[summer daily question] Luogu p7760 [coci2016-2017 5] tuna

Topic link ：P7760 [COCI2016-2017#5] Tuna - Luogu | New ecology of computer science education (luogu.com.cn)

Title Description

The fisherman caught N  Tuna . He is going to sell these fish to a platform .

For every fish , The platform will give two estimates P1,P2. If the difference between these two values does not exceed X, Then take the larger value as the value of the fish ; Otherwise, if the difference exceeds X, Then take another value P3.

Now given N  Value of fish （ There may be a 2  Or 3  individual ）, Find the total value of all fish .

Input format

first line , An integer N, Indicates the number of tuna .

The second line , An integer X.

The next few lines are divided into N  Three parts are used to represent the information of each fish . The format of each part is one of the following ：

- There is only one line in this section , This line contains two integers P1,P2.
- There are two lines in this section , The first line contains two integers P1,P2, The second line contains an integer P3.

Output format

Output the total value of all fish .

Examples #1

The sample input #1

5
2
3 4
2 1
5 3
4 4
4 2

Sample output #1

19

Examples #2

The sample input #2

4
2
3 5
2 8
4
6 5
6 3
7

Sample output #2

22

Examples #3

The sample input #3

3
10
20 50
30
20 40
50
70 20
10

Sample output #3

90

Tips

【 Examples 2 explain 】

The fisherman caught it 4  Tuna , and X  The value of is 2：

Therefore, the total value is 5+4+6+7 = 22.

【 Data scale and agreement 】

about 100%  The data of ,1 <= N <= 20,1 <= X <= 10,1 <= P1,P2,P3 <= 100.

【 Tips and instructions 】

The title is translated from  COCI 2016-2017 CONTEST #5 T1 Tuna.

The score of this question is according to COCI Original question setting , Full marks  50.

AC code:

#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
	int n;
	cin>>n;
	int x;
	cin>>x;
	int sum=0;
	while(n--)
	{
		int p1,p2;
		cin>>p1>>p2;
		if(abs(p1-p2)>x)
		{
			int p3;
			cin>>p3;
			sum+=p3;
		}
		else
		{
			sum+=max(p1,p2);
		}
	}
	cout<<sum;
	
	return 0;
}