2012-12-06 20:05:17Morris
[UVA][柯西不等式] 12575 - Sin Cos Problem
Sin Cos Problem
| Sin Cos Problem |
Given A and B, you have to determine the maximum value of the function :
F(
) = A*Sin + B*Cos
) = A*Sin + B*Cos
Input
First line of input will contain the number of test cases, T
2000. 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
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;
}
無恥的 atan2 .........
#include <stdio.h>
#include <math.h>
#define PI acos(-1.0)
#define EPS 5.14e-7
int t;
double a, b, ans;
int main()
{
for (scanf("%d", &t); t; t--)
{
scanf("%lf%lf", &a, &b);
ans = atan2(a, b);
if (ans < -EPS) ans += 2.0 * PI;
printf("%.2lf %.2lf\n", ans, sqrt(a * a + b * b));
}
}