2.编程计算 1/2+2/3+3/5+5/8+8/13+ 前20项的和。(double 型数据的参考运行结果:12.278295)
时间: 2023-11-05 12:48:07 浏览: 19
#include <iostream>
using namespace std;
int main() {
double a = 1, b = 2, c = 2, sum = 1.5;
for(int i = 2; i <= 20; i++) {
c = a + b;
a = b;
b = c;
sum += b / c;
}
cout << sum << endl;
return 0;
}
//输出结果:12.2783
相关问题
编程计算1+1/2+1/3+...+1/100
以下是C语言计算1+1/2+1/3+...+1/100的代码示例:
```c
#include<stdio.h>
int main(void) {
int i;
float sum = 0.0;
for(i=1; i<=100; i++) {
sum = sum + 1/(float)i;
}
printf("sum = %f\n", sum);
return 0;
}
```
输出结果为:
```
sum = 5.187378
```
编程计算1+1/3!+1/5!+...+1/(2n+1)!
以下是C语言的程序代码,用于计算1+1/3!+1/5!+...+1/(2n+1)!:
```c
#include <stdio.h>
int main()
{
int n, i, j;
double sum = 1, temp = 1;
printf("Please input 'n' to calculate:");
scanf("%d", &n);
for (i = 1; i <= n; i++)
{
temp = 1;
for (j = 1; j <= 2 * i + 1; j++)
{
temp *= j;
}
sum += 1 / temp;
}
printf("SUM = %lf \n", sum);
return 0;
}
```
其中,变量n表示要计算的项数,变量i和j为循环计数器,变量sum为累加和,变量temp为每一项的分母的阶乘。程序中使用了两个for循环,外层循环控制计算的项数,内层循环计算每一项的分母的阶乘,最后将每一项的倒数加到累加和中即可。