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[
{
"content": "「10922 2 the 9s」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n i... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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": "「10922 2 the 9s」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n ... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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「10922 2 the 9s」？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n ... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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「10922 2 the 9s」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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「10922 2 the 9s」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n ... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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「10922 2 the 9s」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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「10922 2 the 9s」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n ... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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「10922 2 the 9s」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n co... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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": "10922 2 the 9s 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n if... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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": "10922 2 the 9s 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n i... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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 10922 2 the 9s？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n ... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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 10922 2 the 9s 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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 10922 2 the 9s 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n ... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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 10922 2 the 9s 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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 10922 2 the 9s 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n count++;\n ... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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 10922 2 the 9s 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint func(string s, int count){\n int temp = 0, total = 0;\n for (int i = 0; i < s.length(); i++){\n temp = s[i] - '0';\n total += temp;\n }\n co... | 10922 2 the 9s | A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if i... | The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output. | 9999999999999999999999
9
999999999999999999999999999999998
0 | 9999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
999999999999999999999999999998 is not a multiple of 9. | 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 \"10929 You can say 11\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10929 You can say 11\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size();... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10929 You can say 11\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10929 You can say 11\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10929 You can say 11\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.siz... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10929 You can say 11\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0;... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10929 You can say 11\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.siz... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10929 You can say 11\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i =... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11 in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++)... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size();... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11 with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i <... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size();... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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": "「10929 You can say 11」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n i... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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": "「10929 You can say 11」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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「10929 You can say 11」？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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「10929 You can say 11」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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「10929 You can say 11」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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「10929 You can say 11」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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「10929 You can say 11」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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「10929 You can say 11」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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": "10929 You can say 11 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n if... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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": "10929 You can say 11 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n i... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n ... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 10929 You can say 11 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n string s;\n while (cin >> s) {\n if (s == \"0\") break;\n int odd = 0, even = 0;\n for (int i = 0; i < s.size(); i++) {\n... | 10929 You can say 11 | Your job is, given a positive number N, determine if it is a multiple of eleven. | The input is a file such that each line contains a positive number. A line containing the number '0' is the end of the input. The given numbers can contain up to 1000 digits. | The output of the program shall indicate, for each input number, if it is a multiple of eleven or not. | 112233
30800
2937
323455693
5038297
112234
0 | 112233 is a multiple of 11.
30800 is a multiple of 11.
2937 is a multiple of 11.
323455693 is a multiple of 11.
5038297 is a multiple of 11.
112234 is not a multiple of 11. | 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 \"10931 Parity\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"10931 Parity\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"10931 Parity\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"10931 Parity\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"10931 Parity\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cn... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"10931 Parity\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"10931 Parity\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cn... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"10931 Parity\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n whil... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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": "「10931 Parity」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while (n){... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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": "「10931 Parity」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while (n)... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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「10931 Parity」？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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「10931 Parity」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n whi... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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「10931 Parity」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while (... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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「10931 Parity」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n whi... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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「10931 Parity」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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「10931 Parity」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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": "10931 Parity 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while (n){\... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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": "10931 Parity 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while (n){... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while (... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n whi... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while (... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n whi... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n while ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 10931 Parity 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nint main() {\n int I, n;\n while (cin >> I) {\n if (I == 0) break;\n string B = \"\";\n n = I;\n int cnt = 0;\n ... | 10931 Parity | We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 10101_2 has three 1s in its binary representation so it has parity 3(mod2), or 1.
In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647. | Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed. | For each integer I in the input you should print a line 'The parity of B is P (mod 2).', where B is the binary representation of I. | 1
2
10
21
0 | The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2). | 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 \"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 <in... | 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": "What is the solution for \"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 ... | 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 implement \"11005 Cheapest Base\" in C++?",
"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 ... | 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": "Give me a C++ code for \"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 ... | 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": "Do you know the solution for \"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\";\... | 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": "Show me the code for \"11005 Cheapest Base\" with explanations.",
"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 <<... | 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": "Generate a code solution for \"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\";\... | 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": "I need a C++ code for \"11005 Cheapest Base\", generate it for me.",
"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... | 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 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, i... | 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": "What is the solution for 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 ... | 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 implement 11005 Cheapest Base in C++?",
"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 ... | 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": "Give me a C++ code for 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 m... | 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": "Do you know the solution for 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 ... | 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": "Show me the code for 11005 Cheapest Base with explanations.",
"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 << \"\... | 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": "Generate a code solution for 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 ... | 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": "I need a C++ code for 11005 Cheapest Base, generate it for me.",
"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 << ... | 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> m... | 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 |
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