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
");
35             continue;
36         }
37         if( k==n ){
38             printf("1
");
39             continue;
40         }
41         int64 delta = k-1;
42         int64 Index = n-delta;
43         
44         if( Index==1 ){
45             printf("1
");
46             continue;
47         }
48         if( Index==2 ){
49             printf("2
");
50             continue;
51         }
52         
53         //printf("%lld
",(FastPow2(2,Index-1)%mod+FastPow2(2,Index-3)*(Index-2)%mod)%mod);
54         printf("%I64d
",(FastPow(2,Index-1)%mod+FastPow(2,Index-3)*(Index-2)%mod)%mod);
55     }
56     return 0;
57 }
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