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
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]
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 #define PI acos(-1.0) 17 #define max(a,b) (a) > (b) ? (a) : (b) 18 #define min(a,b) (a) < (b) ? (a) : (b) 19 #define ll long long 20 #define eps 1e-10 21 #define MOD 1000000007 22 #define N 1000000 23 #define inf 1e12 24 ll n,k; 25 struct Matrix{ 26 ll mp[3][3]; 27 }; 28 Matrix Mul(Matrix a,Matrix b){ 29 Matrix res; 30 for(ll i=0;i<2;i++){ 31 for(ll j=0;j<2;j++){ 32 res.mp[i][j]=0; 33 for(ll k=0;k<2;k++){ 34 res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%MOD+MOD)%MOD; 35 } 36 } 37 } 38 return res; 39 } 40 Matrix fastm(Matrix a,ll b){ 41 Matrix res; 42 memset(res.mp,0,sizeof(res.mp)); 43 for(ll i=0;i<2;i++){ 44 res.mp[i][i]=1; 45 } 46 while(b){ 47 if(b&1){ 48 res=Mul(res,a); 49 } 50 a=Mul(a,a); 51 b>>=1; 52 } 53 return res; 54 } 55 int main() 56 { 57 ll t; 58 scanf("%I64d",&t); 59 while(t--){ 60 scanf("%I64d%I64d",&n,&k); 61 if(k>n){ 62 printf("0 "); 63 continue; 64 } 65 ll tmp=n-k+1; 66 if(tmp==1){ 67 printf("1 "); 68 continue; 69 } 70 if(tmp==2){ 71 printf("2 "); 72 continue; 73 } 74 if(tmp==3){ 75 printf("5 "); 76 continue; 77 } 78 79 Matrix ttt; 80 ttt.mp[0][0]=4; 81 ttt.mp[0][1]=-4; 82 ttt.mp[1][0]=1; 83 ttt.mp[1][1]=0; 84 85 ttt=fastm(ttt,tmp-3); 86 87 Matrix cnt; 88 cnt.mp[0][0]=5; 89 cnt.mp[1][0]=2; 90 ttt=Mul(ttt,cnt); 91 printf("%I64d ",ttt.mp[0][0]); 92 93 } 94 return 0; 95 }