conversations listlengths 2 2 | Question Name stringclasses 49
values | Question Description stringclasses 49
values | Input Requirements stringclasses 49
values | Output Requirements stringclasses 49
values | Sample Input stringclasses 49
values | Sample Output stringclasses 49
values | Difficulty stringclasses 1
value | Question Source stringclasses 1
value | Question Source URL stringclasses 1
value | Answer Source stringclasses 1
value | Answer Source URL stringclasses 1
value | Exam name stringclasses 1
value | Exam Description stringclasses 1
value | CPE Website stringclasses 1
value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
[
{
"content": "你知道「11005 Cheapest Base」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, int>... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫「11005 Cheapest Base」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, ... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成「11005 Cheapest Base」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, int... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要「11005 Cheapest Base」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map ... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "11005 Cheapest Base 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, int> mp;... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "11005 Cheapest Base 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, int> mp... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "如何用C++實作 11005 Cheapest Base？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, int>... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "給我一個 11005 Cheapest Base 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, ... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "你知道 11005 Cheapest Base 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, int>... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫 11005 Cheapest Base 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, ... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成 11005 Cheapest Base 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map <int, int... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要 11005 Cheapest Base 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <map>\nusing namespace std;\n \nint main() {\n int T, cost, Q, x;\n cin >> T;\n for (int Case = 1; Case <= T; Case++){\n if (Case > 1) cout << \"\\n\";\n map ... | 11005 Cheapest Base | When printing text on paper we need ink. But not every character needs the same amount of ink to print: letters such as 'W', 'M' and '8' are more expensive than thinner letters as 'i', 'c' and '1'. In this problem we will evaluate the cost of printing numbers in several bases.
As you know, numbers can be expressed in ... | The input has less than 25 test cases. The first line of input file denotes this number of test cases. The description of every test case is given below:
The first 4 lines of every case contain 9 integers each: the costs of the 36 characters in the order given above. Then follows the number of queries on a line by its... | For every case in the input, print one line 'Case X:', without the quotes, where X is the case number starting from 1.
For every query within a case print one line 'Cheapest base(s) for number Y:' followed by the cheapest base(s) in increasing order, separated by one space. Y is the query in decimal format. Print a bl... | 2
10 8 12 13 15 13 13 16 9
11 18 24 21 23 23 23 13 15
17 33 21 23 27 26 27 19 4
22 18 30 30 24 16 26 21 21
5
98329921
12345
800348
14
873645
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
4
0
1
10
100 | Case 1:
Cheapest base(s) for number 98329921: 24
Cheapest base(s) for number 12345: 13 31
Cheapest base(s) for number 800348: 31
Cheapest base(s) for number 14: 13
Cheapest base(s) for number 873645: 22
Case 2:
Cheapest base(s) for number 0: 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 3... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to solve \"11063 B2-Sequence\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "What is the solution for \"11063 B2-Sequence\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to implement \"11063 B2-Sequence\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Give me a C++ code for \"11063 B2-Sequence\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Do you know the solution for \"11063 B2-Sequence\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Show me the code for \"11063 B2-Sequence\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Generate a code solution for \"11063 B2-Sequence\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "I need a C++ code for \"11063 B2-Sequence\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to solve 11063 B2-Sequence?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "What is the solution for 11063 B2-Sequence?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n c... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to implement 11063 B2-Sequence in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ci... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Give me a C++ code for 11063 B2-Sequence.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Do you know the solution for 11063 B2-Sequence.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Show me the code for 11063 B2-Sequence with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Generate a code solution for 11063 B2-Sequence.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "I need a C++ code for 11063 B2-Sequence, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++)... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「11063 B2-Sequence」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「11063 B2-Sequence」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "如何用C++實作「11063 B2-Sequence」？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "給我一個「11063 B2-Sequence」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "你知道「11063 B2-Sequence」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫「11063 B2-Sequence」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成「11063 B2-Sequence」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要「11063 B2-Sequence」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >>... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "11063 B2-Sequence 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "11063 B2-Sequence 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "如何用C++實作 11063 B2-Sequence？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "給我一個 11063 B2-Sequence 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "你知道 11063 B2-Sequence 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫 11063 B2-Sequence 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成 11063 B2-Sequence 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >> b[i];\n ... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要 11063 B2-Sequence 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n#define maxn 20005\n \nint main() {\n int n, Case = 1;\n while (cin >> n) {\n int b[n], sum[maxn];\n for (int i = 0; i < n; i++) {\n cin >>... | 11063 B2-Sequence | A B2-Sequence is a sequence of positive integers 1 ≤ b₁ < b₂ < b₃... such that all pairwise sums bᵢ + bⱼ, where i ≤ j, are different.
