Dowemo

CS Course

Time limit: 4000/2000 ms ( java/others ) memory limit: 32768/32768 k ( java/others )
Total submission ( s ): 0 accepted submission ( s ): 0

Problem Description
Little a has come to college and majored in computer and science.

Today he's learned bit-operations in algorithm lessons, and he got a problem homework.

Here's the problem:

A query only contains a positive integer p, which means you
Are asked to answer the _ result of bit-operations ( and, or, xor ) of all the

Input
There are no more than 15 test cases.

Each test case begins with two positive integers n and p
In a line, indicate the number of and the number of.

2≤n, q≤105.

Output
For each query three non-negative indicates the

Sample Input
3 3
1 1 1
1
2
3

Sample Output
1 1 0
1 1 0
1 1 0


#include <cstdio>


#include <cstring>


#include <cmath>


#include <algorithm>


#include <vector>


#include <string>


#include <map>


#include <set>


#include <cassert>


using namespace std;


#define rep(i,a,n) for (long long i=a;i<n;i++)


#define per(i,a,n) for (long long i=n-1;i>=a;i--)


#define pb push_back


#define mp make_pair


#define all(x) (x).begin(),(x).end()


#define fi first


#define se second


#define SZ(x) ((long long)(x).size())


typedef vector<long long> VI;


typedef long long ll;


typedef pair<long long,long long> PII;


const ll mod=1e9+7;


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;}


//head



long long _,n;


namespace linear_seq


{


 const long long N=10010;


 ll res[N],base[N],_c[N],_md[N];



 vector<long long> Md;


 void mul(ll *a,ll *b,long long k)


 {


 rep(i,0,k+k) _c[i]=0;


 rep(i,0,k) if (a[i]) rep(j,0,k)


 _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;


 for (long long i=k+k-1;i>=k;i--) if (_c[i])


 rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;


 rep(i,0,k) a[i]=_c[i];


 }


 long long solve(ll n,VI a,VI b)


 { //a 系数 b 初值 b[n+1]=a[0]*b[n]+...


//printf("%dn",SZ(b));


 ll ans=0,pnt=0;


 long long k=SZ(a);


 assert(SZ(a)==SZ(b));


 rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;


 Md.clear();


 rep(i,0,k) if (_md[i]!=0) Md.push_back(i);


 rep(i,0,k) res[i]=base[i]=0;


 res[0]=1;


 while ((1ll<<pnt)<=n) pnt++;


 for (long long p=pnt;p>=0;p--)


 {


 mul(res,res,k);


 if ((n>>p)&1)


 {


 for (long long i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;


 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;


 }


 }


 rep(i,0,k) ans=(ans+res[i]*b[i])%mod;


 if (ans<0) ans+=mod;


 return ans;


 }


 VI BM(VI s)


 {


 VI C(1,1),B(1,1);


 long long L=0,m=1,b=1;


 rep(n,0,SZ(s))


 {


 ll d=0;


 rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;


 if (d==0) ++m;


 else if (2*L<=n)


 {


 VI T=C;


 ll c=mod-d*powmod(b,mod-2)%mod;


 while (SZ(C)<SZ(B)+m) C.pb(0);


 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;


 L=n+1-L; B=T; b=d; m=1;


 }


 else


 {


 ll c=mod-d*powmod(b,mod-2)%mod;


 while (SZ(C)<SZ(B)+m) C.pb(0);


 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;


 ++m;


 }


 }


 return C;


 }


 long long gao(VI a,ll n)


 {


 VI c=BM(a);


 c.erase(c.begin());


 rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;


 return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));


 }


};



int main()


{


 while(~scanf("%I64d", &n))


 { printf("%I64dn",linear_seq::gao(VI{1,5,11,36,95,281,781,2245,6336,18061, 51205},n-1));


 }


}









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