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[
{
"content": "How to solve \"10226 Hardwood Species\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip th... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10226 Hardwood Species\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s)... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10226 Hardwood Species\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s);... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10226 Hardwood Species\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); ... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10226 Hardwood Species\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10226 Hardwood Species\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n g... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10226 Hardwood Species\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10226 Hardwood Species\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n ... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the bl... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); //... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // ... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // S... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s)... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getli... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s)... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n ge... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「10226 Hardwood Species」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blank li... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「10226 Hardwood Species」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blank l... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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「10226 Hardwood Species」？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blan... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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「10226 Hardwood Species」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the b... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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「10226 Hardwood Species」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blank... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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「10226 Hardwood Species」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the b... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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「10226 Hardwood Species」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blan... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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「10226 Hardwood Species」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "10226 Hardwood Species 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blank lin... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "10226 Hardwood Species 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blank li... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blank... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the b... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blank... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the b... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip the blan... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 10226 Hardwood Species 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\n#include <map>\nusing namespace std;\n \nint main() {\n int n;\n string s;\n cin >> n;\n getline(cin, s); // Clear the buffer\n getline(cin, s); // Skip... | 10226 Hardwood Species | The problem involves analyzing a list of tree species observed by satellite imaging and calculating the percentage of each species in the total population. The input consists of multiple test cases, each containing a list of tree species (one per line), and the output should display each species followed by its percent... | The input starts with an integer indicating the number of test cases. Each test case contains a list of tree species (one per line), with a blank line separating consecutive test cases. No species name exceeds 30 characters, and there are no more than 10,000 species or 1,000,000 trees. | For each test case, output the name of each species in alphabetical order, followed by its percentage of the total population (to four decimal places). Separate the output of consecutive test cases with a blank line. | 1
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow | Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.... | One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\"?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i)... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005;... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\".",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005;... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10235 Simply Emirp\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i;... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp in C++?",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 10... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j <... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「10235 Simply Emirp」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n p... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「10235 Simply Emirp」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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「10235 Simply Emirp」？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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「10235 Simply Emirp」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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「10235 Simply Emirp」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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「10235 Simply Emirp」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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「10235 Simply Emirp」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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「10235 Simply Emirp」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "10235 Simply Emirp 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n p[... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "10235 Simply Emirp 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n p... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 10235 Simply Emirp 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\nusing namespace std;\nint p[1000005];\n \nint main() {\n p[1] = 1;\n for (int i = 2; i < 1000005; i++){\n if (p[i] == 0){\n for (int j = i+i; j < 1000005; j+=i){\n ... | 10235 Simply Emirp | The problem involves determining whether a given number N is a prime number, an emirp (a prime that remains prime when its digits are reversed), or not a prime at all. The solution requires checking the primality of N and its reversed form, then classifying N accordingly. | The input consists of multiple lines, each containing an integer N (1 < N < 1,000,000). | For each input number N, output one of the following:
1. 'N is not prime.' if N is not a prime.
2. 'N is prime.' if N is a prime but not an emirp.
3. 'N is emirp.' if N is an emirp (both N and its reversed form are primes, and they are different). | 17
18
19
179
199
