *c*, but *x*[*i*]<=≠<=*c*. For example, the secret string is "0427", the opponent's try is "0724", then the answer is 2 bulls and 2 cows (the bulls are "0" and "2", the cows are "4" and "7"). If the secret string is "0123", the opponent's try is "0330", then the answer is 1 bull and 1 cow. In this problem you are to guess the string *s* that the system has chosen. You only know that the chosen string consists of 4 distinct digits. You can make queries to the testing system, each query is the output of a single 4-digit string. The answer to the query is the number of bulls and number of cows. If the system's response equals "4 0", that means the interaction with your problem is over and the program must terminate. That is possible for two reasons — the program either guessed the number *x* or made an invalid action (for example, printed letters instead of digits). Your program is allowed to do at most 50 queries. You can hack solutions of other participants providing a 4-digit string containing distinct digits — the secret string.

To read answers to the queries, the program must use the standard input. The program will receive pairs of non-negative integers in the input, one pair per line. The first number in a pair is a number of bulls and the second one is a number of cows of the string *s* and the string *x**i* printed by your program. If the system response equals "4 0", then your solution should terminate. The testing system will let your program read the *i*-th pair of integers from the input only after your program displays the corresponding system query in the output: prints value *x**i* in a single line and executes operation flush.

The program must use the standard output to print queries. Your program must output requests — 4-digit strings *x*1,<=*x*2,<=..., one per line. After the output of each line the program must execute flush operation. The program should read the answer to the query from the standard input. Your program is allowed to do at most 50 queries.

[ "0 1\n2 0\n1 1\n0 4\n2 1\n4 0\n" ]

[ "8000\n0179\n3159\n3210\n0112\n0123" ]

The secret string *s* in the example is "0123".

1,000

[ { "input": "0123", "output": "20" }, { "input": "1234", "output": "20" }, { "input": "9876", "output": "20" }, { "input": "7158", "output": "20" }, { "input": "7590", "output": "20" }, { "input": "7325", "output": "20" }, { "input": "7524", "output": "20" }, { "input": "7269", "output": "20" }, { "input": "7802", "output": "20" }, { "input": "7436", "output": "20" }, { "input": "7190", "output": "20" }, { "input": "7390", "output": "20" }, { "input": "2548", "output": "20" }, { "input": "2193", "output": "20" }, { "input": "2491", "output": "20" }, { "input": "2469", "output": "20" }, { "input": "2659", "output": "20" }, { "input": "2405", "output": "20" }, { "input": "2058", "output": "20" }, { "input": "2580", "output": "20" }, { "input": "2316", "output": "20" }, { "input": "2516", "output": "20" }, { "input": "8796", "output": "20" }, { "input": "8534", "output": "20" }, { "input": "9067", "output": "20" }, { "input": "8712", "output": "20" }, { "input": "9023", "output": "20" }, { "input": "8645", "output": "20" }, { "input": "8623", "output": "20" }, { "input": "8923", "output": "20" }, { "input": "8567", "output": "20" }, { "input": "8756", "output": "20" }, { "input": "0351", "output": "20" }, { "input": "9863", "output": "20" }, { "input": "0518", "output": "20" }, { "input": "0263", "output": "20" }, { "input": "0462", "output": "20" }, { "input": "0429", "output": "20" }, { "input": "0629", "output": "20" }, { "input": "0374", "output": "20" }, { "input": "3209", "output": "20" } ]

1,483,003,948

1,648

Python 3

OK

TESTS

41

62

4,608,000

from sys import stdout from itertools import permutations var = [] for i in range(10): print(str(i) * 4) stdout.flush() a, b = map(int, input().split()) if a: var.append(str(i)) for t in permutations(var): print(''.join(t)) stdout.flush() a, b = map(int, input().split()) if a == 4: break

