cf C. Quiz

http://codeforces.com/contest/337/problem/C

得到的分数为：(2^1+2^2+...+2^X)*k + m-X*k = (2^(X+1)-2)*k + m-X*k；

x的确定：max(0, m - (n - n mod k) / k * (k-1) - n mod k)；

为了得到的分数尽可能少，让满足k次的情况发生在前面。

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <cstring>

 4 #include <algorithm>

 5 #define ll long long

 6 using namespace std;

 7 const ll mod=1e9+9;

 8 

 9 ll n,m,k,ans,x;

10 

11 ll pow(ll a,ll b)

12 {

13     if(b==0) return 1;

14     ll xx=pow(a,b/2);

15     xx=(xx*xx)%mod;

16     if(b&1) xx=(xx*a)%mod;

17     return xx;

18 }

19 

20 ll deal1(int n,int m,int k)

21 {

22     ll x=max(0,m-(n-n%k)/k*(k-1)-n%k);

23     return (mod+m-x*k+(pow(2,x+1)-2)*k)%mod;

24 }

25 int main()

26 {

27     cin>>n>>m>>k;

28 

29     cout<<deal1(n,m,k)<<endl;

30     return 0;

31 }

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