传送门
考虑从左往右每次填k个
设E[i]表示如果宝藏在i的期望步数
有E[i+k]=E[i]+1(i<=n−k)
E[i+k−n]=p+(1−p)(E[i]+1)(otherwise)
这样可以手动消元O(n)做
实际上要求的是∑E[1...n]
考虑可以把%k相同的归到一类
然后就发现只需要求E[1...k]就可以了
每个E[i]有个系数
相当于是x−>ax+b这样的函数
发现前n%k个和剩下的系数不一样
分别把2种函数设成f1,f2
考虑每次的f1,f2的变化和E的两种关系式有关
设为A,B,即E[i+k]=A(E[i]),E[i+k−n]=B(E[i])
发现每次f1,f2变化为
f1<−f1+∑i=1n/kf2(Ai)
f2<−f1+∑i=1n/k−1f2(Ai)
于是考虑维护A,B
以A为例,考虑A的意义,
即需要找到A′使E[i+n%k]=A′(E[i])
把下标变换算一下可以得到A′=A−n/k+1(B−1)
B′类似算一下就可以了
题解写的很口胡请见谅
#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
static char ibuf[RLEN],*ib,*ob;
(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
char ch=gc();
int res=0,f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
struct node{
int x,y;
node(int _x=0,int _y=0):x(_x),y(_y){}
friend inline node operator +(cs node &a,cs node &b){
return node(add(a.x,b.x),add(a.y,b.y));
}
friend inline node operator +(cs node &a,cs int &b){
return node(a.x,add(a.y,b));
}
friend inline node operator -(cs node &a,cs node &b){
return node(dec(a.x,b.x),dec(a.y,b.y));
}
friend inline node operator -(cs node &a,cs int &b){
return node(a.x,dec(a.y,b));
}
friend inline node operator *(cs node &a,cs node &b){
return node(mul(a.x,b.x),add(mul(a.y,b.x),b.y));
}
int calc(int a){
return add(mul(a,x),y);
}
};
inline node Inv(node x){
int t=Inv(x.x);
x.x=t,x.y=dec(0,mul(x.y,t));
return x;
}
inline node ksm(node a,int b){
node res(1,0);
for(;b;b>>=1,a=a*a)if(b&1)res=res*a;
return res;
}
inline int gcd(int a,int b){
return b?gcd(b,a%b):a;
}
inline node S(node a,int b){
if(!b)return node(0,0);
if(a.x==1)return node(b,1ll*b*(b+1)/2%mod*a.y%mod);
node x=ksm(a,b+1)-a,p;
p.x=mul(x.x,Inv(a.x-1));
p.y=mul(mul(dec(p.x,b),Inv(dec(a.x,1))),a.y);
return p;
}
inline int solve(int n,int k,node A,node B,node f1,node f2){
if(k==0){return f2.calc(mul(dec(0,A.y),Inv(dec(A.x,1))));}
int k1=n%k,t=n/k;
node F1=f1+S(A,t)*f2;
node F2=f1+S(A,t-1)*f2;
node a=Inv(B)*ksm(Inv(A),t-1),b=Inv(B)*ksm(Inv(A),t);
return add(solve(k,k1,a,b,node(F1.x,0),node(F2.x,0)),add(mul(k1,F1.y),mul(k-k1,F2.y)));
}
int n,k,p;
node f1,f2,A,B;
int main(){
int T=read();
while(T--){
n=read(),k=read(),p=Inv(2);
int g=gcd(n,k);n/=g,k/=g;
f1=node(1,0),f2=node(1,0),A=node(1,1),B=node(dec(1,p),1);
cout<<mul(solve(n,k,A,B,f1,f2),Inv(n))<<'
';
}
}