Polynomial summation AC-Hangzhou Electric

Polynomial summation

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)

Total submission (s): 50228 Accepted Submission (s): 29277

Problem Description

The polynomial is described as follows:

1-1/2 + 1/3-1/4 + 1/5-1/6 + ...

Now ask you to find out the first n of the polynomial.

Input

The input data consists of 2 lines, the first is a positive integer m (m<100), the number of test instances, the second row contains m positive integers, and for each integer (may be set to n,n<1000), the first n of the polynomial.

Output

For each test instance n, the number of the first n items of the output polynomial is required. The output of each test instance is one row, and the result retains 2 decimal places.

Sample Input

2

1 2

Sample Output

1.00

0.50

Author

Lcy

Source

C Language Programming exercises (II.)

#include <stdio.h>

#include <math.h>

int main ()

{

int m,j,i,n;

float sum;

scanf ("%d", &n);

while (n--)

{

scanf ("%d", &m);

for (i=1,sum=0;i<=m;i++)

{

J=pow ( -1, (i+1));

sum+=j* (1.0/i);

}

printf ("%.2\n", sum);

}

}

Polynomial summation AC-Hangzhou Electric