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 }
