Codeforces Round #196 (Div. 2) / 337C Quiz (贪心&快速幂取模)

C. Quiz
http://codeforces.com/problemset/problem/337/C

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Manao is taking part in a quiz. The quiz consists of n consecutive questions. A correct answer gives one point to the player. The game also has a counter
of consecutive correct answers. When the player answers a question correctly, the number on this counter increases by 1. If the player answers a question incorrectly, the counter is reset, that is, the number on it reduces to 0. If after an answer the counter
reaches the number k, then it is reset, and the player's score is doubled. Note that in this case, first 1 point is added to the player's score, and then
the total score is doubled. At the beginning of the game, both the player's score and the counter of consecutive correct answers are set to zero.

Manao remembers that he has answered exactly m questions correctly. But he does not remember the order in which the questions came. He's trying to figure
out what his minimum score may be. Help him and compute the remainder of the corresponding number after division by 1000000009 (109 + 9).

Input

The single line contains three space-separated integers n, m and k (2 ≤ k ≤ n ≤ 109; 0 ≤ m ≤ n).

Output

Print a single integer — the remainder from division of Manao's minimum possible score in the quiz by 1000000009 (109 + 9).

Sample test(s)

input

5 3 2

output

3

input

5 4 2

output

6

思路：

用贪心的思想做，把连对放前面。

然后就是自己在纸上画画找规律了。

反思：多找几组数据，不要漏了某些情况。

完整代码：

/*30ms,0KB*/

#include<cstdio>

const int mod = 1000000009;

__int64 pow(__int64 a, __int64 b)///a^b % mod
{
	__int64 r = 1, base = a;
	while (b)
	{
		if (b & 1)
			r = r * base % mod;
		base = base * base % mod;
		b >>= 1;
	}
	return r;
}

int main()
{
	__int64 n, m, k, temp, ans;
	scanf("%I64d%I64d%I64d", &n, &m, &k);
	if (n < k * (n - m + 1))
		printf("%I64d", m);
	else
	{
		temp = n / k - n + m;
		ans = ((k * (pow(2, temp + 1) - 2 - temp)) % mod  + m + mod) % mod;
		printf("%I64d", ans);
	}
	return 0;
}