BZOJ 2527
题目大意:自己看..
解题报告
JeremyGuo欺骗我…她说这道题复杂度可以有[latex]O(N\times log_2(N))[/latex]想了好久发现是假的..
首先这题当然可以用可持久化线段树之类来做, 但是CDQ+树状数组明显更好写。
代码
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cstdlib> #include <vector> using namespace std; typedef long long LL; const int MAXN = 300030; inline int getInt () { int ret = 0; char ch; bool flg = false; while((ch = getchar()) < '0' || ch > '9') flg |= (ch == '-'); ret = ch-'0'; while((ch = getchar()) >= '0' && ch <= '9') ret = ret*10+(ch-'0'); return flg ? -ret : ret; } int N, M, K; int O[MAXN], P[MAXN]; int oper[MAXN][3]; LL C[MAXN]; inline int lowbit (int a) {return a&-a;} inline void add (int p, int v) { while(p <= M) C[p] += v, p += lowbit(p); } inline LL query (int p) { LL ret = 0; while(p) ret += C[p], p -= lowbit(p); return ret; } int ans[MAXN]; vector<int> pos[MAXN]; inline void runOper (int i, int t) { if(oper[i][0] <= oper[i][1]) add(oper[i][0], t*oper[i][2]), add(oper[i][1]+1, -t*oper[i][2]); else { add(1, t*oper[i][2]), add(oper[i][1]+1, -t*oper[i][2]); add(oper[i][0], t*oper[i][2]); } } void solve (vector<int> & vec, int l, int r) { if(l == r) { runOper(l, 1); for(int i = 0; i<(int)vec.size(); i++) ans[vec[i]] = l; return ; } int mid = (l+r)>>1; for(int i = l; i<=mid; i++) runOper(i, 1); vector<int> lef, rig; for(int p = 0; p<(int)vec.size(); p++) { LL sum = 0; for(int i = 0; i<(int)pos[vec[p]].size(); i++) { sum += query(pos[vec[p]][i]); if(sum >= P[vec[p]]) break; } if(sum >= P[vec[p]]) lef.push_back(vec[p]); else rig.push_back(vec[p]); } for(int i = l; i<=mid; i++) runOper(i, -1); solve(lef, l, mid); solve(rig, mid+1, r); } vector<int> vec; int main () { N = getInt(), M = getInt(); for(int i = 1; i<=M; i++) pos[O[i] = getInt()].push_back(i); for(int i = 1; i<=N; i++) P[i] = getInt(); K = getInt(); for(int i = 1; i<=K; i++) oper[i][0] = getInt(), oper[i][1] = getInt(), oper[i][2] = getInt(); oper[K+1][0] = oper[K+1][1] = oper[K+1][2] = 1; for(int i = 1; i<=N; i++) vec.push_back(i); solve(vec, 1, K+1); for(int i = 1; i<=N; i++) if(ans[i] > K) puts("NIE"); else printf("%d\n", ans[i]); }
BZOJ 4103
题目大意:给你数组A和B, p个询问, 每次询问可重集合[latex]S={A_i \, xor \, B_j \mid u\le i\le d, l\le j \le r}[/latex]中第k大的元素。
解题报告
一开始直接偷懒暴力, 写完暴力发现是错的, 自己zz了。
然后发现在PST(或者说可持久化Trie, 其实也算是PST的变种吧)上暴力就好了, 只是复杂度略有点玄。
代码
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cstdlib> #include <vector> using namespace std; typedef long long LL; const int MAXM = 300030; const int MAXN = 1010; inline int getInt () { int ret = 0; char ch; bool flg = false; while((ch = getchar()) < '0' || ch > '9') flg |= (ch == '-'); ret = ch-'0'; while((ch = getchar()) >= '0' && ch <= '9') ret = ret*10+(ch-'0'); return flg ? -ret : ret; } int N, M; namespace PST { const int MAXD = MAXM*32; struct Node { int ch[2]; int sz; } d[MAXD]; int dtot = 0; int roots[MAXM], tot; unsigned ins; void insert (int & u, int r, int b, bool l) { if(l != (bool)(ins&(1u<<b))) {u = r; return ;} d[u = ++dtot].sz = d[r].sz+1; if(!b) return ; insert(d[u].ch[0], d[r].ch[0], b-1, false); insert(d[u].ch[1], d[r].ch[1], b-1, true); } inline void hInsert (unsigned val) { ins = val, ++tot; insert(roots[tot], roots[tot-1], 31, false); } inline int size (int u) {return d[u].sz;} } int A[MAXN], B[MAXM]; int bufl[MAXN], bufr[MAXN]; int main () { N = getInt(), M = getInt(); for(int i = 