029: Tao Tao picks apples

029: Tao Tao picks apples

Total time limit :

1000ms

Memory limit :

65536kB

describe

There is an apple tree in the Tao Tao's yard , Every autumn the trees will bear 10 An apple . When the apple is ripe , Tao Tao would run to pick apples . Tao Tao has a 30 Centimeter high bench , When she can't pick the apple directly by hand , Will step on the bench and try again .

Now we know 10 The height of an apple to the ground , And the maximum height that can be reached when the handle of pottery is extended , Please help Tao Tao calculate the number of apples she can pick . Suppose she ran into an apple , The apple will fall .

Input

Including two lines of data . The first line contains 10 individual 100 To 200 Between （ Include 100 and 200） The integer of （ In centimeters ） respectively 10 The height of an apple to the ground , Two adjacent integers are separated by a space . The second line only includes one 100 To 120 Between （ contain 100 and 120） The integer of （ In centimeters ）, It means the maximum height that can be reached when the handle of pottery is extended .

Output

Including a line , This line contains only one integer , The number of apples that pottery can pick .

The sample input

100 200 150 140 129 134 167 198 200 111
110

Sample output

5

source

NOIP2005 The rematch Popularization group The first question is

  We can look at the source , wow ！ First question of popularization group , So difficult , But see the year and “ The rematch ” After two words , I don't think so , The current popularity group is much more difficult than the previous popularity Group , So don't make a fuss .

  The key to this problem is to judge , We can judge whether the height of each apple is less than or equal to the maximum height of mischief plus 30cm, If so , Add one to the value of a variable , repeat 10 Time , Finally, output this variable , Is the most naughty can pick .

Code ：

#include <iostream>
using namespace std;
int main()
{
	int a[10];
	for(int i=0;i<10;i++)
	  cin>>a[i];
	int b,sum=0;
	cin>>b;
	for(int i=0;i<10;i++)
	  if(b+30>=a[i])
	    sum++;
	cout<<sum;
} 

Only a short 14 Line code , Come on ！

I'll see a lot of questions in the future , Don't panic , We should examine the questions carefully , Maybe the problem is long , The topic is simple ？