CodeForces 992C Nastya and a Wardrobe (law, fast power)

 

http://codeforces.com/problemset/problem/992/C

 

 

 

 

Meaning of the questions:

Give you two numbers x, k, k k + 1 representatives month, x can be doubled every month, the month after an increase at the beginning of a 50% chance x minus 1, an increase of k + 1 month after the result is divided by the sum of the number of species in each case.

The subject of the request is the result of growth k + 1 month after.

 

We can launch his growth process according to the third sample

　　　　3

　　6　　　　5

12　　11　　10　　9

|　　    |　   　|　　　|

24　   22　　 20　　18

From this we deduce easily draw each case the resulting constitute a arithmetic sequence, with a tolerance of 2

The results are: (2-n-*. 1) * 2 ^K + 1'd

Note that special circumstances sentence 0, 0 when the answer is 0, direct output

Also seeking a power of 2, when rapid power, pay attention to take the remainder

 

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <string>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <sstream>
13 const int INF=0x3f3f3f3f;
14 typedef long long LL;
15 const int mod=1e9+7;
16 //const double PI=acos(-1);
17 #define Bug cout<<"---------------------"<<endl
18 const int maxm=1e6+10;
19 const int maxn=1e5+10;
20 using namespace std;
21 
22 LL POW(LL a,LL b )
23 {
24     LL ans = 1;
25     a = a % mod;
26     while(b) 
27     {
28         if( b % 2 == 1 )
29             ans = ( ans * a ) % mod;
30         a = ( a * a ) % mod;
31         b /= 2;
32     }
33     return ans;
34 }
35 
36 int main()
37 {
38     LL n,k;
39     scanf("%lld %lld",&n,&k);
40     LL ans=0;
41     if(n!=0)
42     {
43         n=n%mod;
44         ans=(((2*n-1)*POW(2,k)+mod)%mod+1+mod)%mod;
45     }
46     printf("%d\n",ans);
47     return 0;
48 }