CodeForces 337C Quiz(1等比数列找规律）

Quiz

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Appoint description:  System Crawler  (Oct 22, 2016 3:24:05 PM)jehad  (Aug 17, 2013 11:29:52 PM)

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

Input

5 3 2

Output

3

Input

5 4 2

Output

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.

题目大意：

给出要做题总和，以及答对的题数，以及连续答对个数要求，求最少得分；

游戏规则

每答对一题加一分，若连续答对k题后总分翻一倍；

解题思路：

将n道题按每堆k题分成x堆；

然后计算答错题数目；

若答错题数目大于等于x，则说明可以使其都不连续k个答对,最后总分为m;

否则连续答对k题的个数为m-n+n/k;

若使得分最少，则连续情况放在前面，符合等比数列形式

代码：

#include 
#include  
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;  
}