CoderForces-Round60D(1117) Magic Gems

Reziba has many magic gems. Each magic gem can be split into

Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is

How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is

Input

The input contains a single line consisting of

Output

Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is

Examples

input

Copy

4 2

output

Copy

input

Copy

3 2

output

Copy

In the first example each magic gem can split into

Let

The total configurations you can have is:

Hence, answer is

题解：

考虑 f[n].
对于一个位置,我们可以放 1 空间.
转移方程即为 f[0]=f[1]=f[2]=…f[m−1]=1.
直接转移是 O(n) 的,无法通过,需要矩阵优化.

CoderForces-Round60D(1117) Magic Gems

也可以用杜教BM，求线性递推式；

参考代码：(矩阵快速幂)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define Mod 1000000007
 5 const double PI = acos(-1.0);
 6 const double eps = 1e-6;
 7 const int INF = 0x3f3f3f3f;
 8 const int N = 100 + 5;
 9 struct Matrix {
10   ll n , m;
11   ll grid[N][N];
12   Matrix () { n = m = 0; memset(grid , 0 , sizeof(grid)); }
13 };
14 
15 Matrix mul(Matrix a,Matrix b) 
16 {
17     Matrix c;
18     c.n = a.n;c.m = b.m;
19     for(ll i=1;i<=c.n;++i)
20         for(ll j=1;j<=c.m;++j) 
21         {
22             ll cnt = 0;
23             for(ll k=1;k<=a.m;++k) 
24             {
25                 c.grid[i][j] = (c.grid[i][j] + a.grid[i][k] * b.grid[k][j]);
26                 cnt++;
27                 if(cnt % 8 == 0) c.grid[i][j] %= Mod;
28             }
29           c.grid[i][j] %= Mod;
30         }
31   return c;
32 }
33 Matrix QuickMul(Matrix a ,ll k) 
34 {
35     if(k == 1) return a;
36     Matrix mid = QuickMul(a ,(k >> 1));
37     if(k & 1) return mul(mul(mid , mid),a);
38     else return mul(mid , mid);
39 }
40 ll n , m;
41 int main() 
42 {
43     cin >> n >> m;
44     if(n < m) {return puts("1") , 0;}
45     if(n == m) return puts("2") , 0;
46     Matrix basic; basic.n = m; basic.m = 1;
47     for(ll i=1;i<=m;++i) basic.grid[i][1] = (i == m) ? 2 : 1;//{1,1,1...1,m}T
48     Matrix base; base.n = base.m = m;
49     
50     for(ll i = 1; i <= m - 1; i++) base.grid[i][i + 1] = 1;
51     base.grid[m][1] = base.grid[m][m] = 1;
52     
53     Matrix ans = mul(QuickMul(base , n - m) , basic);
54     cout << ans.grid[m][1] << endl;
55     return 0;
56 }

View Code

杜教BM

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define rep(i,a,n) for (int i=a;i<n;i++)
 4 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 5 #define pb push_back
 6 #define mp make_pair
 7 #define all(x) (x).begin(),(x).end()
 8 #define fi first
 9 #define se second
10 #define SZ(x) ((int)(x).size())
11 typedef vector<int> VI;
12 typedef long long ll;
13 typedef pair<int,int> PII;
14 const ll mod=1000000007;
15 ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1) { if(b&1)res=res*a%mod; a=a*a%mod;  }  return res;  }
16 ll _,n,m,dp[321];
17 namespace linear_seq {
18     const int N=10010;
19     ll res[N],base[N],_c[N],_md[N];
20     vector<ll> Md;
21     void mul(ll *a,ll *b,int k)
22     {
23         rep(i,0,k+k) _c[i]=0;
24         rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]= (_c[i+j]+a[i]*b[j])%mod;
25         for (int i=k+k-1;i>=k;i--)  if (_c[i])
26             rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
27         rep(i,0,k) a[i]=_c[i];
28     }
29     int solve(ll n,VI a,VI b)
30     {
31         ll ans=0,pnt=0;
32         int k=SZ(a);
33         assert(SZ(a)==SZ(b));
34         rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
35         Md.clear();
36         rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
37         rep(i,0,k) res[i]=base[i]=0;
38         res[0]=1;
39         while ((1ll<<pnt)<=n) pnt++;
40         for (int p=pnt;p>=0;p--)
41         {
42             mul(res,res,k);
43             if ((n>>p)&1)
44             {
45                 for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
46                 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
47             }
48         }
49         rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
50         if (ans<0) ans+=mod;
51         return ans;
52     }
53     VI BM(VI s) {
54         VI C(1,1),B(1,1);
55         int L=0,m=1,b=1;
56         rep(n,0,SZ(s)) {
57             ll d=0;
58             rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
59             if (d==0) ++m;
60             else if (2*L<=n) {
61                 VI T=C;
62                 ll c=mod-d*powmod(b,mod-2)%mod;
63                 while (SZ(C)<SZ(B)+m) C.pb(0);
64                 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
65                 L=n+1-L; B=T; b=d; m=1;
66             } else {
67                 ll c=mod-d*powmod(b,mod-2)%mod;
68                 while (SZ(C)<SZ(B)+m) C.pb(0);
69                 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
70                 ++m;
71             }
72         }
73         return C;
74     }
75     int gao(VI a,ll n) {
76         VI c=BM(a);
77         c.erase(c.begin());
78         rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
79         return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
80     }
81 };
82 int main()
83 {
84     scanf("%lld%lld",&n,&m);
85     vector<int> v;
86     for(int i=1;i<m;++i) v.push_back(1);
87     for(ll i=1;i<=m;++i) dp[i]=i+1,v.push_back(dp[i]);
88     for(int i=m+1;i<=10;++i) dp[i]=dp[i-1]+dp[i-m],v.push_back(dp[i]);
89 
90     printf("%lld\n",linear_seq::gao(v,n-1)%mod);
91     return 0;
92 }

View Code

D. Magic Gems