1 条题解

  • 0
    @ 2025-4-7 21:28:48

    C :

    #include <stdio.h>
    int main()
    {
    	int n;
    	double m=0;
    	scanf("%d",&n);
    
    	if(n>1000000)
    	{
    		m += (n-1000000)*0.01;
    		n = 1000000;
    	}
    	if(n>600000)
    	{
    		m += (n-600000)*0.015;
    		n = 600000;
    	}
    	if(n>400000)
    	{
    		m += (n-400000)*0.03;
    		n = 400000;
    	}
    	if(n>200000)
    	{
    		m += (n-200000)*0.05;
    		n = 200000;
    	}
    	if(n>100000)
    	{
    		m += (n-100000)*0.075;
    		n = 100000;
    	}
    	m += n*0.1;
    	printf("%d\n",(int)m);
    	return 0;
    }
    

    C++ :

    #include<iostream>
    using namespace std;
    int main()
    {
    double num,score,i,j;
    int m;
      cin>>num;
      if(num<=100000)
        score=num*10/100;
      else if(num>100000&&num<=200000)
        score=10000+(num-100000)*7.5/100;
      else if(num>200000&&num<=400000)
        score=10000+7500+(num-200000)*5/100;
      else if(num>400000&&num<=600000)
        score=10000+7500+10000+(num-400000)*3/100;
      else if(num>600000&&num<=1000000)
        score=10000+7500+10000+6000+(num-600000)*1.5/100;
      else
        score=10000+7500+10000+6000+6000+(num-1000000)*1/100;
        cout<<(int)score<<endl;
        //system("pause");
        return 0;
    }
        
        
    
    

    Java :

    import java.text.DecimalFormat;
    import java.util.*;
    public class Main {
        public static void main(String args[]) {
            Scanner cin = new Scanner(System.in);
            DecimalFormat df = new DecimalFormat("0");
            int i;
            i = cin.nextInt();
            double b = 0;
            if(i <= 100000)
            {
                b += i*0.1;
            }
            else if(i > 100000 && i <= 200000)
            {
                b += 10000;
                b += (i-100000)*0.075;
            }
            else if(i > 200000 && i <= 400000)
            {
                b += 10000 + 7500;
                b += (i-200000)*0.05;
            }
            else if(i > 400000 && i <= 600000)
            {
                b += 17500 + 10000;
                b += (i - 400000)*0.03;
            }
            else if(i > 600000 && i <= 1000000)
            {
                b += 27500 + 6000;
                b += (i - 600000)*0.015;
            }
            else {
                b += 33500 + 6000;
                b += (i - 1000000)*0.01;
            }
            System.out.println(df.format(b));
        }
    }
    

    Python :

    a = float(input())
    b = 0
    if a - 10 ** 5 > 0:
        b += 10000
        if a - 200000 > 0:
            b += 7500
            if a - 400000 > 0:
                b += 10000
                if a - 600000 > 0:
                    b += 6000
                    if a - 1000000 > 0:
                        b += 6000 + (a - 1000000) * .01
                    else:
                        b += (a - 600000) * .015
                else:
                    b += (a - 400000) * .03
            else:
                b += (a - 200000) * .05
        else:
            b += (a - 10 ** 5) * .075
    else:
        b += a * .1
    print int(round(b))
    
    • 1

    C语言程序设计教程(第三版)课后习题5.8

    信息

    ID
    1405
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者