| 17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >>... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2]... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] ... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >>... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\" with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] ... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >>... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10242 Fourth Point!!\", generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!!?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!!?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >>... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!! in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> ... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!!.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!!.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!! with explanations.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!!.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!!, generate it for me.",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「10242 Fourth Point!!」要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3]){\n ... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "「10242 Fourth Point!!」的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3]){\n... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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「10242 Fourth Point!!」？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3])... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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「10242 Fourth Point!!」的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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「10242 Fourth Point!!」的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3]){... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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「10242 Fourth Point!!」的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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「10242 Fourth Point!!」的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3])... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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「10242 Fourth Point!!」的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3]... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "10242 Fourth Point!! 要怎麼解? ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3]){\n ... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=0 | Yui Huang's Website | https://yuihuang.com/cpe-level-1/ | null | null | null |
[
{
"content": "10242 Fourth Point!! 的解答是什麼？",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3]){\n ... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!!？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3]){... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!! 的C++程式碼。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!! 的解法嗎？ ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3]){... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!! 的程式碼並附帶說明。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!! 的程式碼解答。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3] >> y[3])... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 10242 Fourth Point!! 的C++程式碼，幫我生成一個。 ",
"role": "user"
},
{
"content": "#include <iostream>\n#include <iomanip>\nusing namespace std;\n \nint main() {\n double x[4], y[4];\n double x1, y1, x2, y2, x3, y3;\n while (cin >> x[0] >> y[0] >> x[1] >> y[1] >> x[2] >> y[2] >> x[3]... | 10242 Fourth Point!! | Given the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram, the task is to find the coordinates of the fourth point. The solution involves identifying the common point of the two sides and using vector addition to determine the missing point. | Each line of input contains eight floating-point numbers representing the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. The coordinates are in meters, accurate to the nearest millimeter, and range between -10000 and +10000. The input continues until the end of file. | For each input line, output the (x,y) coordinates of the fourth point of the parallelogram, formatted to three decimal places and separated by a space. | 0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
| 1.000 0.000
-2.500 -2.500
0.151 -0.398
| One star | Online Judge | https://onlinejudge.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 \"10252 Common Permutation\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nstring a, b;\nint cnta[26], cntb[26];\n \nint main() {\n while (getline(cin, a) && getline(cin, b)){\n fill(cnta, cnta+26, 0);\n ... | 10252 Common Permutation | Given two strings of lowercase letters, the task is to find the longest string that can be formed by selecting characters common to both strings, where the frequency of each character in the result is the minimum of its frequencies in the two input strings. The result should be the lexicographically smallest possible p... | The input consists of multiple pairs of lines. Each pair contains two strings (a and b) of lowercase letters, each up to 1000 characters long. The input continues until there are no more pairs. | For each pair of strings, output a single line containing the longest common permutation string in alphabetical order. If multiple permutations exist, choose the lexicographically smallest one. | pretty
women
walking
down
the
street
| e
nw
tet
| One star | Online Judge | https://onlinejudge.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 \"10252 Common Permutation\"?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nstring a, b;\nint cnta[26], cntb[26];\n \nint main() {\n while (getline(cin, a) && getline(cin, b)){\n fill(cnta, cnta+... | 10252 Common Permutation | Given two strings of lowercase letters, the task is to find the longest string that can be formed by selecting characters common to both strings, where the frequency of each character in the result is the minimum of its frequencies in the two input strings. The result should be the lexicographically smallest possible p... | The input consists of multiple pairs of lines. Each pair contains two strings (a and b) of lowercase letters, each up to 1000 characters long. The input continues until there are no more pairs. | For each pair of strings, output a single line containing the longest common permutation string in alphabetical order. If multiple permutations exist, choose the lexicographically smallest one. | pretty
women
walking
down
the
street
| e
nw
tet
| One star | Online Judge | https://onlinejudge.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 \"10252 Common Permutation\" in C++?",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nstring a, b;\nint cnta[26], cntb[26];\n \nint main() {\n while (getline(cin, a) && getline(cin, b)){\n fill(cnta, cnta+2... | 10252 Common Permutation | Given two strings of lowercase letters, the task is to find the longest string that can be formed by selecting characters common to both strings, where the frequency of each character in the result is the minimum of its frequencies in the two input strings. The result should be the lexicographically smallest possible p... | The input consists of multiple pairs of lines. Each pair contains two strings (a and b) of lowercase letters, each up to 1000 characters long. The input continues until there are no more pairs. | For each pair of strings, output a single line containing the longest common permutation string in alphabetical order. If multiple permutations exist, choose the lexicographically smallest one. | pretty
women
walking
down
the
street
| e
nw
tet
| One star | Online Judge | https://onlinejudge.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 \"10252 Common Permutation\".",
"role": "user"
},
{
"content": "#include <iostream>\n#include <algorithm>\nusing namespace std;\n \nstring a, b;\nint cnta[26], cntb[26];\n \nint main() {\n while (getline(cin, a) && getline(cin, b)){\n fill(cnta, cnta+26... | 10252 Common Permutation | Given two strings of lowercase letters, the task is to find the longest string that can be formed by selecting characters common to both strings, where the frequency of each character in the result is the minimum of its frequencies in the two input strings. The result should be the lexicographically smallest possible p... | The input consists of multiple pairs of lines. Each pair contains two strings (a and b) of lowercase letters, each up to 1000 characters long. The input continues until there are no more pairs. | For each pair of strings, output a single line containing the longest common permutation string in alphabetical order. If multiple permutations exist, choose the lexicographically smallest one. | pretty
women
walking
down
the
street
| e
nw
tet
| One star | Online Judge | https://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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