HDU4602+推导公式

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 1 /*

 2 推导公式

 3 a[n] = 2^(n-1) + (n-2)*2^(n-3)

 4 */

 5 #include<stdio.h>

 6 #include<math.h>

 7 #include<algorithm>

 8 using namespace std;

 9 typedef long long int64;

10 const int64 mod = 1e9+7;

11 

12 int64 FastPow( int64 n,int64 m ){//n^m

13     int64 sum = 1;

14     while( m>=1 ){

15         if( m%2==1 ){

16             sum *= n;

17             sum %= mod;

18         }

19         n *= n;

20         n %= mod;

21         m/=2;

22     }

23     return sum;

24 }

25 

26 int main(){

27     int T;

28     scanf("%d",&T);

29     while( T-- ){

30         int64 n,k;

31         //scanf("%lld%lld",&n,&k);

32         scanf("%I64d%I64d",&n,&k);

33         if( k>n ){

34             printf("0\n");

35             continue;

36         }

37         if( k==n ){

38             printf("1\n");

39             continue;

40         }

41         int64 delta = k-1;

42         int64 Index = n-delta;

43         

44         if( Index==1 ){

45             printf("1\n");

46             continue;

47         }

48         if( Index==2 ){

49             printf("2\n");

50             continue;

51         }

52         

53         //printf("%lld\n",(FastPow2(2,Index-1)%mod+FastPow2(2,Index-3)*(Index-2)%mod)%mod);

54         printf("%I64d\n",(FastPow(2,Index-1)%mod+FastPow(2,Index-3)*(Index-2)%mod)%mod);

55     }

56     return 0;

57 }

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