先计算有规律变化的分母

#include<cstdio>
#include<cmath>
int count(int n);
int main(){
    int n;
    scanf("%d", &n);
    double sum = 0;
    for(int i = 1; i <= n; i++){
        double temp = 1.0 / count(i);
        sum += temp;
    }
    printf("%.3f", sum);

    return 0;
}
int count(int n){
    int temp = 0;
    for(int i = 1; i <= n; i++){
        temp += pow(-1, i-1) * (2*i-1) * 1.0;
    }
    return temp;

}