Mahmoud and Ehab and yet another xor task
存在的元素的方案数都是一样的, 啊, 我好菜啊。
离线之后用线性基取check存不存在,然后计算答案。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} int n, q, a[N], ans[N]; vector<int> base; vector<PII> qus[N]; int ok(int v) { for(auto& x : base) v = min(v, v ^ x); return v; } int main() { scanf("%d%d", &n, &q); for(int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= q; i++) { int l, x; scanf("%d%d", &l, &x); qus[l].push_back(mk(x, i)); } int way = 1; for(int i = 1; i <= n; i++) { int val = ok(a[i]); if(!val) way = 1LL * way * 2 % mod; else base.push_back(val); for(auto& q : qus[i]) ans[q.se] = ok(q.fi) ? 0 : way; } for(int i = 1; i <= q; i++) printf("%d ", ans[i]); return 0; } /* */