hdu 4602 递推关系矩阵快速幂模

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2362    Accepted Submission(s): 937

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

 

Source

2013 Multi-University Training Contest 1

 

递推公式：f(n)=4*f(n-1)-4*f(n-2)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

typedef __int64 LL;
const int Mod=;

struct Matrix
{
    LL a[][];
};

Matrix mult_mod(Matrix A,Matrix B)
{
    int i,j,k;
    Matrix C;
    for(i=;i<=;i++)
    {
        for(j=;j<=;j++)
        {
            C.a[i][j]=;
            for(k=;k<=;k++)
            {
                C.a[i][j]=(C.a[i][j]+A.a[i][k]*B.a[k][j]%Mod+Mod)%Mod;
            }
        }
    }
    return C;
}

Matrix Matrix_pow_mod(Matrix A,int n)
{
    struct Matrix ret;
    ret.a[][]=ret.a[][]=;
    ret.a[][]=ret.a[][]=;
    while(n)
    {
        if(n&) ret=mult_mod(ret,A);
        A=mult_mod(A,A);
        n>>=;
    }
    return ret;
}

LL deal(int n)
{
    struct Matrix A;
    A.a[][]=;A.a[][]=-;A.a[][]=;A.a[][]=;
    A=Matrix_pow_mod(A,n);
    return (A.a[][]*%Mod+A.a[][]*%Mod)%Mod;
}

int main()
{
    int t,n,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&k);
        if(k > n) {printf("0\n");continue;}
        n=n-k+;
        if(n==) {printf("1\n");continue;}
        if(n==) {printf("2\n");continue;}
        if(n==) {printf("5\n");continue;}
        printf("%I64d\n",deal(n-));
    }
    return ;
}