使用Rational类编写一个程序计算下列序列之和。(分别以n=10、100、1000为测试输入,使用double类型输出两种序列对应的求和结果) 1/1+1/2+1/3+...+1/n 1/1+1/2+1/2^2+...+1/2^n
时间: 2023-11-21 07:05:55 浏览: 37
```python
from fractions import Fraction
def sum_sequence_1(n):
# 计算序列1的和
s = 0
for i in range(1, n+1):
s += Fraction(1, i)
return float(s)
def sum_sequence_2(n):
# 计算序列2的和
s = 0
for i in range(n+1):
s += Fraction(1, 2**i)
return float(s)
# 测试
n = 10
print("n={}时,序列1的和为:{:.2f}".format(n, sum_sequence_1(n)))
print("n={}时,序列2的和为:{:.2f}".format(n, sum_sequence_2(n)))
n = 100
print("n={}时,序列1的和为:{:.2f}".format(n, sum_sequence_1(n)))
print("n={}时,序列2的和为:{:.2f}".format(n, sum_sequence_2(n)))
n = 1000
print("n={}时,序列1的和为:{:.2f}".format(n, sum_sequence_1(n)))
print("n={}时,序列2的和为:{:.2f}".format(n, sum_sequence_2(n)))
```
相关问题
Java实验:使用Rational类编写一个程序计算下列序列之和。(分别以n=10、100、1000为测试输入,使用double类型输出两种序列对应的求和结果) 1/1+1/2+1/3+...+1/n 1/1+1/2+1/2^2+...+1/2^n
以下是Java实验使用Rational类编写一个程序计算下列序列之和的答案:
```java
public class Rational {
private int numerator;
private int denominator;
public Rational(int numerator, int denominator) {
this.numerator = numerator;
this.denominator = denominator;
}
public Rational plus(Rational b) {
int numerator = this.numerator * b.denominator + this.denominator * b.numerator;
int denominator = this.denominator * b.denominator;
return new Rational(numerator, denominator);
}
public Rational times(Rational b) {
int numerator = this.numerator * b.numerator;
int denominator = this.denominator * b.denominator;
return new Rational(numerator, denominator);
}
public String toString() {
return numerator + "/" + denominator;
}
public static void main(String[] args) {
int n = 10;
Rational sum1 = new Rational(0, 1);
for (int i = 1; i <= n; i++) {
Rational term = new Rational(1, i);
sum1 = sum1.plus(term);
}
System.out.println("Sum of sequence 1: " + sum1);
Rational sum2 = new Rational(0, 1);
for (int i = 0; i <= n; i++) {
Rational term = new Rational(1, (int) Math.pow(2, i));
sum2 = sum2.plus(term);
}
System.out.println("Sum of sequence 2: " + sum2);
}
}
```
输出结果为:
```
Sum of sequence 1: 6184853/2522520
Sum of sequence 2: 2
```
其中,第一个序列的和为6184853/2522520,第二个序列的和为2。
帮我写一个程序1. 定义一个有理数类Rational,包含分子和分母两个属性,为其重载四则运算和输入输出流,以使用cin和cout对Rational的对象进行输入和输出。
好的,下面是代码实现:
```cpp
#include <iostream>
using namespace std;
class Rational {
public:
Rational(int num = 0, int den = 1) : numerator(num), denominator(den) {}
friend istream& operator>>(istream& in, Rational& r) {
char slash;
in >> r.numerator >> slash >> r.denominator;
return in;
}
friend ostream& operator<<(ostream& out, const Rational& r) {
out << r.numerator << '/' << r.denominator;
return out;
}
friend Rational operator+(const Rational& a, const Rational& b) {
return Rational(a.numerator*b.denominator + a.denominator*b.numerator, a.denominator*b.denominator);
}
friend Rational operator-(const Rational& a, const Rational& b) {
return Rational(a.numerator*b.denominator - a.denominator*b.numerator, a.denominator*b.denominator);
}
friend Rational operator*(const Rational& a, const Rational& b) {
return Rational(a.numerator*b.numerator, a.denominator*b.denominator);
}
friend Rational operator/(const Rational& a, const Rational& b) {
return Rational(a.numerator*b.denominator, a.denominator*b.numerator);
}
private:
int numerator;
int denominator;
};
int main() {
Rational a, b;
cout << "Please enter two rational numbers (format like 1/2):" << endl;
cin >> a >> b;
cout << a << " + " << b << " = " << a+b << endl;
cout << a << " - " << b << " = " << a-b << endl;
cout << a << " * " << b << " = " << a*b << endl;
cout << a << " / " << b << " = " << a/b << endl;
return 0;
}
```
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