## p171ex7: Payment

Given N (0 ≤ N ≤ 1 000 000), you are to count how many times N falls into any of the 6 ranges given below.

 Range A 0 .. 9999 Range B 10000 .. 19999 Range C 20000 .. 29999 Range D 30000 .. 39999 Range E 40000 .. 49999 Range F 50000 .. 1000000

### Input

Input consists of bunch of integers (0 ≤ N ≤ 1000000) which will be terminated by '-1'.
There will always be less than 10000 integers.

### Output

Output the number of occurences of N in the ranges A .. F. Output '0' if N was never found for that particular range.
Do not include -1 into your calculations.

### Sample Input

`1`
`9999`
`10001`
`1000000`
`29999`
`-1`

### Sample Output

`2`
`1`
`1`
`0`
`0`
`1`

Point Value: 3
Time Limit: 2.00s
Memory Limit: 16M

Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3

• (0/0)
What does it mean to output the number of times N is in a range?

Like the number 1, isn't it ONLY in Range A once? And doesn't appear in any other range? Why would the output be 2?

• (3/0)
You are to output exactly 6 integers, no matter how many integers are in the input.

In the sample case, there are two integers in Range A: 1 and 9999.

• (0/1)
I'm confused, what is the program supposed to do?

• (2/0)
You're supposed to make a program that counts the amount of times a number that you input is in a certain range, The first output is the amount of times a number falls into the range of 'A', second output tells you 'B' and so on and so forth.

• (0/2)
he asked what's wrong not what to use

• (0/2)
Oh my god param, you proved me wrong!!! Good job!!

• (0/3)
is that necessary..? and still doesnt work..?!?!

• (0/3)
Jay, try using 'longint' for the input. An 'Integer' can only go up to 32767 or go down upto -32768.