Description
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 Input
5 3 2
3
5 4 2
6
Hint
Sample 1. Manao answered 3 questions out of 5, and his score would double for each two consecutive correct answers. If Manao had answered the first, third and fifth questions, he would have scored as much as 3 points.
Sample 2. Now Manao answered 4 questions. The minimum possible score is obtained when the only wrong answer is to the question 4.
Also note that you are asked to minimize the score and not the remainder of the score modulo 1000000009. For example, if Manao could obtain either 2000000000 or 2000000020 points, the answer is 2000000000 mod 1000000009, even though2000000020 mod 1000000009 is a smaller number.
題目大意：
給出要做題總和，以及答對的題數，以及連續答對個數要求，求最少得分；
遊戲規則
每答對一題加一分，若連續答對k題後總分翻一倍；
解題思路：
將n道題按每堆k題分成x堆；
然後計算答錯題數目；
若答錯題數目大於等於x，則說明可以使其都不連續k個答對,最後總分爲m;
否則連續答對k題的個數爲m-n+n/k;
若使得分最少，則連續情況放在前面，符合等比數列形式
代碼：
#include<stdio.h>
#include<math.h>
using namespace std;
long long Pow(long long int a,int b)
{
long long sum=1;
while(b)
{
if(b&1)
sum=a*sum%1000000009;
b/=2;
a=a*a%1000000009;
}
return sum;
}
int main()
{
long long n,m,k;
long long x,y,z,num,sum;
while(~scanf("%lld%lld%lld",&n,&m,&k))
{
sum=0;
x=n-n/k;
num=m-x;
if(num<=0)
{
printf("%lld\n",m%1000000009);
continue;
}
sum=(Pow(2,num)*k-k)*2%1000000009;
num=m-k*num%1000000009;
printf("%lld\n",(num+sum)%1000000009);
}
return 0;
}