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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>usingnamespacestd;#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())typedefvector<longlong> VI;typedeflonglong ll;typedef pair<longlong,longlong> 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;}//headlonglong _,n;namespace linear_seq
{
 constlonglong N=10010;
 ll res[N],base[N],_c[N],_md[N];
 vector<longlong> Md;
 void mul(ll *a,ll *b,longlong 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 (longlong 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];
 }
 longlong 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;
 longlong 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 (longlong p=pnt;p>=0;p--)
 {
 mul(res,res,k);
 if ((n>>p)&1)
 {
 for (longlong 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);
 longlong 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;
 elseif (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;
 }
 longlong 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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