int led_number=0; int led_number1=0; int led_number2=8; int r1=0; int r2=0;

时间: 2024-04-11 07:29:24 浏览: 8
您好!对于您提供的代码片段,我可以看出您定义了一些变量,并为它们赋予了一些初始值。其中包括: - `led_number`:一个整数变量,初始值为0。 - `led_number1`:一个整数变量,初始值为0。 - `led_number2`:一个整数变量,初始值为8。 - `r1`:一个整数变量,初始值为0。 - `r2`:一个整数变量,初始值为0。 请问还有其他问题吗?
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#!/usr/bin/env python3 import os from Crypto.Util.number import getPrime, getRandomRange def isSquare(a, p): return pow(a, (p-1)//2, p) != p-1 class SquareRNG(object): def __init__(self, p, sa, sb): assert sa != 0 and sb != 0 (self.p, self.sa, self.sb) = (p, sa, sb) self.x = 0 def int(self, nbits): v, s = 0, 1 for _ in range(nbits): self.x = (self.x + 1) % p s += pow(self.sa, self.x, self.p) * pow(self.sb, self.x, self.p) s %= self.p v = (v << 1) | int(isSquare(s, self.p)) return v def bool(self): self.x = (self.x + 1) % self.p t = (pow(self.sa, self.x, self.p) + pow(self.sb, self.x, self.p)) t %= self.p return isSquare(t, self.p) p = getPrime(256) sb1 = int(input("Bob's seed 1: ")) % p sb2 = int(input("Bob's seed 2: ")) % p for _ in range(77): sa = getRandomRange(1, p) r1 = SquareRNG(p, sa, sb1) print("Random 1:", hex(r1.int(32))) r2 = SquareRNG(p, sa, sb2) print("Random 2:", hex(r2.int(32))) guess = int(input("Guess next bool [0 or 1]: ")) if guess == int(r1.bool()): print("OK!") else: print("NG...") break else: print("Congratz!") print(os.getenv("FLAG", "nek0pts{*** REDACTED ***}"))

以下是您提供的Python代码的翻译版本: ```python #!/usr/bin/env python3 import os from Crypto.Util.number import getPrime, getRandomRange def isSquare(a, p): return pow(a, (p-1)//2, p) != p-1 class SquareRNG(object): def __init__(self, p, sa, sb): assert sa != 0 and sb != 0 (self.p, self.sa, self.sb) = (p, sa, sb) self.x = 0 def int(self, nbits): v, s = 0, 1 for _ in range(nbits): self.x = (self.x + 1) % p s += pow(self.sa, self.x, self.p) * pow(self.sb, self.x, self.p) s %= self.p v = (v << 1) | int(isSquare(s, self.p)) return v def bool(self): self.x = (self.x + 1) % self.p t = (pow(self.sa, self.x, self.p) + pow(self.sb, self.x, self.p)) t %= self.p return isSquare(t, self.p) p = getPrime(256) sb1 = int(input("Bob's seed 1: ")) % p sb2 = int(input("Bob's seed 2: ")) % p for _ in range(77): sa = getRandomRange(1, p) r1 = SquareRNG(p, sa, sb1) print("Random 1:", hex(r1.int(32))) r2 = SquareRNG(p, sa, sb2) print("Random 2:", hex(r2.int(32))) guess = int(input("Guess next bool [0 or 1]: ")) if guess == int(r1.bool()): print("OK!") else: print("NG...") break else: print("Congratz!") print(os.getenv("FLAG", "nek0pts{*** REDACTED ***}")) ``` 请注意,这是直接的翻译,可能需要根据您的具体需求进行进一步调整和优化。如果您有任何其他问题,请随时提问。

