1 条题解

  • 0
    @ 2025-4-7 21:29:25

    C :

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

    C++ :

    #include<iostream>
    #include<iostream>
    using namespace std;
    int main()
    {int a;cin>>a;float d;
    if(a<=100000)d=a*0.1;
    else if(a<=200000)d=10000+(a-100000)*0.075;
    else if(a<=400000)d=10000+7500+(a-200000)*0.05;
    else if(a<=600000)d=27500+(a-400000)*0.03;
    else if(a<=1000000)d=33500+(a-600000)*0015;
    else d+=39500+(d-1000000)*0.01;
    cout<<int (d)<<endl;return 0;}
    
    

    Pascal :

    var
    n:longint;
    x:real;
    begin
     readln(n);
     if n<=100000 then x:=n*0.1;
     if (n>100000)and(n<=200000) then x:=n*0.1;
     if (n>200000)and(n<=400000) then x:=n*0.05;
     if (n>400000)and(n<=600000) then x:=n*0.03;
     if (n>600000)and(n<=1000000) then x:=n*0.015;
     if n>1000000 then x:=n*0.01;
     writeln(x:0:0);
    end.
    
    
    
    
    
    
    • 1

    信息

    ID
    1417
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    (无)
    递交数
    0
    已通过
    0
    上传者