poj 2000 Gold Coins

/*
Gold Coins
Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 19148		Accepted: 11946

Description
The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On
each of the next two days (the second and third days of service), the knight receives two gold coins. On each of the
next three days (the fourth, fifth, and sixth days of service), the knight receives three gold coins. On each of the
next four days (the seventh, eighth, ninth, and tenth days of service), the knight receives four gold coins. This pattern
of payments will continue indefinitely: after receiving N gold coins on each of N consecutive days, the knight will
receive N+1 gold coins on each of the next N+1 consecutive days, where N is any positive integer.

Your program will determine the total number of gold coins paid to the knight in any given number of days (starting
from Day 1).

Input
The input contains at least one, but no more than 21 lines. Each line of the input file (except the last one) contains
data for one test case of the problem, consisting of exactly one integer (in the range 1..10000), representing the number
of days. The end of the input is signaled by a line containing the number 0.

Output
There is exactly one line of output for each test case. This line contains the number of days from the corresponding
line of input, followed by one blank space and the total number of gold coins paid to the knight in the given number
of days, starting with Day 1.

Sample Input

10
6
7
11
15
16
100
10000
1000
21
22
0

Sample Output

10 30
6 14
7 18
11 35
15 55
16 61
100 945
10000 942820
1000 29820
21 91
22 98

Source
Rocky Mountain 2004*/

#include <cstdio>
using namespace std;

int main()
{
int s = 1, value = 1;
int day = 0;
int money[10005] = {0};
for(int i = 1; i <= 10000; ++i)
{
if(s)
{
money[i] = money[i - 1] + value;
--s;
}
else
{
s = value;
++value;
money[i] = money[i - 1] + value;
}
}
while(true)
{
scanf("%d", &day);
if(!day) break;
printf("%d %d\n" ,day, money[day]);
}
return 0;
}