Chapter 2 control structure and function (programming problem)

7-10 h0053. find the root of the quadratic equation

Use the formula x1 = (-b + sqrt(b* b-4* a* c))/(2* a), x2 = (-b - sqrt(b* b-4* a* c))/(2* a) Find the quadratic equation of one variable ax^2 + bx + c =0 The root of the , among a It's not equal to 0.

Input format :

The first line is the number of equations to be solved n.
rest n Each row contains three floating-point numbers a, b, c（ They are separated by spaces ）, Express the equations respectively ax^2 + bx + c =0 The coefficient of .

Output format :

The output shares n That's ok , Each row is the root of an equation ：
If it's two real roots , The output ：x1=...;x2 = ...
If two real roots are equal , The output ：x1=x2=...
If it's two imaginary roots , The output ：x1= Real component + Imaginary part i; x2= Real component - Imaginary part i

All real parts shall be accurate to the decimal point 5 position , Numbers 、 There is no space between symbols .
x1 and x2 The order of ：x1 The real part of >Re The real part of ||(x1 The real part of ==x2 The real part of &&x1 The imaginary part of >=x2 The imaginary part of )

sample input :

3
1.0 3.0 1.0
2.0 -4.0 2.0
1.0 2.0 8.0

sample output :

Here is the corresponding output . for example ：

x1=-0.38197;x2=-2.61803
x1=x2=1.00000
x1=-1.00000+2.64575i;x2=-1.00000-2.64575i

import scala.io.StdIn
import math.sqrt
object Main {
  def main(args: Array[String]): Unit = {
      var n = StdIn.readInt();
      for (i <- 1 to n) {
          var l = StdIn.readLine();
          var s = l.split(" ");
          var a = s(0).toDouble;
          var b = s(1).toDouble;
          var c = s(2).toDouble;
//           println(a+" "+b+" "+c);
          var part1 : Double = (0 - b) / (2 * a);
//           println(part1);
          var pd = b * b - 4 * a * c;
          if (pd > 0) {
              var part2 : Double = sqrt(pd) / (2 * a);
              printf("x1=%.5f;x2=%.5f\n" ,part1+part2 ,part1-part2);
          }
          else if(pd == 0) {
              printf("x1=x2=%.5f\n",part1);
          }
          else {
              var part2 : Double = sqrt(-pd) / (2 * a);
              printf("x1=%.5f+%.5fi;x2=%.5f%.5fi\n",part1,part2,part1,-part2);
          }
      }
  }
}

 

7-12 h0016. Precision movement

Xiao Ming Learning modeling . She is interested in models with moving parts . As her first task , She made a size 2×n Rectangular box , It consists of two parallel guide rails and rectangular strips on each rail . The size of the short strip 1×a, Size of the strip 1×b. There is a plug at each end of the long rod , The short one is always between the two plugs .

As long as the short rod is between the stops , The rod can move along the guide rail , One rod at a time . therefore , Xiao Ming moves one of the rectangular bars and moves it , And the other one stays in place . first , Both bars are aligned with the side of the box , Xiao Ming wants them to align with the other side with as little movement as possible . In order to achieve the goal , At least how many times does he have to move ？

 

Input format :

Enter... On one line 3 It's an integer a、b and c(1<=a<b<=n<=10^7), Separated by spaces .

Output format :

Enter... For each group , Output the minimum number of times Xiaoming needs to move in one line .

sample input :

1 3 6
2 4 9

sample output :

5
7

import scala.io.StdIn
object Main{
    def main(args: Array[String]): Unit={
        while(true){
            var l = StdIn.readLine();
            var s = l.split(" ");
            var a = s(0).toInt;
            var b = s(1).toInt;
            var c = s(2).toInt;
            var ans : Int = 0;
                var x : Int = a;
                var y : Int = b;
                var z : Int = b - a;
                while(x < c || y < c) {
                    if (x < c) {
                        x += z;
                        ans += 1;
                    }
                    if (y < c) {
                        y += z;
                        ans += 1;
                    }
                }
            println(ans)
        }
    }
}

 

7-13 h0010.sum

This topic requires reading 1 It's an integer n, Then the output 1~n The sum of all integers between .

Input format :

Enter... On one line 1 The absolute value is no more than 10000 The integer of n.

Output format :

Enter... For each group , Output in one line 1~n The sum of all integers between .

sample input :

-3
3

sample output :

Method 1 ：

import scala.io.StdIn


object Main {
    def main(args: Array[String]): Unit = {
        while(true){
            var n : Int = StdIn.readInt()
          var s : Int = 0
          if(n < 1){
              for(i<- Range(n,2)) s += i
              println(s)
          }

            else {
                for (i <- 1 to n) s += i
                println(s)
            }
        }
  }
}

Method 2 ：

x > 0 1~x  (1+x)*x/2

x <0 -x~1 -((1+x)*x/2 - 1)

import scala.io.StdIn

object Main {
    
    def main(args:Array[String]):Unit={
        
        while(true) {
            var sum : Int = 0
            var x = StdIn.readInt()
            if (x >= 0)
                sum = (x + 1) * x / 2
            else if(x < 0) {
                x = -x
                sum = -((x + 1) * x / 2) + 1
            }
            println(sum)
        }
    }
    
}

7-14 h0080. The diamond

Enter an odd number n, Output a by ‘*’ Composed of n Step hollow diamond .

Input format :

An odd number n.

Output format :

Output a by ‘*’ Composed of n Step solid diamond .

