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2013
11-09

POJ 1045 Bode Plot [解题报告] Java

Bode Plot

问题描述 :

Consider the AC circuit below. We will assume that the circuit is in steady-state. Thus, the voltage at nodes 1 and 2 are given by v1 = VS coswt and v2 = VRcos (wt + q ) where VS is the voltage of the source, w is the frequency (in radians per second), and t is time. VR is the magnitude of the voltage drop across the resistor, and q is its phase.



You are to write a program to determine VR for different values of w. You will need two laws of electricity to solve this problem. The first is Ohm’s Law, which states v2 = iR where i is the current in the circuit, oriented clockwise. The second is i = C d/dt (v1-v2) which relates the current to the voltage on either side of the capacitor. “d/dt”indicates the derivative with respect to t.

输入:

The input will consist of one or more lines. The first line contains three real numbers and a non-negative integer. The real numbers are VS, R, and C, in that order. The integer, n, is the number of test cases. The following n lines of the input will have one real number per line. Each of these numbers is the angular frequency, w.

输出:

For each angular frequency in the input you are to output its corresponding VR on a single line. Each VR value output should be rounded to three digits after the decimal point.

样例输入:

1.0 1.0 1.0 9
0.01
0.031623
0.1
0.31623
1.0
3.1623
10.0
31.623
100.0

样例输出:

0.010
0.032
0.100
0.302
0.707
0.953
0.995
1.000
1.000

解题代码:

import java.util.Scanner;

 public class Main {

     static double vs = 0;
     static double c = 0;
     static double r = 0;
     static double w = 0;
     static int times = 0;

     public static void main(String[] args) {
         Scanner scan = new Scanner(System.in);
         vs = scan.nextDouble();
         r = scan.nextDouble();
         double r2 = r * r;
         double vsr = vs * r;
         c = scan.nextDouble();
         double c2 = c * c;
         times = scan.nextInt();
         for (int i = 0; i < times; i++) {
             w = scan.nextDouble();
             System.out.printf("%.3f\n", vsr / Math.sqrt(r2 + 1 / (w * w * c2)));
         }
     } 
}

  1. #include <cstdio>

    int main() {
    //answer must be odd
    int n, u, d;
    while(scanf("%d%d%d",&n,&u,&d)==3 && n>0) {
    if(n<=u) { puts("1"); continue; }
    n-=u; u-=d; n+=u-1; n/=u;
    n<<=1, ++n;
    printf("%dn",n);
    }
    return 0;
    }

  2. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术,也不懂什么是算法,从而也不要求程序员懂什么算法,做程序从来不考虑性能问题,只要页面能显示出来就是好程序,这是国内的现状,很无奈。