Hdu4602 Partition Matrix Multiplication

Formula

A [1] = 2
A [n] = 2 * A [n-1] + 2 ^ (n-2)

Here N is n-k in the question

When K> N outputs 0

K = N output 1

N-k is equal to 1, because 2 ^ (n-2) This is 1/2 bad representation

Matrix used for quick power

[2, 2] [A [2] ([0, 2]) ^ (n-k-2) * [1] to get the first element of the column vector is the desired a [n ];

This recursion can directly use the Matrix to quickly power fruit...

Partition

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)

Total submission (s): 349 accepted submission (s): 168

Problem descriptiondefine 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.

Inputthe 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 ).

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

Sample Input

24 25 5

 

Sample output

51

 

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cstdlib>#define MAXN 20#define MOD 1000000007using namespace std;struct Matrix{    int size;    long long element[MAXN][MAXN];    void setSize(int);    Matrix operator* (Matrix);    Matrix power(int);};void Matrix::setSize(int a){    for (int i=0; i<a; i++)        for (int j=0; j<a; j++)            element[i][j]=0;    size = a;}Matrix Matrix::operator* (Matrix param){    Matrix product;    product.setSize(size);    for (int i=0; i<size; i++)        for (int j=0; j<size; j++)            for (int k=0; k<size; k++)                product.element[i][j]=(1LL*element[i][k]*param.element[k][j]+product.element[i][j])%MOD;    return product;}Matrix Matrix::power(int exp){    Matrix res,A;    A=*this;    res.setSize(size);    for(int i=0;i<size;i++)        res.element[i][i]=1;    while(exp)    {        if(exp&1)            res=res*A;        exp>>=1;        A=A*A;    }    return res;}Matrix a,b;int bas=2;int main(){    int cs;    cin>>cs;    a.setSize(2);    a.element[1][1]=a.element[0][0]=a.element[0][1]=2;    a.element[1][0]=0;    while(cs--)    {        int n,k;        cin>>n>>k;        if(k>n) printf("0\n");        else if(k==n) printf("1\n");        else        {            n=n-k;            if(n==1) printf("2\n");            else            {                b=a.power(n-2);                int mm[]={5,1};                long long ans=0;                for(int k=0;k<2;k++)                    ans=(ans+1LL*b.element[0][k]*mm[k])%MOD;                cout<<ans<<endl;            }        }    }    return 0;}