C++ 下列程序实现:x=1-1/22+1/333- 1/4444+1/55555......+1/999999999表达式的求值。
时间: 2024-01-29 22:02:45 浏览: 85
以下是C++实现该表达式求值的程序:
```cpp
#include <iostream>
#include <cmath>
int main() {
double x = 0;
double denominator = 2;
int sign = -1;
for (int i = 2; i <= 999999999; i++) {
x += sign * (1 / denominator);
denominator = denominator * 10 + i % 10;
sign *= -1;
}
std::cout << "x is " << x << std::endl;
return 0;
}
```
相关问题
C++编写程序计算下列公式的值:y = x - x^3/3!+ x^5/5!- x^7/7!+ ⋯,精确到第n项
#include <stdio.h>
#include <math.h>
double factorial(int n) {
double result = 1;
for (int i = 1; i <= n; i++) {
result *= i;
}
return result;
}
double calculate(double x, int n) {
double result = x;
int sign = -1;
for (int i = 2; i <= n; i += 2) {
double term = pow(x, i) / factorial(i);
if (sign == -1) {
result += term;
} else {
result -= term;
}
sign = -sign;
}
return result;
}
int main() {
double x;
int n;
printf("请输入x和n:");
scanf("%lf %d", &x, &n);
double y = calculate(x, n);
printf("y = %lf", y);
return 0;
}
C++编写程序计算下列公式的值:y = x - x^3/3!+ x^5/5!- x^7/7!+ ⋯,精确到第n项,
其中x为用户输入的实数,n为用户输入的正整数。
提示:阶乘可以用循环计算,递归计算或者利用数学库函数计算。可以定义一个函数来计算阶乘。同时,可以定义一个函数来计算每一项的值。最后,在主函数中利用循环计算所有项的和,直到计算到第n项为止。
以下是示例代码:
#include <stdio.h>
double factorial(int n)
{
double result = 1.0;
for (int i = 1; i <= n; i++)
{
result *= i;
}
return result;
}
double term(double x, int n)
{
double sign = (n % 2 == 0) ? 1.0 : -1.0;
double power = 2 * n + 1;
double denominator = factorial(power);
double numerator = sign * pow(x, power);
return numerator / denominator;
}
double series(double x, int n)
{
double sum = x;
for (int i = 1; i <= n; i++)
{
sum += term(x, i);
}
return sum;
}
int main()
{
double x;
int n;
printf("Enter x: ");
scanf("%lf", &x);
printf("Enter n: ");
scanf("%d", &n);
double y = series(x, n);
printf("y = %.10f\n", y);
return 0;
}
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