Solution UVA: 11413 - Fill the Containers

 

Last Submissions
Problem	Verdict	Lang	Time	Best	Rank	Submit Time
| discuss11413 - Fill the Containers	Accepted	C++11	0.000	0.000	1972	5 mins ago

Suggest:

- Use binary search to determine the capacity (Once the capacity is determined, simulate and calculate the number of containers)

Note: I use a binary search algorithm along with a while + a for.

#include<bits/stdc++.h>

using namespace std;

int n, m;

int a[1000111];

int main(){

    while(cin >> n >> m){

       

        m = min(n, m);

       

        for(int i=1; i<=n; i++)

            cin >> a[i];

       

        int r = 0;

        for(int i=1; i<=n; i++)

            r += a[i];  

       

       

        int l= 1;

       

        while(r-l+1 >= 3){

           

            int mid = (r + l) / 2;

            int s = 0;

            int container= 0;

            bool bl = false;

           

            for(int i=1; i<=n; i++)

            if (a[i] + s <=  mid){

                s += a[i];  

                bl = true;

            }

            else{

                if (not bl) {

                    container = m + 1;

                    break;  

                }

                if (bl) container++;

                bl = false;

                s = 0;

                i--;

            }

           

            if (bl) container++;

           

        //  cout<<"mid container: "<<mid<<" "<<container<<"\n";

           

            if (container > m) l = mid;

            else

                r = mid;

        }

       

        for(int mid=l; mid<=r; mid++){

           

            int s = 0;

            int container= 0;

            bool bl = false;

           

            for(int i=1; i<=n; i++)

            if (a[i] + s <=  mid){

                s += a[i];  

                bl = true;

            }

            else{

                if (not bl) {

                    container = m + 1;

                    break;  

                }

                if (bl) container++;

                bl = false;

                s = 0;

                i--;

            }

           

            if (bl) container++;

           

            if (container <= m){

                cout << mid <<"\n";

                break;

            }          

           

        }

       

       

    }

   

       

}

Solution UVA: 10114 - Loansome Car Buyer

 

Problem	Verdict	Lang	Time	Best	Rank	Submit Time
| discuss10114 - Loansome Car Buyer	Accepted	Python	0.020	0.000	5752	1 mins ago

Suggest:

You only need forcus: (for output 1)

Consider the first example below of borrowing $15,000 for 30 months. As the buyer drives off the

lot, he still owes $15,000, but the car has dropped in value by 10% to $13,950. After 4 months, the

buyer has made 4 payments, each of $500, and the car has further depreciated 3% in months 1 and 2

and 0.2% in months 3 and 4. At this time, the car is worth $13,073.10528 and the borrower only owes

$13,000 

AND note

month_pay = loan / months (15000 / 30 = 500) 

I lost many time because think down_payment is month_payment ($500)

while True:

    months, down_pay, loan, n = map(float, input().split())

    months = int(months)

    n = int(n)

    if months < 0:

        break

    depreciations = [-1]*(months+1)  

    for _ in range(n):

        month, depreciation = map(float, input().split())

        month = int(month)

        depreciations[month] = depreciation

   

    car_money = (down_pay + loan)*(1-depreciations[0])

    owe = loan  

    if (owe < car_money):

        print("0 months")

    else:

        month_pay = loan / months

        for i in range(1, months+1):

            if depreciations[i] == -1:

                depreciations[i] = depreciations[i-1]

            owe -= month_pay

            car_money -= car_money * depreciations[i]

           

            if (owe < car_money):

                if (i>1):

                    print(i, 'months')

                else:

                    print(i, 'month')

                break      

   

# 30 500.0 15000.0 3

# 0 .10

# 1 .03

# 3 .002

# 12 500.0 9999.99 2

# 0 .05

# 2 .1

# 60 2400.0 30000.0 3

# 0 .2

# 1 .05

# 12 .025

# -99 0 17000 1

I lost many time because down_payment is month_payment

Trust its is diff