RationalNumber类C++

下面是一个简单的实现RationalNumber类的C++代码: ```c++ #include <iostream> using namespace std; class RationalNumber { private: int numerator; // 分子 int denominator; // 分母 public: RationalNumber() { numerator = 0; denominator = 1; } RationalNumber(int numerator, int denominator) { this->numerator = numerator; this->denominator = denominator; simplify(); // 约分 } // 获取分子 int getNumerator() const { return numerator; } // 获取分母 int getDenominator() const { return denominator; } // 赋值运算符重载 RationalNumber& operator=(const RationalNumber& other) { if (this != &other) { numerator = other.numerator; denominator = other.denominator; } return *this; } // 加法运算符重载 RationalNumber operator+(const RationalNumber& other) { int newNumerator = numerator * other.denominator + other.numerator * denominator; int newDenominator = denominator * other.denominator; return RationalNumber(newNumerator, newDenominator); } // 减法运算符重载 RationalNumber operator-(const RationalNumber& other) { int newNumerator = numerator * other.denominator - other.numerator * denominator; int newDenominator = denominator * other.denominator; return RationalNumber(newNumerator, newDenominator); } // 乘法运算符重载 RationalNumber operator*(const RationalNumber& other) { int newNumerator = numerator * other.numerator; int newDenominator = denominator * other.denominator; return RationalNumber(newNumerator, newDenominator); } // 除法运算符重载 RationalNumber operator/(const RationalNumber& other) { int newNumerator = numerator * other.denominator; int newDenominator = denominator * other.numerator; return RationalNumber(newNumerator, newDenominator); } // 约分 void simplify() { int gcd = getGCD(numerator, denominator); numerator /= gcd; denominator /= gcd; if (denominator < 0) { numerator *= -1; denominator *= -1; } } // 获取最大公约数 int getGCD(int a, int b) { if (b == 0) { return a; } return getGCD(b, a % b); } // 输出有理数 void print() const { if (denominator == 1) { cout << numerator << endl; } else { cout << numerator << "/" << denominator << endl; } } }; int main() { RationalNumber r1(2, 3); RationalNumber r2(3, 4); RationalNumber r3 = r1 + r2; RationalNumber r4 = r1 - r2; RationalNumber r5 = r1 * r2; RationalNumber r6 = r1 / r2; cout << "r1 = "; r1.print(); cout << "r2 = "; r2.print(); cout << "r1 + r2 = "; r3.print(); cout << "r1 - r2 = "; r4.print(); cout << "r1 * r2 = "; r5.print(); cout << "r1 / r2 = "; r6.print(); return 0; } ``` 在该代码中,定义了一个 `RationalNumber` 类,包含了分子和分母两个私有成员变量,以及一些公有成员函数,包括构造函数、赋值运算符重载、加法、减法、乘法、除法运算符重载、约分、获取最大公约数、输出有理数等。在 `main` 函数中,创建了两个有理数对象 `r1` 和 `r2`,并对其进行了加、减、乘、除等运算,最后输出了结果。

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Create a class called Rational for performing arithmetic with fractions. Use integer variables to represent the private data of the class – the numerator and the denominator. Provide a constructor that enables an object of this class to be initialized when it’s declared. The constructor should store the fraction in reduced form. For example, the fraction would be stored in the object as 1 in the numerator and 2 in the denominator. In order to compute the reduced form, you need to write a reduction function which uses the Euclidean algorithm to get the greatest common divisor (GCD) of the numerator and denominator first and then divides GCD to get the reduced numerator and denominator. Provide public member functions that perform each of the following tasks:(a) Subtract a Rational number from the other Rational number. The result should be stored in reduced form. (b) Divide a Rational number by the other Rational number. The result should be stored in reduced form. (c) Print Rational numbers in the form a/b, where a is the numerator and b is the denominator. (d) Compare two Rational numbers to make sure which one is smaller or they are equal. (1 for the first number, 2 for the second number and 0 if they are equal) Please also write a main function to prompt the user to input two Rational numbers . Subtract one number from the other from these two numbers using (a) and then print the result using (c). Divide one number from the other from these two numbers using (b) and then print the result using (c). Compare these two Rational numbers using (d) and indicate which one is smaller or they are equal.用c++ 寫出和上面不一樣的版本,要用using namespace std; 的版本,並且要有註解

Kars is tired and resentful of the narrow mindset of his village since they are content with staying where they are and are not trying to become the perfect life form. Being a top-notch inventor, Kars wishes to enhance his body and become the perfect life form. Unfortunately, n of the villagers have become suspicious of his ideas. The i -th villager has a suspicion of ai on him. Individually each villager is scared of Kars, so they form into groups to be more powerful. The power of the group of villagers from l to r be defined as f(l,r) where f(l,r)=|al−al+1|+|al+1−al+2|+…+|ar−1−ar|. Here |x−y| is the absolute value of x−y . A group with only one villager has a power of 0 . Kars wants to break the villagers into exactly k contiguous subgroups so that the sum of their power is minimized. Formally, he must find k−1 positive integers 1≤r1<r2<…<rk−1<n such that f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) is minimised. Help Kars in finding the minimum value of f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) . Input The first line contains a single integer t (1≤t≤100) — the number of test cases. The description of test cases follows. The first line of each test case contains two integers n,k (1≤k≤n≤100) — the number of villagers and the number of groups they must be split into. The second line of each test case contains n integers a1,a2,…,an (1≤ai≤500) — the suspicion of each of the villagers. Output For each test case, output a single integer — the minimum possible value of sum of power of all the groups i. e. the minimum possible value of f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) . Example inputCopy 3 4 2 1 3 5 2 6 3 1 9 12 4 7 2 12 8 1 9 8 2 3 3 1 8 7 7 9 2 outputCopy 4 11 2 Note In the first test case, we will group the villagers with suspicion (1,3,5,2) into (1,3,5) and (2) . So, f(1,3)+f(4,4)=(|1−3|+|3−5|)+0=4+0=4 . In the second test case, we will group the villagers with suspicion (1,9,12,4,7,2) into (1),(9,12),(4,7,2) . So, f(1,1)+f(2,3)+f(4,6)=0+3+8=11 .

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