hdu 4602 Partition（矩阵快速幂乘法）

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=, we have
  =+++
  =++
  =++
  =++
  =+
  =+
  =+
  =
totally  ways. Actually, we will have f(n)=(n-) 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 (n-) ways. In the example above, number  occurs for  times, while number  only occurs once.

Input

The first line contains a single integer T(≤T≤), indicating the number of test cases.
Each test case contains two integers n and k(≤n,k≤).

 

Output

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

Sample Input

Sample Output

 

Source

2013 Multi-University Training Contest 1

题目大意：将一个数 n 拆分，问所有的拆分组合中 K 出现了几次。

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]

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 1000000
 #define inf 1e12
 ll n,k;
 struct Matrix{
    ll mp[][];
 };
 Matrix Mul(Matrix a,Matrix b){
    Matrix res;
    for(ll i=;i<;i++){
       for(ll j=;j<;j++){
          res.mp[i][j]=;
          for(ll k=;k<;k++){
             res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%MOD+MOD)%MOD;
          }
       }
    }
    return res;
 }
 Matrix fastm(Matrix a,ll b){
    Matrix res;
    memset(res.mp,,sizeof(res.mp));
    for(ll i=;i<;i++){
       res.mp[i][i]=;
    }
    while(b){
       if(b&){
          res=Mul(res,a);
       }
       a=Mul(a,a);
       b>>=;
    }
    return res;
 }
 int main()
 {
    ll t;
    scanf("%I64d",&t);
    while(t--){
       scanf("%I64d%I64d",&n,&k);
       if(k>n){
          printf("0\n");
          continue;
       }
       ll tmp=n-k+;
       if(tmp==){
          printf("1\n");
          continue;
       }
       if(tmp==){
          printf("2\n");
          continue;
       }
       if(tmp==){
          printf("5\n");
          continue;
       }

       Matrix ttt;
       ttt.mp[][]=;
       ttt.mp[][]=-;
       ttt.mp[][]=;
       ttt.mp[][]=;

       ttt=fastm(ttt,tmp-);

       Matrix cnt;
       cnt.mp[][]=;
       cnt.mp[][]=;
       ttt=Mul(ttt,cnt);
       printf("%I64d\n",ttt.mp[][]);

    }
     return ;
 }