Title: Interactive Bulls and Cows (Easy) Time Limit: None seconds Memory Limit: None megabytes Problem Description: This problem is a little bit unusual. Here you are to implement an interaction with a testing system. That means that you can make queries and get responses in the online mode. Please be sure to use the stream flushing operation after each query's output in order not to leave part of your output in some buffer. For example, in C++ you've got to use the fflush(stdout) function, in Java — call System.out.flush(), and in Pascal — flush(output). Bulls and Cows (also known as Cows and Bulls or Pigs and Bulls or Bulls and Cleots) is an old code-breaking paper and pencil game for two players, predating the similar commercially marketed board game Mastermind. On a sheet of paper, the first player thinks a secret string. This string consists only of digits and has the length 4. The digits in the string must be all different, no two or more equal digits are allowed. Then the second player tries to guess his opponent's string. For every guess the first player gives the number of matches. If the matching digits are on their right positions, they are "bulls", if on different positions, they are "cows". Thus a response is a pair of numbers — the number of "bulls" and the number of "cows". A try can contain equal digits. More formally, let's the secret string is *s* and the second player are trying to guess it with a string *x*. The number of "bulls" is a number of such positions *i* (1<=≤<=*i*<=≤<=4) where *s*[*i*]<==<=*x*[*i*]. The number of "cows" is a number of such digits *c* that *s* contains *c* in the position *i* (i.e. *s*[*i*]<==<=*c*), *x* contains *c*, but *x*[*i*]<=≠<=*c*. For example, the secret string is "0427", the opponent's try is "0724", then the answer is 2 bulls and 2 cows (the bulls are "0" and "2", the cows are "4" and "7"). If the secret string is "0123", the opponent's try is "0330", then the answer is 1 bull and 1 cow. In this problem you are to guess the string *s* that the system has chosen. You only know that the chosen string consists of 4 distinct digits. You can make queries to the testing system, each query is the output of a single 4-digit string. The answer to the query is the number of bulls and number of cows. If the system's response equals "4 0", that means the interaction with your problem is over and the program must terminate. That is possible for two reasons — the program either guessed the number *x* or made an invalid action (for example, printed letters instead of digits). Your program is allowed to do at most 50 queries. You can hack solutions of other participants providing a 4-digit string containing distinct digits — the secret string. Input Specification: To read answers to the queries, the program must use the standard input. The program will receive pairs of non-negative integers in the input, one pair per line. The first number in a pair is a number of bulls and the second one is a number of cows of the string *s* and the string *x**i* printed by your program. If the system response equals "4 0", then your solution should terminate. The testing system will let your program read the *i*-th pair of integers from the input only after your program displays the corresponding system query in the output: prints value *x**i* in a single line and executes operation flush. Output Specification: The program must use the standard output to print queries. Your program must output requests — 4-digit strings *x*1,<=*x*2,<=..., one per line. After the output of each line the program must execute flush operation. The program should read the answer to the query from the standard input. Your program is allowed to do at most 50 queries. Demo Input: ['0 1\n2 0\n1 1\n0 4\n2 1\n4 0\n'] Demo Output: ['8000\n0179\n3159\n3210\n0112\n0123'] Note: The secret string *s* in the example is "0123".

```python from sys import stdout from itertools import permutations var = [] for i in range(10): print(str(i) * 4) stdout.flush() a, b = map(int, input().split()) if a: var.append(str(i)) for t in permutations(var): print(''.join(t)) stdout.flush() a, b = map(int, input().split()) if a == 4: break ```

3

63

B

Settlers' Training

PROGRAMMING

1,200

[ "implementation" ]

B. Settlers' Training

2

256

In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly *n* soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease. Every soldier has a rank — some natural number from 1 to *k*. 1 stands for a private and *k* stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank. To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the *n* soldiers are present. At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank *k* are present, exactly one soldier increases his rank by one. You know the ranks of all *n* soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank *k*.

The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains *n* numbers in the non-decreasing order. The *i*-th of them, *a**i*, represents the rank of the *i*-th soldier in the defense building (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=*k*).

Print a single integer — the number of golden coins needed to raise all the soldiers to the maximal rank.

[ "4 4\n1 2 2 3\n", "4 3\n1 1 1 1\n" ]

[ "4", "5" ]

In the first example the ranks will be raised in the following manner: 1 2 2 3 → 2 2 3 4 → 2 3 4 4 → 3 4 4 4 → 4 4 4 4 Thus totals to 4 training sessions that require 4 golden coins.

1,000

