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;
}