HDU 4602 Partition （矩阵乘法）

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 797    Accepted Submission(s): 322

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

 

Recommend

liuyiding

 

思路：

列出了 n=5 时 5,4,3,2,1 出现的次数为 1 2 5 12 28
f[n+1]=3*f[n]-f[n-1]-f[n-2]-..f[1]
f[n]=3*f[n-1]-f[n-2]-..f[1]
==> f[n+1]=4*f[n]-4*f[n-1]

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int mod=;

struct Matrix{
    long long arr[][];
};

Matrix init,unit;
int n,k;
long long num[][]={{,-},{,}};

void Init(){
    for(int i=;i<;i++)
        for(int j=;j<;j++){
            init.arr[i][j]=num[i][j];
            unit.arr[i][j]=(i==j)?:;
        }
}

Matrix Mul(Matrix a,Matrix b){
    Matrix c;
    for(int i=;i<;i++)
        for(int j=;j<;j++){
            c.arr[i][j]=;
            for(int k=;k<;k++)
                c.arr[i][j]=(c.arr[i][j]%mod+a.arr[i][k]*b.arr[k][j]%mod+mod)%mod;
            c.arr[i][j]%=mod;
        }
    return c;
}

Matrix Pow(Matrix a,Matrix b,int k){
    while(k){
        if(k&){
            b=Mul(a,b);
        }
        a=Mul(a,a);
        k>>=;
    }
    return b;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&k);
        Init();
        if(k>n){
            printf("0\n");
            continue;
        }
        int tmp=n-k+;
        if(tmp==){
            printf("1\n");
            continue;
        }
        if(tmp==){
            printf("2\n");
            continue;
        }
        if(tmp==){
            printf("5\n");
            continue;
        }
        Matrix res=Pow(init,unit,tmp-);
        //long long ans=((res.arr[0][0]%mod*5)%mod+(res.arr[0][1]%mod*2)%mod)%mod;
        long long ans=(res.arr[][]*+res.arr[][]*)%mod;
        cout<<ans<<endl;
    }
    return ;
}