hdu-4602-Partition(矩阵快速幂)

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=4602

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3371    Accepted Submission(s): 1301


Problem Description

Define f(n) as the number of ways to perform n in format of the sum of some positive integers. For instance, when n=4, we have

  4=1+1+1+1

  4=1+1+2

  4=1+2+1

  4=2+1+1

  4=1+3

  4=2+2

  4=3+1

  4=4

totally 8 ways. Actually, we will have f(n)=2(n-1) after observations.

Given a pair of integers n and k, your task is to figure out how many times that the integer k occurs in such 2(n-1) ways. In the example above, number 1 occurs for 12 times, while number 4 only occurs once.

 

Input

The first line contains a single integer T(1≤T≤10000), indicating the number of test cases.

Each test case contains two integers n and k(1≤n,k≤109).

 

Output

Output the required answer modulo 109+7 for each test case, one per line.

 

Sample Input

2
4 2
5 5

 

Sample Output

5
1

题解：

最近在做矩阵快速幂的题，这个题也推出了序列是1,2,5,12,28,64.....但可惜的是没有推出它的递推公式，看网上的解法有用公式的有找规律的，在这里就说一下矩阵快速幂吧。

先给出递推公式f(N)=2*f(N-1)+2^(N-3);这里的N是根据题目中的n和k得出来的。n和序列的对应是：a
=1,a[n-1]=2,a[n-3]=5,a[n-4]=12,a[n-5]=28  ......。那么在n中的k出现的次数是多少呢？那就要看看k是n减几了。所以这里的N=n-k;

然后就是构造矩阵了：

【f(n),2^(N-2)】=【f(n-1),2^(N-3)】*【(2,1) (0,2)】

这里求N的时候到了N-3的值，所以N要从3开始。接下来就是利用快速幂求就行了。关于矩阵快速幂不会自行解决吧。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
typedef long long LL;
const LL MOD=1e9+7;
struct Matrix
{
LL a[2][2];//矩阵大小根据需求修改
Matrix()
{
memset(a,0,sizeof(a));
}
void init()
{
for(int i=0; i<2; i++)
for(int j=0; j<2; j++)
a[i][j]=(i==j);
}
Matrix operator + (const Matrix &B)const
{
Matrix C;
for(int i=0; i<2; i++)
for(int j=0; j<2; j++)
C.a[i][j]=(a[i][j]+B.a[i][j])%MOD;
return C;
}
Matrix operator * (const Matrix &B)const
{
Matrix C;
for(int i=0; i<2; i++)
for(int k=0; k<2; k++)
for(int j=0; j<2; j++)
C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%MOD;
return C;
}
Matrix operator ^ (const LL &t)const
{
Matrix A=(*this),res;
res.init(); ///矩阵的单位矩阵初始化
LL p=t;
while(p)
{
if(p&1)res=res*A;
A=A*A;
p>>=1;
}
return res;
}
}base,ans;
LL q_pow(LL a,LL b)
{
LL res=1;
while(b)
{
if(b&1)res=res*a%MOD;
a=a*a%MOD;
b=b>>1;
}
return res;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
LL n,k;
scanf("%lld%lld",&n,&k);
if(k>n||k<=0)
{
printf("0\n");
continue;
}
if(n-k+1==1)
{
printf("1\n");
continue;
}
if(n-k+1==2)
{
printf("2\n");
continue;
}
if(n-k+1==3)
{
printf("5\n");
continue;
}
n=n-k;
base.a[0][0]=2;base.a[0][1]=1;
base.a[1][0]=0;base.a[1][1]=2;
ans=base^(n-1);
cout<<(2*ans.a[0][0]+ans.a[0][1])%MOD<<endl;
}
return 0;
}