Codeforces Round #661(Div.3) B. Gifts Fixing

2178 단어
You have n gifts and you want to give all of them to children. Of course, you don't want to offend anyone, so all gifts should be equal between each other. The i -th gift consists of\(a_i\)candies and\(b_i\)oranges.
During one move, you can choose some gift 1≤i≤n and do one of the following operations:
  • eat exactly one candy from this gift (decrease ai by one);
  • eat exactly one orange from this gift (decrease bi by one);
  • eat exactly one candy and exactly one orange from this gift (decrease both ai and bi by one).

  • Of course, you can not eat a candy or orange if it's not present in the gift (so neither ai nor bi can become less than zero).
    As said above, all gifts should be equal. This means that after some sequence of moves the following two conditions should be satisfied: a1=a2=⋯=an and b1=b2=⋯=bn(and ai equals bi is not necessary).
    Your task is to find the minimum number of moves required to equalize all the given gifts.
    You have to answer t independent test cases.
    Input
    ~~
    Output
    For each test case, print one integer: the minimum number of moves required to equalize all the given gifts.
    Example
    Input
    Copy
    5
    3
    3 5 6
    3 2 3
    5
    1 2 3 4 5
    5 4 3 2 1
    3
    1 1 1
    2 2 2
    6
    1 1000000000 1000000000 1000000000 1000000000 1000000000
    1 1 1 1 1 1
    3
    10 12 8
    7 5 4
    

    Output
    Copy
    6
    16
    0
    4999999995
    7
    

    모든 a는 초기의 a에서 가장 작은 값으로 낮춰야 한다. b는 같은 이치이다.따라서 매gift를 두루 돌아다니며 max(a[i]-mina, b[i]-minb)로 답안을 업데이트하면 된다(ai가mina로 5회, 비가minb로 4회 조작해야 한다고 가정하면 b의 4회는 5회에 모두 5회 포함하면 된다).
    #include 
    using namespace std;
    int a[55], b[55], n;
    int main()
    {
    	int t;
    	cin >> t;
    	while(t--)
    	{
    		cin >> n;
    		int mina = 0x3f3f3f3f, minb = 0x3f3f3f3f;
    		long long ans = 0;
    		for(int i = 1; i <= n; i++)
    		{
    			cin >> a[i];
    			mina = min(mina, a[i]);
    		}	
    		for(int i = 1; i <= n; i++)
    		{
    			cin >> b[i];
    			minb = min(minb, b[i]);
    		}
    		for(int i = 1; i <= n; i++)
    		{
    			ans += 1ll * max(a[i] - mina, b[i] - minb);
    		}
    		cout << ans << endl;
    	}
    	return 0;
    }
    

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