标签:include double cond sign add oid nbsp 博客 begin
考虑枚举 B[ i ], 分成50000 / B[ i ] 段每段用bitset处理, 手写bitset的话复杂度n ^ 2 / 32。
训练的时候我没有手写bitset, 靠卡常加开o3优化卡过去了。
其实有一种方法能用手写bitset做到n ^ 2 / 32 + n * log(n), 感觉很巧妙呀。
博客在这里https://blog.csdn.net/luricheng/article/details/77006185
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #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 = 5e4 + 1; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; 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;} mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int n, m, q; int T, A[N], B[N], C[N]; int ret[N]; bitset<N> bit, tmp; bitset<N> ans; int main() { scanf("%d", &T); while(T--) { memset(C, 0, sizeof(C)); memset(ret, 0, sizeof(ret)); bit.reset(); ans.reset(); int maxVal = 1; scanf("%d%d%d", &n, &m, &q); for(int i = 1; i <= n; i++) { scanf("%d", &A[i]); C[A[i]] ^= 1; chkmax(maxVal, A[i]); } for(int i = 1; i <= maxVal; i++) bit[i] = C[i]; for(int i = 1; i <= m; i++) { scanf("%d", &B[i]); } sort(B + 1, B + 1 + m); for(int i = 1; i <= m; i++) { if(B[i] <= 1200) { for(int j = 0; j < B[i]; j++) { int now = j, c = 0; while(now <= maxVal) { c ^= C[now]; now += B[i]; } if(c) { if(ans[j]) ans[j] = 0; else ans[j] = 1; } } } else { for(int L = 0, R = B[i] - 1; L <= maxVal; L += B[i], R += B[i]) { R = min(maxVal, R); tmp = (bit << (N - 1 - R)) >> (N - (R - L + 1)); ans ^= tmp; } } } while(q--) { int k; scanf("%d", &k); printf("%d\n", (int)ans[k]); } } return 0; } /* */
HDU 6085 Rikka with Candies bitset
标签:include double cond sign add oid nbsp 博客 begin
原文地址:https://www.cnblogs.com/CJLHY/p/11107838.html