Codeforces Round #196 (Div. 2) C. Quiz

C. Quiz

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

Note

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：Manao 在进行测试，一共要回答n个题目，答对一题得一分，如果连续答对k到题，那么他得到的分数加倍。答错一题或者是加倍后，连续答对题目的个数清零重新开     始计数。现在给定n和k，还有m（Manao一共答对的题目数），求最少Manao可以得多分。

题目分析：首先讨论能否加倍。

                    假设，全部能加倍，那么一共会加倍(n/k)次，那么如果全部不能加倍，需要答错(n/k)到题目

                    所以可以根据（n-m）和（n/k）的大小，来判断Manao的分数能否加倍。

                    1）n/k < n-m（不能加倍）那么此时，Manao最少只能得分m

                    2）n/k >= n-m （能加倍）如果能加倍，n一共会被分为三个部分（这种分发保证得分最少）

                          第一个部分 连续答对X到题目（有加倍）如果有加倍，当然趁得分少的时候加倍了

                          第二个部分 没有加倍，且没连续k道题目，那么第k到题目答错这是最优的一种排法

                          第三个部分 最后连续答对题目，但是没有加倍    答错的题目排在前面，也是最优的，因为答错的题目要破坏加倍 

                    将n分为每份大小为k的区域（n/k），那么还剩下（n%k）（这n%k不会加倍）

                    那么如果有加倍，只能在这（n/k）个区域的最前面开始加倍，然后后面的每个区域的最后一个总是答错（这里有(n-m)个答错的题，所以这些区域得分为（n-m）*（k-1）），那么连续加倍（（m-（n-m）*（k-1））/ k）次。

下面是代码：

#include <stdio.h>#define mod 1000000009int pow_mod(long long a,long long n,long long m){    long long ret = 1;    while(n)    {        if(n&1) ret = ret*a%m;        a = (a*a)%m;        n >>= 1;    }    return (int)ret%m;}int main(){    int n,m,k;    scanf("%d%d%d",&n,&m,&k);    int s1 = n/k;    int s2 = n-m;    if(s1-s2 < 0) printf("%d\n",m);    else    {        int exp = s1 - s2;        long long ans = 0;        ans = pow_mod(2,exp+1,mod) - 2;        ans = (ans*k + mod)%mod;        ans = (ans + (long long) s2 * (k-1) + n%k) % mod;        printf("%I64d\n",ans);    }    return 0;}