Your task is to determine if a given sequence is a B2-Sequence or not. | Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element bᵢ is an integer such that bᵢ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF). | For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line. | 4
1 2 4 8
4
3 7 10 14 | Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence. | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to solve \"11150 Cola\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "What is the solution for \"11150 Cola\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to implement \"11150 Cola\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Give me a C++ code for \"11150 Cola\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Do you know the solution for \"11150 Cola\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Show me the code for \"11150 Cola\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Generate a code solution for \"11150 Cola\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "I need a C++ code for \"11150 Cola\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to solve 11150 Cola?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "What is the solution for 11150 Cola?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to implement 11150 Cola in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Give me a C++ code for 11150 Cola.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Do you know the solution for 11150 Cola.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Show me the code for 11150 Cola with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Generate a code solution for 11150 Cola.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "I need a C++ code for 11150 Cola, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「11150 Cola」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N == ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「11150 Cola」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N ==... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "如何用C++實作「11150 Cola」？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "給我一個「11150 Cola」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans +=... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "你知道「11150 Cola」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫「11150 Cola」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans +=... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成「11150 Cola」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要「11150 Cola」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "11150 Cola 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N == 2... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "11150 Cola 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N == ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "如何用C++實作 11150 Cola？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "給我一個 11150 Cola 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans +=... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "你知道 11150 Cola 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫 11150 Cola 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans +=... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成 11150 Cola 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ans += (N... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要 11150 Cola 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int N;\n while (cin >> N){\n int ans = N;\n while (N >= 3){\n ans += N / 3;\n N = N / 3 + N % 3;\n }\n ... | 11150 Cola | You see the following special offer by the convenience store:
"A bottle of Choco Cola for every 3 empty bottles returned"
Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.
The figure below shows the case where N = 8. Method 1 is the sta... | Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200). | For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them. | 8 | 12 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to solve \"11321 Sort! Sort!! And Sort!!!\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "What is the solution for \"11321 Sort! Sort!! And Sort!!!\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to implement \"11321 Sort! Sort!! And Sort!!!\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n i... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Give me a C++ code for \"11321 Sort! Sort!! And Sort!!!\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Do you know the solution for \"11321 Sort! Sort!! And Sort!!!\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Show me the code for \"11321 Sort! Sort!! And Sort!!!\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } el... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Generate a code solution for \"11321 Sort! Sort!! And Sort!!!\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "I need a C++ code for \"11321 Sort! Sort!! And Sort!!!\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n }... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to solve 11321 Sort! Sort!! And Sort!!!?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (y... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "What is the solution for 11321 Sort! Sort!! And Sort!!!?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if (... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "How to implement 11321 Sort! Sort!! And Sort!!! in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Give me a C++ code for 11321 Sort! Sort!! And Sort!!!.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Do you know the solution for 11321 Sort! Sort!! And Sort!!!.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Show me the code for 11321 Sort! Sort!! And Sort!!! with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "Generate a code solution for 11321 Sort! Sort!! And Sort!!!.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "I need a C++ code for 11321 Sort! Sort!! And Sort!!!, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } els... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「11321 Sort! Sort!! And Sort!!!」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (y % 2))... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「11321 Sort! Sort!! And Sort!!!」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (y % 2)... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "如何用C++實作「11321 Sort! Sort!! And Sort!!!」？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (y %... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "給我一個「11321 Sort! Sort!! And Sort!!!」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "你知道「11321 Sort! Sort!! And Sort!!!」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (y % ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫「11321 Sort! Sort!! And Sort!!!」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成「11321 Sort! Sort!! And Sort!!!」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % 2) * (y %... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要「11321 Sort! Sort!! And Sort!!!」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\nint N, M;\n \nbool cmp(int x, int y){\n if ((x % M) != (y % M)){\n return (x % M) < (y % M);\n } else {\n if ((x % ... | 11321 Sort! Sort!! And Sort!!! | Hmm! Here you are asked to do a simple sorting. You will be given N numbers and a positive integer M. You will have to sort the N numbers in ascending order of their modulo M value. If there is a tie between an odd number and an even number (that is their modulo M value is the same) then the odd number will precede the... | The input file contains 20 sets of inputs. Each set starts with two integers N (0 < N ≤ 10000) and M (0 < M ≤ 10000) which denotes how many numbers are within this set. Each of the next N lines contains one number each. These numbers should all fit in 32-bit signed integer. Input is terminated by a line containing two ... | For each set of input produce N + 1 lines of outputs. The first line of each set contains the value of N and M. The next N lines contain N numbers, sorted according to the rules mentioned above. Print the last two zeroes of the input file in the output file also. | 15 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 0 | 15 3
15
9
3
6
12
13
7
1
4
10
11
5
2
8
14
0 0 | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.