Barney goes to Work

Computer Science Level 3

Barney is a busy guy. He can only afford to come to his office on 100 days.

Goliath National Bank decided to pay him $ a i \$a_i if he comes on day i i and fine him $ b i \$b_i if he misses day i i .

This being the lists < a i > <a_i> and < b i > <b_i> , for n = 1 0 4 n = 10^4 days, what is the minimum amount that Barney would have to pay Goliath National Bank at the end of the year?

  • Input Format: Line 1 contains all a i a_i sequentially from i = 1 i = 1 to 1 0 4 10^4 and Line 2 contains all the corresponding b i b_i


The answer is 392268547939.

Agnishom Chattopadhyay by Agnishom Chattopadhyay , 22, India

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1 solution

Barney’s earning = p a y w o r k − f i n e m i s s \text{Barney's earning} = pay_{work} - fine_{miss} = p a y w o r k + f i n e w o r k − f i n e w o r k − f i n e m i s s = pay_{work} + fine_{work} - fine_{work} - fine_{miss} = ( p a y + f i n e ) w o r k − f i n e w o r k + m i s s = (pay + fine)_{work} - fine_{work + miss}

maximise pay + fine to find the answer. 

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#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
    unsigned long int a[10000], b[10000], c[10000];
    unsigned long int max_sum = 0, sum_b = 0;
    for(int i = 0; i < 10000; i++){
        scanf("%lu", &a[i]);
    }
    for(int i = 0; i < 10000; i++){
        scanf("%lu", &b[i]);
        sum_b += b[i];
        c[i] = a[i] + b[i];
    }
    sort(c, c+10000);
    for(int i = 10000 - 100; i < 10000; i++) max_sum += c[i];
    printf("sum_b = %lu\nmax sum = %lu", sum_b, max_sum);
}

max earning = max_sum - sum_b = 106882249890 - 499150797829 = -392268547939

Poor Barney!

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