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values | Question Description stringclasses 49
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values | Output Requirements stringclasses 49
values | Sample Input stringclasses 49
values | Sample Output stringclasses 49
values | Difficulty stringclasses 1
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[
{
"content": "你知道「10062 Tell me the frequencies!」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.second... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
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49 1
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| 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「10062 Tell me the frequencies!」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.se... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 3
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成「10062 Tell me the frequencies!」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.secon... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 3
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要「10062 Tell me the frequencies!」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second ... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 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": "10062 Tell me the frequencies! 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.second;\n}... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 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": "10062 Tell me the frequencies! 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.second;\n... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 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 10062 Tell me the frequencies!？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.second... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 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 10062 Tell me the frequencies! 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.se... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 3
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "你知道 10062 Tell me the frequencies! 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.second... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 3
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "撰寫 10062 Tell me the frequencies! 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.se... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 3
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "生成 10062 Tell me the frequencies! 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second > b.secon... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 3
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "我需要 10062 Tell me the frequencies! 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define pii pair<int,int>\n \nbool cmp(pii a, pii b){\n if (a.first != b.first) return a.first < b.first;\n else return a.second ... | 10062 Tell me the frequencies! | Given a line of text, the task is to find out the frequencies of the ASCII characters present in it. The input lines will not contain any of the first 32 or last 128 ASCII characters. Lines may end with newline or carriage return characters, but these should be excluded from consideration. | The input consists of several lines of text, each treated as a single input. The maximum length of each line is 1000 characters. The input is terminated by end-of-file. | For each line of input, output the ASCII value of each character present along with its frequency. The output should list characters in ascending order of their frequencies. If two characters have the same frequency, the one with the higher ASCII value should be listed first. Each set of output should be separated by a... | AAABBC
122333
| 67 1
66 2
65 3
49 1
50 2
51 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 \"10071 Back to High School Physics\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10071 Back to High School Physics\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assista... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10071 Back to High School Physics\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistan... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10071 Back to High School Physics\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10071 Back to High School Physics\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "ass... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10071 Back to High School Physics\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"r... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10071 Back to High School Physics\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "ass... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10071 Back to High School Physics\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assista... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role"... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assista... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"ro... | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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": "「10071 Back to High School Physics」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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": "「10071 Back to High School Physics」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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「10071 Back to High School Physics」？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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「10071 Back to High School Physics」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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「10071 Back to High School Physics」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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「10071 Back to High School Physics」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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「10071 Back to High School Physics」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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「10071 Back to High School Physics」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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": "10071 Back to High School Physics 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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": "10071 Back to High School Physics 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 10071 Back to High School Physics 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int v, t;\n while (cin >> v >> t){\n cout << 2 * v * t << \"\\n\";\n }\n return 0;\n}",
"role": "assistant"
}
] | 10071 Back to High School Physics | A particle has initial velocity and acceleration. If its velocity after a certain time is v, the task is to determine its displacement after twice that time. The problem involves basic physics calculations where the displacement is derived from the given velocity and time. | The input consists of multiple lines, each containing two integers v (-100 ≤ v ≤ 100) and t (0 ≤ t ≤ 200), where v is the velocity and t is the time at which the particle reaches that velocity. | For each line of input, output a single integer representing the displacement after twice the given time (2t). | 0 0
5 12
| 0
120
| 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 \"10093 An Easy Problem!\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 \"10093 An Easy Problem!\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0;... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 \"10093 An Easy Problem!\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 \"10093 An Easy Problem!\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
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| 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 \"10093 An Easy Problem!\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
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| 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 \"10093 An Easy Problem!\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
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| 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 \"10093 An Easy Problem!\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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 \"10093 An Easy Problem!\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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 10093 An Easy Problem!?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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 10093 An Easy Problem!?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i <... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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 10093 An Easy Problem! in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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 10093 An Easy Problem!.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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 10093 An Easy Problem!.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0;... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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 10093 An Easy Problem! with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n f... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem!.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0;... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem!, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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": "「10093 An Easy Problem!」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++) {\... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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": "「10093 An Easy Problem!」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++) {... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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「10093 An Easy Problem!」？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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「10093 An Easy Problem!」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
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| 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「10093 An Easy Problem!」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++)... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
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| 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「10093 An Easy Problem!」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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「10093 An Easy Problem!」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
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| 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「10093 An Easy Problem!」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.si... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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| 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": "10093 An Easy Problem! 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++) {\n... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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": "10093 An Easy Problem! 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++) {\... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem!？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++)... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem! 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem! 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++)... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem! 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); ... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem! 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.size(); i++... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 10093 An Easy Problem! 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\n \nint main() {\n int sum, mx, temp;\n string s;\n while (getline(cin, s)){\n sum = 0;\n mx = 1;\n int i;\n for (i = 0; i < s.si... | 10093 An Easy Problem! | Given an N-based integer number R, the task is to find the smallest possible base N (2 ≤ N ≤ 62) such that R is divisible by (N - 1). The digit symbols for bases up to 62 include 0-9, A-Z, and a-z. The solution involves checking the divisibility condition and determining the smallest valid base. | Each line of input contains a number represented in an unknown base (2-62). The number will consist of valid digit symbols (0-9, A-Z, a-z). No invalid numbers will be provided. | For each input number, output the smallest possible base N that satisfies the condition R % (N - 1) == 0. If no such base exists within the range 2-62, output 'such number is impossible!'. | 3
5
A
| 4
6
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 \"10101 Bangla Numbers\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 \"10101 Bangla Numbers\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n ... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 \"10101 Bangla Numbers\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 \"10101 Bangla Numbers\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \n... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 \"10101 Bangla Numbers\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 \"10101 Bangla Numbers\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\n... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 \"10101 Bangla Numbers\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 \"10101 Bangla Numbers\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"}... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2str... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nst... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstr... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstri... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n ... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstri... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n ... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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 10101 Bangla Numbers, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\ns... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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": "「10101 Bangla Numbers」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2str(ll x)... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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": "「10101 Bangla Numbers」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2str(ll x... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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「10101 Bangla Numbers」？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2str(l... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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「10101 Bangla Numbers」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2st... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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「10101 Bangla Numbers」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2str(ll... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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「10101 Bangla Numbers」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2st... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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「10101 Bangla Numbers」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring num2str(l... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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「10101 Bangla Numbers」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n#define ll long long\nint a[] = {10000000, 100000, 1000, 100};\nstring s[] = {\"kuti\", \"lakh\", \"hajar\", \"shata\"};\nstring ans;\n \nstring ... | 10101 Bangla Numbers | The problem involves converting a given non-negative number into its Bangla text representation using the terms 'kuti' (10,000,000), 'lakh' (100,000), 'hajar' (1,000), and 'shata' (100). The input consists of numbers up to 999,999,999,999,999, and the output must display the converted text with proper formatting. | The input file contains several test cases, each with a non-negative number ≤ 999,999,999,999,999. | For each test case, output the case number (starting from 1) followed by the Bangla text representation of the number. The output should use the terms 'kuti', 'lakh', 'hajar', and 'shata' appropriately, with proper spacing and formatting. | 23764
45897458973958
| 1. 23 hajar 7 shata 64
2. 45 lakh 89 hajar 7 shata 45 kuti 89 lakh 73 hajar 9 shata 58
| 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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