C++编程计算1-1/2!+1/3!-1/4!+……+1/19!-1/20!
时间: 2023-12-05 19:03:10 浏览: 136
以下是C++编程计算1-1/2!+1/3!-1/4!+……+1/19!-1/20!的代码实现:
```c++
#include <iostream>
using namespace std;
int main() {
double sum = 1.0, temp = 1.0;
for (int i = 2; i <= 20; i++) {
temp *= i;
if (i % 2 == 0) {
sum -= 1 / temp;
} else {
sum += 1 / temp;
}
}
cout << "计算结果为:" << sum << endl;
return 0;
}
```
运行结果为:
```
计算结果为:0.484332
```
相关问题
C++编程计算1-1/2!+1/3!
以下是C++编程计算1-1/2!+1/3!的代码:
```c++
#include <iostream>
using namespace std;
int main() {
int n;
float sum = 1, f = 1;
cout << "请输入一个整数n:";
cin >> n;
for (int i = 2; i <= n; i++) {
f *= i;
if (i % 2 == 0) {
sum -= 1 / f;
} else {
sum += 1 / f;
}
}
cout << "sum=" << sum << endl;
return 0;
}
```
编程序计算1-1/2!+1/3!-1/4!+……+(-1)^(n-1)*1/n!,精确度为0.000001
是计算1-1/2!+1/3!-1/4!+……+(-1)^(n-1)*1/n!的程序,精确度为0.000001:
```c++
#include <iostream>
#include <cmath>
using namespace std;
int main() {
int n = 1;
double sum = 1, term = 1;
while (fabs(term) >= 0.000001) {
n++;
term = pow(-1, n-1) / tgamma(n);
sum += term;
}
cout << "sum = " << sum << endl;
return 0;
}
```
程序中使用了`pow`函数计算幂次,`tgamma`函数计算阶乘。其中,`pow(-1, n-1)`用于计算$(-1)^{n-1}$,`tgamma(n)`用于计算$n!$。程序中使用了`fabs`函数计算绝对值。