题解 | 牛牛学数列3

#include <stdio.h>

int main() {
    int n;
    float sum;
    float count = 0.0;
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        float denominator =0.0;
        int sign =1;
    for (int j = 1; j <= i; j++) {  // j表示第几个奇数
            int odd_number = 2 * j - 1;  // 第j个奇数
            denominator += sign * odd_number;
            sign = -sign;
    }
        count += 1.0 / denominator;
    }
    printf("%.3f",count);
    return 0;
}