Sin Cos Problem

Given A and B, you have to determine the maximum value of the function :

Input 

First line of input will contain the number of test cases, T2000. Then there follows T lines, each containing two integers A and B separated by a single space. A and B will fit in a signed 32bit integer.

Output 

For each case, print one line containing two single space separated real values rounded to two decimal places. The first one is the lowest non-negative value of ( is in Radian) for which the F() gives maximum value and the second one is the maximum value.

Note: Pi is considered to be arccos(- 1).

Sample Input 

4
1 1
-1 1
1 -1
-1 -1

Sample Input 

0.79 1.41
5.50 1.41
2.36 1.41
3.93 1.41

Problem Setter: Muhammad Ridowan
Alternate Solution: Zobayer Hasan

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注意 0 0 的輸入，請輸出 0.00 0.00

#include <stdio.h>
#include <math.h>
#define eps 1e-2
int main() {
    int t, i;
    scanf("%d", &t);
    double A, B, pi = acos(-1);
    while(t--) {
        scanf("%lf %lf", &A, &B);
        if(A == 0 && B == 0) {
            puts("0.00 0.00");
            continue;
        }
        double tmp1, tmp2;
        double F = sqrt(A*A+B*B);
        tmp1 = atan(A/B);
        for(; tmp1 <= 2*pi; tmp1+=pi/2) {
            if(fabs(A*sin(tmp1)+B*cos(tmp1)-F) < eps && tmp1 >= 0)
                break;
        }
        printf("%.2lf %.2lf\n", tmp1, F);
    }
    return 0;
}

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