In Chinese, according to the pronunciation, it can be divided into flat tone and oblique tone. Assuming we mark the flat tone as 0 and the oblique tone as 1.
A seven-character couplet consists of two sentences, each containing exactly seven characters.
There are three constraints for a seven-character couplet:
- A: 二四不同二六同: In each sentence, the second and fourth characters must have different tones, while the second and sixth characters must have the same tone.
- B: 仄起平收: The end of the first sentence must be a rising tone, while the end of the second sentence must be a level tone.
- C: 上下相對: The tone of the second, fourth, and sixth characters in both the first and second sentences must be opposite.
Given N pairs of couplets, represented by 0 and 1 to denote "ping" and "ze", please output which rules they violate. If none of the rules are violated, please output "None."
Sample Inputs/Outputs
| Sample Input(s) | Sample Output(s) |
|---|---|
| Input a positive integer N (1 <= N <= 30), representing the number of couplets. Following that are 2*N lines, each containing 7 numbers. The numbers are either 0 or 1. | For each couplet, output a line indicating which rules it violates. If it follows all three rules, output "None." |
| 1 1 1 0 0 0 1 1 1 0 0 0 1 1 0 | AC |
| 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 | None |
| 2 0 1 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 0 0 0 1 0 0 0 0 0 1 1 | AB ABC |
Thought Process
You can use a vector to store the two lines of each couplet and use the index (data in the vector) to make the judgments according to the rules of the problem.
Sample Code－ZeroJudge G275: Couplets
#include <iostream>
#include <vector>
using namespace std;
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
int N;
cin >> N;
for (int i = 0; i<N; i++)
{
vector<int>one;
vector<int>two;
one.push_back(-1);
two.push_back(-1);
bool A = false, B = false, C = false;
for (int j = 0; j<7; j++)
{
int tmp;
cin >> tmp;
one.push_back(tmp);
}
for (int j = 0; j<7; j++)
{
int tmp;
cin >> tmp;
two.push_back(tmp);
}
if (one[2] != one[4] && one[2] == one[6] && two[2] != two[4] && two[2] == two[6]) A = true;
if (one[7] == 1 && two[7] == 0) B = true;
if (one[2] != two[2] && one[4] != two[4] && one[6] != two[6]) C = true;
bool none = true;
if (!A)
{
none = false;
cout << "A";
}
if (!B)
{
none = false;
cout << "B";
}
if (!C)
{
none = false;
cout << "C";
}
if (none) cout << "None";
cout << "\n";
}
}
//ZeroJudge G275
//Dr. SeanXD