C/code/求公式近似值formula.c

#include <stdio.h>

double calculateFactorial(int n) {
    if (n == 0) {
        return 1;
    } else {
        return n * calculateFactorial(n - 1);
    }
}

int main() {
    double x, result = 1.0;
    int n;
    
    scanf("%lf %d", &x, &n);
    int i,j;
    for (i = 1; i <= n; i++) {
        double term = 1.0;
        for (j = 1; j <= i; j++) {
            term *= x / j;
        }
        result += term;
    }
    
    printf("%.6f\n", result);
    
    return 0;
}