Codeforces 392C Yet Another Number Sequence 题解&代码
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这道题!我推了四个小时!公式!
不是这题真的好打击智商啊…实话,我看着别人十几分钟推完一脸轻松就觉得该AFO
好悲伤啊…数学能力果然差没得救
(i+1)^k=C(k,k)(i^k)+C(k,k-1)(i^(k-1))……+C(k,0)(i^0)
Ai(k)=Fi×(i^k)=F(i-1)(i^k)+F(i-2)(i^k)
设:
u(i,k)=F(i)(i^k)=∑C(k,j)((i-1)^j)F(i-1)+∑C(k,i)((i-1)^j)F(i-2)=∑C(k,j)u(i-1,j)+∑C(k,i)v(i-1,j)
v(i,k)=F(i-1)(i^k)=∑C(k,j)((i-1)^j)F(i-1)=∑C(k,j)u(i-1,j)
Si(n)=Si(n-1)+u(n,k)
目的是求出Si(n),设出一个(2k+3)^2的初始矩阵和单位矩阵(表示u(size:k+1),v(size:k+1)和s(size:1)),然后矩阵快速幂就好啦
#include<iostream>#include<stdio.h>#define MOD 1000000007LL#define LL long longusing namespace std;const int maxn=45;LL s,ans;int k,n,c[maxn][maxn];struct matrix{ LL a[maxn*2][maxn*2]; };matrix base,m;matrix mul(matrix x,matrix y){ matrix ans; for(int i=0;i<=n;i++) for(int j=0;j<=n;j++) { ans.a[i][j]=0; for(int k=0;k<=n;k++) ans.a[i][j]+=((x.a[i][k])*(y.a[k][j]))%MOD; ans.a[i][j]%=MOD; } return ans;}void pow(LL k){ while(k) { if(k&1)m=mul(m,base); base=mul(base,base); k>>=1; }}int main(void){ cin>>s>>k; n=k*2+2; for(int i=0;i<=k;i++) { c[i][0]=c[i][i]=1; for(int j=1;j<i;j++) c[i][j]=(c[i-1][j-1]+c[i-1][j])%MOD; } for(int i=0;i<=k;i++) for(int j=0;j<=i;j++) base.a[i][j]=base.a[i][j+k+1]=base.a[i+k+1][j]=c[i][j]; base.a[n][k]=base.a[n][n]=1; for(int i=0;i<=n;i++)m.a[i][i]=1; pow(s); for(int i=0;i<n;i++)ans+=m.a[n][i]; cout<<ans%MOD<<endl; return 0;}
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