360 2018 Autumn Recruitment Written Exam Questions

360 2018 Autumn Recruitment Written Exam Questions

(I'll update the code later...)

1. Sell chalk

Xiaoming has m colored chalks and n white chalks, of which a colored chalk and b white chalks can be sold for x yuan, c white chalks can be sold for y yuan, d colored pencils can be sold for z yuan, and chalk does not have to be sold. After all, how to maximize the benefits?

Example:

Input:
m=5, n=5, a=1, b=2, c=3, d=3, x=2, y=1, z=3,
output:
7

Parse:

Search recursively . There are three ways to sell:
[a , b] —> x
[0, c] —> y
[d, 0] —> z
Choose the one with the most benefits from the three ways each time, and the rest Chalk continues to search recursively until it is sold out or not.

c++ code implementation:

#include <iostream>
#include <vector>
#include <map>
using namespace std;

int sell(vector<int>& objects,vector<vector<int>>& special,map<vector<int>,int>& answer)
{
    if(answer.find(objects)!=answer.end())
        return answer[objects];
    int result = 0, j = 0;
    for(vector<int> s : special) {
        vector<int> rest(objects.begin(),objects.end());
        for( j=0; j<objects.size(); j++) {
            int diff = rest[j]-s[j];
            if(diff<0)
                break;
            rest[j] = diff;
        }
        if(j==objects.size()) {
            int money = s[j] + sell(rest,special,answer);
            result = money > result ? money : result;
        }
    }
    answer[objects] = result;
    //没卖完的，返回0
    return result;
}

int shoppingOffer(vector<int>& objects,vector<vector<int>>& special)
{
    map<vector<int>,int> answer;
    return sell(objects,special,answer);
}

int main()
{
    int m,n,a,b,c,d,x,y,z;
    cin>>m>>n>>a>>b>>c>>d>>x>>y>>z;
    vector<vector<int>> special = {{a,b,x},{0,c,y},{d,0,z}};
    vector<int> objects = {m,n};
    cout<<shoppingOffer(objects,special)<<endl;
    return 0;
}

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2. Array exchange

There are two arrays, namely a[a1, a2, …, an] and b[b1, b2, …, bm], sum represents the sum of all elements of the array, and at most two pairs of numbers in the a and b arrays are exchanged , so that; The absolute value of the difference between the sums of the two swapped arrays is the smallest?

Example:

Input:
n=4, a= [1, 3, 7, 9], m=3, b=[2, 10, 12]
Output:
0
Explanation: Swap (3, 2) and (9, 12) pairs , the sum of the elements of the last two arrays is both 22. The absolute value of the difference is 0.

Parse:

Let’s look at this problem first, exchanging a pair of elements of two arrays so that the absolute value of the elements and the difference between the two arrays is the smallest . This problem has one less restriction than the above problem: the number of exchange of elements is not limited .
Array a, the sum of elements is denoted as sumA; array b, the sum of elements is denoted as sumB.
If sumA==sumB, no swap is required.
Without loss of generality, we assume sumA > sumB and delta = sumA-sumB.
Suppose a[i] is exchanged with b[j], then: sumA'=sumA-a[i]+b[j], sumB'=sumB-b[j]+a[i]
delta' = sumA'-sumB ' = delta - 2*(a[i]-b[j]).
Therefore, to make the sum of two array elements as equal as possible, delta' should be as equal to 0 as possible, and a[i]-b[j] should be as close as possible to delta/2.
However, there is a limit to this question, at most two exchanges. The method has not yet been figured out.

C++ code implementation:
Swap a pair of elements of two arrays so that the absolute value of the elements and the difference between the two arrays is the smallest

#include <iostream>
#include <math.h>
using namespace std;

int change(int a[],int n,int b[],int m,int sumA,int sumB)
{
    int delta = 0;
    bool bigger = true;
    if(sumA>sumB){
        delta = sumA - sumB;
        bigger = true;
    }
    else{
        delta = sumB - sumA;
        bigger = false;
    }
    int diff = 0, x=0;
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            diff = a[i]-b[j];
            x = diff > 0 ? diff : (~diff+1); //abs(diff)
            if((x*2 <= delta)&&((bigger && diff>0)||(!bigger && diff<0))){
               a[i] ^= b[i];
               b[i] = a[i]^b[i];
               a[i] = a[i]^b[i];   //交换a[i] b[j]
               delta -= 2*x;
               if(delta==0)
                    return 0;
            }
        }
    }
    return delta;
}

int main()
{
    int n,m,sumA=0,sumB=0;
    cin>>n;
    int a[n];
    for(int i=0; i<n; i++){
        cin>>a[i];
        sumA += a[i];
    }

    cin>>m;
    int b[m];
    for(int i=0; i<m; i++){
        cin>>b[i];
        sumB += b[i];
    }
    if(sumA==sumB)
        cout<<0;
    else
        cout<<change(a,n,b,m,sumA,sumB);
    return 0;
}

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