[ { "input": "4 4\n1 2 2 3", "output": "4" }, { "input": "4 3\n1 1 1 1", "output": "5" }, { "input": "3 3\n1 2 3", "output": "2" }, { "input": "1 1\n1", "output": "0" }, { "input": "1 5\n1", "output": "4" }, { "input": "1 5\n4", "output": "1" }, { "input": "2 6\n2 5", "output": "4" }, { "input": "6 10\n1 1 3 4 9 9", "output": "10" }, { "input": "7 7\n1 1 1 1 1 1 7", "output": "11" }, { "input": "10 10\n1 1 1 3 3 4 7 8 8 8", "output": "11" }, { "input": "10 13\n1 1 1 1 1 1 1 1 1 1", "output": "21" }, { "input": "10 13\n2 6 6 7 9 9 9 10 12 12", "output": "11" }, { "input": "17 9\n2 3 4 5 5 5 5 5 6 6 7 7 8 8 8 8 8", "output": "17" }, { "input": "18 24\n3 3 3 4 5 7 8 8 9 9 9 9 10 10 11 11 11 11", "output": "30" }, { "input": "23 2\n1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2", "output": "12" }, { "input": "37 42\n1 1 1 1 1 2 2 2 2 2 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8", "output": "70" }, { "input": "44 50\n38 38 38 38 38 38 38 39 39 39 39 39 39 39 40 40 40 40 40 41 41 41 41 41 41 41 42 42 42 43 43 43 44 44 44 44 45 45 45 46 46 46 46 46", "output": "47" }, { "input": "57 100\n2 2 4 7 8 10 12 12 14 15 16 18 19 21 21 22 25 26 26 33 38 40 44 44 44 45 47 47 50 51 51 54 54 54 54 55 56 58 61 65 67 68 68 70 74 75 78 79 83 86 89 90 92 95 96 96 97", "output": "99" }, { "input": "78 10\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9", "output": "78" }, { "input": "96 78\n20 20 20 20 20 21 21 21 22 23 23 24 24 25 25 27 28 29 30 30 30 32 32 32 33 33 33 33 34 34 35 36 37 37 39 39 41 41 41 41 42 42 43 43 43 44 44 45 46 46 48 48 49 50 51 51 51 52 53 55 55 56 56 56 56 57 58 59 60 61 61 61 62 62 62 63 63 64 64 64 65 65 65 66 66 67 68 69 71 72 72 73 73 75 75 75", "output": "98" }, { "input": "100 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "198" }, { "input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "0" }, { "input": "100 100\n1 1 4 4 5 5 7 9 10 10 11 11 12 12 12 13 14 15 16 16 16 17 18 18 19 20 22 25 26 27 29 32 33 34 34 35 35 35 36 36 37 37 38 39 39 40 41 42 44 44 46 47 47 47 47 50 53 53 53 55 56 56 57 57 58 58 59 59 62 64 64 64 64 68 68 68 69 70 70 71 74 77 77 77 79 80 80 81 84 86 88 88 91 93 94 96 96 99 99 99", "output": "108" }, { "input": "100 100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 7 7 7 7 8 8 8 8 8 9 9 9 9 9 9 9 10 10 10 10 10 11 11 11 11 11 12 12 12 12 12 12 13 13 13 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15 15", "output": "184" }, { "input": "100 100\n20 20 20 21 21 21 21 21 22 23 23 23 23 23 23 24 24 25 25 26 26 26 26 26 27 27 27 27 28 28 28 28 29 29 29 29 29 30 30 30 30 31 32 32 34 34 34 34 34 34 34 34 35 35 35 36 36 37 37 37 37 37 37 38 38 38 39 40 41 41 42 42 42 42 42 43 43 43 44 44 44 44 44 45 45 45 45 45 46 46 46 46 46 47 47 47 48 48 48 50", "output": "150" }, { "input": "100 2\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2", "output": "59" }, { "input": "30 50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 49", "output": "77" }, { "input": "40 20\n5 5 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 20 20 20 20 20 20 20 20 20 20", "output": "31" }, { "input": "81 90\n1 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90", "output": "89" }, { "input": "100 20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 13 13 13 13 13 13 13 13 13", "output": "106" }, { "input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 100", "output": "197" }, { "input": "100 100\n49 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 51", "output": "148" }, { "input": "1 100\n1", "output": "99" }, { "input": "4 3\n1 1 2 2", "output": "4" }, { "input": "10 100\n98 99 99 99 99 99 99 100 100 100", "output": "7" }, { "input": "5 100\n1 2 2 100 100", "output": "100" } ]

1,622,060,653

2,147,483,647

PyPy 3

OK

TESTS

37

218

1,331,200

n,k=list(map(int,input().split())) a=list(map(int,input().split())) c=0 while a!=[k]*n: a.sort() for i in range(1,n): if a[i]!=a[i-1]: a[i-1]+=1 if a[n-1]<k: a[n-1]+=1 c+=1 print(c)

Title: Settlers' Training Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: In a strategic computer game "Settlers II" one has to build defense structures to expand and protect the territory. Let's take one of these buildings. At the moment the defense structure accommodates exactly *n* soldiers. Within this task we can assume that the number of soldiers in the defense structure won't either increase or decrease. Every soldier has a rank — some natural number from 1 to *k*. 1 stands for a private and *k* stands for a general. The higher the rank of the soldier is, the better he fights. Therefore, the player profits from having the soldiers of the highest possible rank. To increase the ranks of soldiers they need to train. But the soldiers won't train for free, and each training session requires one golden coin. On each training session all the *n* soldiers are present. At the end of each training session the soldiers' ranks increase as follows. First all the soldiers are divided into groups with the same rank, so that the least possible number of groups is formed. Then, within each of the groups where the soldiers below the rank *k* are present, exactly one soldier increases his rank by one. You know the ranks of all *n* soldiers at the moment. Determine the number of golden coins that are needed to increase the ranks of all the soldiers to the rank *k*. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100). They represent the number of soldiers and the number of different ranks correspondingly. The second line contains *n* numbers in the non-decreasing order. The *i*-th of them, *a**i*, represents the rank of the *i*-th soldier in the defense building (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=*k*). Output Specification: Print a single integer — the number of golden coins needed to raise all the soldiers to the maximal rank. Demo Input: ['4 4\n1 2 2 3\n', '4 3\n1 1 1 1\n'] Demo Output: ['4', '5'] Note: In the first example the ranks will be raised in the following manner: 1 2 2 3 → 2 2 3 4 → 2 3 4 4 → 3 4 4 4 → 4 4 4 4 Thus totals to 4 training sessions that require 4 golden coins.

```python n,k=list(map(int,input().split())) a=list(map(int,input().split())) c=0 while a!=[k]*n: a.sort() for i in range(1,n): if a[i]!=a[i-1]: a[i-1]+=1 if a[n-1]<k: a[n-1]+=1 c+=1 print(c) ```

3.94302