Get minimum water level of dam without loop-CodePudding

I was given a problem. In the problem I am suppose to take input current month, current year and current water level of a dam from user. There is a scale that from month march to august the water level increase by 150 feet each month and from sep to feb it decreases by 200 feet. Now I am supposed to tell in which month and year the Dam will have no water or 0 feet water level. I have made the below program using loops but I have to it with out loops/recursive function. I got the year by dividing the water level with avg_decrease in water level.You cans see the program that does what I want with loop.

#include<iostream>
using namespace std;
int main(){
    int c_month, c_year, wlevel, avg_decrease;
    cout<<"Enter current month number: ";
    cin>>c_month;
    cout<<"Enter current water level: ";
    cin>>wlevel;
    cout<<"Enter current year: ";
    cin>>c_year;
    avg_decrease = 300; //-25 each month, -300 each year
    cout<<wlevel/avg_decrease<<endl;

    int m = c_month, level = wlevel, y = c_year;
    while(true){
        if(x)
            break;
        for(int i =0;i<=12;i  ){
            if(level < 0){
                x = true;
                break;
                }
            else if(m >= 3 && m<=8){
                level = level  150;
                m  ;
                }
            else{
                level = level -200;
                if(m == 12)
                    m=1;
                else
                    m  ;
            }
        }
        y  ;
    }
    cout<<y<<"\t"<<m<<endl;
}

I want to get the month and year in which the water level is 0 feet which is being printed in the last line without using the loops. I dont know how to implement the above program without using loops. If any one can help, it would be great. Thanks in advance

CodePudding user response：

test this and change code if my logic is wrong

#include<iostream>
using namespace std;

int main(){
    int m, y, lev;
    cout << "Enter current month number: ";
    cin >> m;
     cout << "Enter current year: ";
    cin >> y;
    cout << "Enter current water level: ";
    cin >> lev;
   
    int avg_year    = 300;                  // average year decrease 
    int year_count  = lev/avg_year;         // how many entire years we will be decreasing for sure  
    lev = lev % avg_year;                   // how much level we still have after entire years have pass
    
    if ((lev > 0) &&  (m <= 8) && (m >=2))        // march - aug we are adding  150 each month 
    {
        int delta = (6 - m   2);            // how much times we should add  150
        lev = lev   150*delta;
        m = m   delta;
    }
    
    if((lev > 0) && (m  == 8))             // end of aug (sep = -200)
    {
        m  ;
        lev = lev - 200;
    }
    if((lev > 0) && (m  == 9))             // end of sep (oct = -200)
    {
        m  ;
        lev = lev - 200;
    }
    if((lev > 0) && (m  == 10))             // end of oct (nov = -200)
    {
        m  ;
        lev = lev - 200;
    }
    if((lev > 0) && (m  == 11))             // end of nov (dec = -200)
    {
        m  ;
        lev = lev - 200;
    }
    if((lev > 0) && (m == 12))             // end of dec (jan = -200)
    {
        m = 1;
        year_count  ;
        lev = lev - 200;
    }
    
   
    /*
        in case at the beginning of the program m== 1 AND Level == (801 - 899)
        2 years past and we are at m == 1 with level == (201 - 299)
        februarry gives -200. So at the end of m==2, level == (1 - 99)
        
        when we do march-december we gain  100 level
        
        so in the beginning of the next year jan level will be 101-199
        and since jan takes -200 from level. It is definitely next years jan
        so we are just increasing year count
    */
    if((lev > 0) && (m  == 1))             // enf of jan (feb = -200)
    {
       lev = lev - 200;
       if(lev <= 0)
            m  ;
        else
            year_count  ;
    }
       

   cout << (y   year_count)  << "\t" << m << endl;
}

UPD this is bellow I think the right solution as I would do it with all the loops. I doubt it can be achieved just by plain code with no loops or recursion

#include<iostream>
using namespace std;

int level_func(int m, int y, int lev)
{
    int ar[] = { -200, -200, 150, 150, 150, 150, 150, 150, -200, -200, -200, -200 };    

    int m_count = 0;
    while (lev > 0)
    {
        m  ;
        m_count  ;
        if (m > 12)
            m = 1;
        lev = lev   ar[m - 1];
    } 
    return m_count;
}

int main() {
    int m, y, lev;
    cout << "Enter current month number: ";
    cin >> m;
    cout << "Enter current year: ";
    cin >> y;
    cout << "Enter current water level: ";
    cin >> lev;

    int month_count = level_func(m, y, lev);
    y = y   month_count / 12;
    m = m   month_count % 12;

    if (m > 12)
    {
        y  ;
        m = m - 12;
    }  

    cout << y << "\t" << m << endl;
}

CodePudding user response：

I worked on the solution myself and found a very efficient way to do it. I calculated total months required and got the total years from the month.The code is below

#include<iostream>
using namespace std;
int main(){
    int exact_months,year = 0;
    int c_month, c_year, wlevel, avg_decrease;
    cout<<"Enter current month number: ";
    cin>>c_month;
    cout<<"Enter current water level: ";
    cin>>wlevel;
    cout<<"Enter current year: ";
    cin>>c_year;
    int mn; 
    exact_months = wlevel/25;
    mn = (c_month exact_months);
    if ((c_month exact_months) !=0 && c_month != 1)
        year = 1;
    year = exact_months/12 year;
    cout<<"Date: "<<year c_year<<":"<<mn;
}