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。。

//STATUS:C++_AC_31MS_248KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=100010;
const LL INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int T,n,m;

LL Pow(LL n,int m)
{
    LL ret=1;
    for(;m;m>>=1){
        if(m&1)ret=(ret*n)%MOD;
        n=(n*n)%MOD;
    }
    return ret;
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        if(m>n)
            printf("0
");
        else if(n==m)
            printf("1
");
        else
            printf("%I64d
",(Pow(2,n-m)+(m<n-1?(n-m-1)*Pow(2,n-m-2):0))%MOD);
    }
    return 0;
}