异或序列
题目背景:
分析:莫队
莫队裸题······直接求前缀和,那么相当于一段区间里面sum[x - 1] ^ sum[y] == k的对数,直接莫队,记录一个桶,加入一个新的sum,对答案的贡献是cnt[sum ^ k],减去一个sum,同理减去cnt[sum ^ k]没什么细节,也没什么注意事项,直接写就行了·····
Source:
/* created by scarlyw */ #include <cstdio> #include <string> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <cctype> #include <vector> #include <set> #include <queue> #include <ctime> #include <bitset> inline char read() { static const int IN_LEN = 1024 * 1024; static char buf[IN_LEN], *s, *t; if (s == t) { t = (s = buf) + fread(buf, 1, IN_LEN, stdin); if (s == t) return -1; } return *s++; } ///* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = read(), iosig = false; !isdigit(c); c = read()) { if (c == -1) return ; if (c == '-') iosig = true; } for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int OUT_LEN = 1024 * 1024; char obuf[OUT_LEN], *oh = obuf; inline void write_char(char c) { if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf; *oh++ = c; } template<class T> inline void W(T x) { static int buf[30], cnt; if (x == 0) write_char('0'); else { if (x < 0) write_char('-'), x = -x; for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48; while (cnt) write_char(buf[cnt--]); } } inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); } /* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int MAXN = 150000 + 1; int n, m, k, l, r, bs; int a[MAXN], cnt[MAXN]; long long final_ans[MAXN]; struct query { int l, r, bl, id; query() {} query(int l, int r, int id) : l(l), r(r), bl((l - 1) / bs + 1), id(id) {} inline bool operator < (const query &a) const { return (bl == a.bl) ? (r < a.r) : (bl < a.bl); } } q[MAXN]; inline void read_in() { R(n), R(m), R(k); for (int i = 1; i <= n; ++i) R(a[i]), a[i] ^= a[i - 1]; bs = sqrt(n); for (int i = 1; i <= m; ++i) R(l), R(r), q[i] = query(l, r, i); std::sort(q + 1, q + m + 1); } long long ans; inline void compose_l(int cur) { cur--, cnt[a[cur]]--, ans -= (long long)cnt[a[cur] ^ k]; } inline void compose_r(int cur) { cnt[a[cur]]--, ans -= (long long)cnt[a[cur] ^ k]; } inline void extend_l(int cur) { cur--, ans += (long long)cnt[a[cur] ^ k], cnt[a[cur]]++; } inline void extend_r(int cur) { ans += (long long)cnt[a[cur] ^ k], cnt[a[cur]]++; } inline void solve() { int l = 1, r = 1; cnt[a[1]]++, cnt[a[0]]++, ans = (a[1] == k); for (int i = 1; i <= m; ++i) { while (l < q[i].l) compose_l(l), l++; while (r > q[i].r) compose_r(r), r--; while (l > q[i].l) --l, extend_l(l); while (r < q[i].r) ++r, extend_r(r); final_ans[q[i].id] = ans; } for (int i = 1; i <= m; ++i) W(final_ans[i]), write_char('\n'); } int main() { freopen("xor.in", "r", stdin); freopen("xor.out", "w", stdout); read_in(); solve(); flush(); return 0; }