It is a very water topic.
First, let's look at the topic :( probably describe) Tata family has an apple tree, autumn will bear 10 apples (only bear ten, so little thing),
These apples are highly, and Tao Tao also has a 30 cm bench. Let us ask this question Tao Tao arm length plus the bench thirty centimeters, can
How many apples picked to be.
Ideas: a definition of a first array, the array is preferably larger, then the error array bounds prevented, then the definition of a O, represents the length of the arm Toto, defined sum,
Remember to be initialized to 0, because for a while to add up, so the beginning of the value must be 0. Then enter a loop array 10 are stored apples height, and
Enter o, i.e. the length of the arm Toto. Then the cycle again, and then determine: If the length of the arm plus the bench tata 30 cm greater than the height of apple,
So that Apple can be picked with a sum add up, and finally the output end.
· Code:
#include <the iostream>
#include <cstdio>
the using namespace STD;
int main () {
int n-[110], sum = 0, O; // defines three variables, n array storing the apple height, sum stored
// total how many apples can reach them, o storage Tao Tao handle straight time to reach the maximum height.
for (int i = 0; i <10; i ++) {// cycle
scanf ( "% d", & n [i]); // input apple height
}
Scanf ( "% D", & O); // Input Toto straight handle when the maximum height can be achieved
for (int i = 0; i <10; i ++) {// cycle
if (n [i] <= o + 30) {// If the height is less than the number of apples Toto hand stretched to reach the maximum height // add height stepping on the bench, then Apple will be able to be picked.
sum ++; // use the sum added up}
}
the printf ( "% D", sum); // number of outputs can be picked apples in
return 0; // End
}
I think this method should be regarded as a relatively simple.