Refer to the output sample for specific format .

sample input :

5

sample output :

  *  
 * * 
*   *
 * * 
  *  

Law 1 ：

import scala.io.StdIn
object Main{
  def main(args: Array[String]): Unit = {
    // Print white diamond :   Hollow equilateral triangle + Hollow inverted triangle  =  Hollow diamond 
      var n : Int = StdIn.readInt()
    // Hollow equilateral triangle 
//     6      ==12
//    5 7     ==12
//   4   8    ==12
//  3     9   ==12
// 2       10 ==12
//1         11==12
// 2       10 ==12
//  3      9  ==12
//   4    8   ==12
//    5  7    ==12
//     6      ==12
// Each number represents each * No. is the column number of the row 
// Because the diamond is symmetrical , So consider the upper part first 
// By observing the upper part, it is 6 That's ok 11 Column 
// The left half is 1 Behavior 6、 The first 2 Behavior 5、..............、 The first 6 Behavior 1; Therefore, the external circulation should use for(i <- 6 to (1,-1))
// use i Representative line number ,j Representative column number ; When i=j perhaps j=12-i Should output at this position * Number , Output spaces in other places .
      for (i <- n / 2 + 1 to (1 ,-1)) {
          for (j <- 1 to n) {
              if (i == j || j == n + 1 -i) {
                  print("*")
              }
              else {
                  print(" ")
              }
          }
          println()
      }
    // Inverted triangle 
      for (i <- 2 to n / 2 + 1) {
          for (j <- 1 to n) {
              if (i == j || j == n + 1 -i) {
                  print("*")
              }
              else print(" ")
          }
          println()
      }
  }
}

Law two ：

import scala.io.StdIn

object Main{
  def main(args: Array[String]): Unit = {
      var n = StdIn.readInt()
      var a = n / 2
      
      // The upper part of a prism 
      for (i <- 1 to n / 2 + 1) {
          
          // Output * The blank line before No 
          for (j <- 1 to a) print(" ")
          
          // Output asterisk 
          for (j <- 1 to 2 * i - 1) {
              if (j == 1 || j == 2 * i - 1) print("*")
              else print(" ")
          }
          
          // Output * The blank line after the No 
          for (j <- 1 to a) print(" ")
          println()
          a -= 1
      }
      
      // The bottom half of the diamond 
      a = 1
      for (i <- n / 2 to (1, -1)) {
          
          // Output * The blank line before No 
          for (j <- 1 to a) print(" ")
          
          // Output asterisk 
          for (j <- 1 to 2 * i - 1) {
              if (j == 1 || j == 2 * i - 1) print("*")
              else print(" ")
          }
          
          // Output * The blank line after the No 
          for (j <- 1 to a) print(" ")
          println()
          a += 1
      }
  }
}

7-16 h0157. Fibonacci number

Give you an integer n, Find out its Fibonacci number ？

Input format :

Give... In one line 1 No more than one. 30 The positive integer n.

Output format :

Output the value of Fibonacci number in one line .

sample input :

5

sample output :

1 1 2 3 5 

Law 1 ： 

import scala.io.StdIn

object Main {
  def main(args: Array[String]): Unit = {
      var n = StdIn.readInt()
      def f(x: Int):Int= {
          if (x == 1) 1
          else if (x == 2) 1
          else f(x - 1) + f(x - 2)
      }
      var flag : Int = 0
      for (i <- 1 to n) {
          print(f(i) + " ")
      }
  }
}

  Law two ：

import scala.io.StdIn;
import scala.collection.mutable.ArrayBuffer     //ArrayBuffer Packages make arrays mutable 
object Main {                           // class 
def main(args: Array[String]): Unit = {         //main Method 
        var n = StdIn.readInt();
        def dg(n: Int): Array[Int] = {                  // The type is Int
            def func(n: Int): Int = {                       // Define methods 
                if (n<=2) 1; 
                else func(n - 1) + func(n - 2);  // features <=2, The output is 1,>2 by （n-1）+（n-2）
            }
//             println(func(1))                         // Check the special case value data 
            val array = new ArrayBuffer[Int];               //new An object 
            for (i <- 1 to n) {                         // determine i Value range of 
                array.append(func(i));                  // Go to buffer Add in func Data generated 
            }
            array.toArray;                      //toArray
        }
        dg(n).foreach(x => print(s"$x "));           // Traverse the output 
    }
}

7-17 h0158. The problem of monkeys eating peaches

There is a pile of peaches , The monkey ate half of it on the first day , And one more ！ Later, every day monkeys eat half of them , And then one more . When it comes to the third n days , When you want to eat again （ Haven't eaten yet ）, Only found 1 A peach . problem ： How many peaches are there in the first place ？

Input format :

Give... In one line 1 No more than one. 30 The positive integer n.

Output format :

Output the initial number of peaches in one line .

sample input :

10

sample output :

 Peach =1534

 

Law 1 ：

import scala.io.StdIn
object Main {
    def main(args: Array[String]): Unit = {
        var n = StdIn.readInt();
        val num = f(1,1,n);
        print(" Peach ="+num);
    }
    def f(a : Int ,d : Int ,n : Int): Int = {
        var ans = (a + 1) * 2;
        var dcnt = d + 1;
        if(dcnt == n) return ans;
        else f(ans ,dcnt ,n);
    }
}

  Law two ：

import scala.io.StdIn
object Main {
    def main(args: Array[String]): Unit = {
        var n = StdIn.readInt()
        var sum : Int = 1
        for (i <- 1 to n - 1) {
            sum = (sum + 1) * 2
        }
        println(" Peach =" + sum)
    }
}