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 has learned bit-operations in Algorithm Lessons, and he got a problem as homework.
Here is the problem:
You are giving n non-negative integers a1,a2,⋯,an, and some queries.
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 integers except ap.
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 positive integers and the number of queries.
2≤n,q≤105
Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤109 for each i in range[1,n].
After that there are q positive integers p1,p2,⋯,pqin q lines, 1≤pi≤n for each i in range[1,q].
Output For each query p, output three non-negative integers indicates the result of bit-operations(and, or, xor) of all non-negative integers except ap in a line.
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("%d\n",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("%I64d\n",linear_seq::gao(VI{1,5,11,36,95,281,781,2245,6336,18061, 51205},n-1)); } }