Codeforces 337C Quiz【贪心+快速幂】

最新推荐文章于 2020-11-24 18:41:50 发布

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最新推荐文章于 2020-11-24 18:41:50 发布

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分类专栏：贪心文章标签： Codeforces 337C

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本文探讨了一个有趣的Quiz游戏得分策略问题，玩家需要在有限的正确答案数量下，通过合理的答题顺序安排来最小化最终得分。文章提供了详细的解题思路及AC代码实现。

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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 by1000000009 (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).

Examples

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：

一共有N个题，已知我们能够答对m道题，如果连续答对了k道题之后，已得分数会翻倍，问我们最少能够获得多少分。

思路：

我们假设现在m道题就是连着答对的，那么很显然，越到后期翻倍得到的收益越大。

那么我们将打错的题目插入后边优先能够使得分数尽可能的小。

不难想到，我们在后边每隔k-1道答对的题之间都插入一个打错的题就能够最优。

那么结果能够推出公式：

①插入完之后剩余的正确题数个数：

yu=m-(k-1)*(n-m);

②正确题目个数能够带来多少次翻倍的收益：

zu=yu/k;

③一分一分累计起来的收益：

m-zu*k

④翻倍的收益：

((k*2+k)*2+k)*2................=(k+k*2^2+k*2^3+k*2^4................)

那么这里快速幂一下即可。

注意有减法，取摸注意一下。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
const ll mod=1000000009;
ll kuaisumi(ll a,ll b)
{
    a%=mod;
    ll ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=ans*a;
            ans%=mod;
        }
        a=a*a;
        a%=mod;
        b/=2;
    }
    return ans;
}

int main()
{
    ll n,m,k;
    while(~scanf("%lld%lld%lld",&n,&m,&k))
    {
        ll yu=m-(k-1)*(n-m);
        if(yu<=0)
        {
            printf("%I64d\n",m);
        }
        else
        {
            ll zu=yu/k;
            ll output=((m-zu*k)%mod+mod)%mod;
            output%=mod;
            ll tmp=((kuaisumi(2,zu+1)-2)%mod+mod)%mod;
            tmp=((tmp*k)%mod+mod)%mod;
            output=((output+tmp)%mod+mod)%mod;
            printf("%lld\n",(output+mod)%mod);
        }
    }
}