C语言-自定义函数

解题思路:

注意事项:

参考代码:

#include<iostream>

using namespace std;

#include<math.h>

#include<stdio.h>

double jca(int n)

{

double sum=1;//1

for(int j=1;j<=n;j++)

{

sum *= j;

}

return sum;

}

double mypow(double x, int n)

{

double sum=0,fh=1;

for (int i = 1; i <= n; i++)

{

sum += fh * pow(x, i) / (jca(i));

fh = -fh;

}

// x - x2/2! + x3/3! + ... + (-1)n-1xn/n!

return sum;

}

int main()

{

int n;

double x;

cin >> x >> n;

printf("%.4lf", mypow(x, n));

return 0;

}