hdu 4602 Partition

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;

}