HUD-4602 Partition 排列

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4602

　　把n等效为排成一列的n个点，然后就是取出其中连续的k个点。分两种情况，一种是不包含两端，2^( n−k−2 ) ∗ (n−k−1) ，另一种是包含两端：2 ∗ 2^( n – k − 1)。然后特殊情况特判一下。。

 1 //STATUS:C++_AC_31MS_248KB

 2 #include <functional>

 3 #include <algorithm>

 4 #include <iostream>

 5 //#include <ext/rope>

 6 #include <fstream>

 7 #include <sstream>

 8 #include <iomanip>

 9 #include <numeric>

10 #include <cstring>

11 #include <cassert>

12 #include <cstdio>

13 #include <string>

14 #include <vector>

15 #include <bitset>

16 #include <queue>

17 #include <stack>

18 #include <cmath>

19 #include <ctime>

20 #include <list>

21 #include <set>

22 #include <map>

23 using namespace std;

24 //using namespace __gnu_cxx;

25 //define

26 #define pii pair<int,int>

27 #define mem(a,b) memset(a,b,sizeof(a))

28 #define lson l,mid,rt<<1

29 #define rson mid+1,r,rt<<1|1

30 #define PI acos(-1.0)

31 //typedef

32 typedef __int64 LL;

33 typedef unsigned __int64 ULL;

34 //const

35 const int N=100010;

36 const LL INF=0x3f3f3f3f;

37 const int MOD=1000000007,STA=8000010;

38 const LL LNF=1LL<<60;

39 const double EPS=1e-8;

40 const double OO=1e15;

41 const int dx[4]={-1,0,1,0};

42 const int dy[4]={0,1,0,-1};

43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

44 //Daily Use ...

45 inline int sign(double x){return (x>EPS)-(x<-EPS);}

46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

49 template<class T> inline T Min(T a,T b){return a<b?a:b;}

50 template<class T> inline T Max(T a,T b){return a>b?a:b;}

51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

55 //End

56 

57 int T,n,m;

58 

59 LL Pow(LL n,int m)

60 {

61     LL ret=1;

62     for(;m;m>>=1){

63         if(m&1)ret=(ret*n)%MOD;

64         n=(n*n)%MOD;

65     }

66     return ret;

67 }

68 

69 int main()

70 {

71  //   freopen("in.txt","r",stdin);

72     int i,j;

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

74     while(T--)

75     {

76         scanf("%d%d",&n,&m);

77         if(m>n)

78             printf("0\n");

79         else if(n==m)

80             printf("1\n");

81         else

82             printf("%I64d\n",(Pow(2,n-m)+(m<n-1?(n-m-1)*Pow(2,n-m-2):0))%MOD);

83     }

84     return 0;

85 }