http://acm.hdu.edu.cn/showproblem.php?pid=4602
输入 n 和 k
首先 f(n)中k的个数 等于 f(n-1) 中 k-1的个数
最终等于 f(n-k+1) 中 1 的个数
舍 s(n) = f(n) + f(n-1) + ....+ f(1)
则 f(n) = s(n) - s(n-1)
由于 s(n) = s(n-1) + 2^(n-2) + s(n-1) = 2*(s(n-1)) + 2^(n-2)
= 2^(n-1) + (n-1)*2^(n-2)
= (n+1)*2^(n-2)
代码：
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
#include<stack>
#include<vector>
#include<algorithm>
#include<queue>
#include<stdexcept>
#include<bitset>
#include<cassert>
#include<deque>
#include<numeric>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
const double eps=1e-12;
const int INF=0x3f3f3f3f;
const ll MOD=1000000007;
ll power(ll x,ll y)
{
ll tmp=1;
while(y)
{
if((y&1))
tmp=(tmp*x)%MOD;
x=(x*x)%MOD;
y=y>>1;
}
return tmp;
}
int main()
{
//freopen("data.in","r",stdin);
int T;
cin>>T;
while(T--)
{
ll n,m;
cin>>n>>m;
ll k=n-m+1;
if(k<=0)
{
cout<<"0"<<endl;
continue;
}
if(k==1)
{
cout<<"1"<<endl;
continue;
}
if(k==2)
{
cout<<"2"<<endl;
continue;
}
ll w1=(k+1)*power(2,k-2)%MOD;
--k;
ll w2=(k+1)*power(2,k-2)%MOD;
cout<<(w1-w2+MOD)%MOD<<endl;
}
return 0;
}
