比赛总结：
看错F题意@byf
要勇于打表找规律
题解(不定更新)
A EddyWalker
B EddyWalker2
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 long long LL; 14 typedef pair<int,int> PII; 15 const ll mod=1000000007; 16 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;} 17 // head 18 19 ll n; 20 namespace linear_seq { 21 const int N=10010; 22 ll res[N],base[N],_c[N],_md[N]; 23 24 vector<int> Md; 25 void mul(ll *a,ll *b,int k) { 26 rep(i,0,k+k) _c[i]=0; 27 rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod; 28 for (int i=k+k-1;i>=k;i--) if (_c[i]) 29 rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod; 30 rep(i,0,k) a[i]=_c[i]; 31 } 32 int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+... 33 ll ans=0,pnt=0; 34 int k=SZ(a); 35 assert(SZ(a)==SZ(b)); 36 rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1; 37 Md.clear(); 38 rep(i,0,k) if (_md[i]!=0) Md.push_back(i); 39 rep(i,0,k) res[i]=base[i]=0; 40 res[0]=1; 41 while ((1ll<<pnt)<=n) pnt++; 42 for (int p=pnt;p>=0;p--) { 43 mul(res,res,k); 44 if ((n>>p)&1) { 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 83 int main() { 84 vector<int>v; 85 int t; 86 scanf("%d",&t); 87 int k;LL n; 88 while(t--) 89 { 90 scanf("%d%lld",&k,&n); 91 if(n==-1) 92 { 93 printf("%lld\n",2LL*powmod(k+1,mod-2)%mod); 94 continue; 95 } 96 v.clear(); 97 ll ook=powmod(k,mod-2); 98 v.push_back(1); 99 for(int i=1;i<=2*k+1;i++) 100 { 101 ll pa=0; 102 for(int j=max(0,i-k);j<i;j++) pa=(pa+1ll*v[j]*ook)%mod; 103 v.push_back(int(pa%mod)); 104 } 105 printf("%lld\n",1LL * linear_seq::gao(v,n) % mod); 106 } 107 }