1; i<=N; i++) A[i] = getInt(); for(int i = 1; i<=M; i++) { B[i] = getInt(); PST::hInsert(B[i]); } int P = getInt(); for(int q = 1; q<=P; q++) { int x1 = getInt(), x2 = getInt(); int y1 = getInt(), y2 = getInt(), k = getInt(); int cur = 0; for(int i = x1; i<=x2; i++) { bufl[i-x1+1] = PST::roots[y1-1]; bufr[i-x1+1] = PST::roots[y2]; } int xlen = x2-x1+1; for(int p = 30; p>=0; p--) { int cnt1 = 0; for(int i = 1; i<=xlen; i++) { if(A[i+x1-1]&(1<<p)) cnt1 += PST::size(PST::d[bufr[i]].ch[0])-PST::size(PST::d[bufl[i]].ch[0]); else cnt1 += PST::size(PST::d[bufr[i]].ch[1])-PST::size(PST::d[bufl[i]].ch[1]); } bool flg = true; if(cnt1 < k) k -= cnt1, flg = false; else cur |= (1<<p); for(int i = 1; i<=xlen; i++) { bufl[i] = PST::d[bufl[i]].ch[flg^(bool)(A[i+x1-1]&(1<<p))]; bufr[i] = PST::d[bufr[i]].ch[flg^(bool)(A[i+x1-1]&(1<<p))]; } } printf("%d\n", cur); } }
BZOJ 3489
题目大意:给个序列A, 每次询问一个区间[l, r]内最大的只出现一次的数字。
解题报告
…没想出来, 我果然zz。
首先考虑每个数字会对哪一段区间做出贡献, 对于数字A[i], 设A[i]上一次出现位置为pre[i], 下一次出现位置为suc[i], 那么i可以对所有区间[l, r]满足pre[i]<l≤i并且pre[i]≤r<suc[i]做出贡献, 把区间l和r看成坐标系上的点, 每个贡献就是一个矩形, 用树套树做矩形取max和单点查询操作即可。
这种考虑贡献的思路还是比较好的…
代码
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cstdlib> using namespace std; const int MAXN = 200020; inline int getInt () { int ret = 0; char ch; bool flg = false; while((ch = getchar()) < '0' || ch > '9') flg |= (ch == '-'); ret = ch-'0'; while((ch = getchar()) >= '0' && ch <= '9') ret = ret*10+(ch-'0'); return flg ? -ret : ret; } int N, M; namespace PST { const int MAXD = MAXN*100; int lc[MAXD], rc[MAXD], mxv[MAXD]; // 下标元素01和线段树 int dtot, insl, insr, insv; void insert (int & u, int l, int r) { if(l > insr || r < insl) return ; if(!u) u = ++dtot; if(insl <= l && r <= insr) {mxv[u] = max(insv, mxv[u]); return ;} int mid = (l+r)>>1; insert(lc[u], l, mid), insert(rc[u], mid+1, r); } int qp; int query (int u, int l, int r) { if(!u || l > qp || r < qp) return 0; if(l == r) return mxv[u]; int mid = (l+r)>>1; return max(mxv[u], max(query(lc[u], l, mid), query(rc[u], mid+1, r))); } int uid[MAXN*4]; int insyl, insyr; void insertY (int u, int l, int r) { if(l > insyr || r < insyl) return ; if(insyl <= l && r <= insyr) {insert(uid[u], 1, N); return ;} int mid = (l+r)>>1; insertY(u<<1, l, mid); insertY((u<<1)+1, mid+1, r); } void handleInsert (int x1, int x2, int y1, int y2, int val) { insl = x1, insr = x2, insyl = y1, insyr = y2; insv = val; insertY(1, 1, N); } int qyp; int queryY (int u, int l, int r) { if(l > qyp || r < qyp) return 0; int val = query(uid[u], 1, N); if(l == r) return val; int mid = (l+r)>>1; return max(val, max(queryY(u<<1, l, mid), queryY((u<<1)+1, mid+1, r))); } int handleQuery (int invl, int invr) { qp = invl, qyp = invr; return queryY(1, 1, N); } } int A[MAXN], pre[MAXN], suc[MAXN], cur[MAXN]; int main () { N = getInt(), M = getInt(); for(int i = 1; i<=N; i++) { A[i] = getInt(); pre[i] = cur[A[i]]; cur[A[i]] = i; } for(int i = 1; i<=N; i++) cur[i] = N+1; for(int i = N; i>=1; i--) { suc[i] = cur[A[i]]; cur[A[i]] = i; } for(int i = 1; i<=N; i++) PST::handleInsert(pre[i]+1, i, i, suc[i]-1, A[i]); int lastans = 0; for(int i = 1; i<=M; i++) { int x = getInt(), y = getInt(); int l = min((x+lastans)%N+1, (y+lastans)%N+1); int r = max((x+lastans)%N+1, (y+lastans)%N+1); printf("%d\n", lastans=PST::handleQuery(l, r)); } }
BZOJ 3545/3551
题目大意:给你一个图, 点边都带权, 每次询问一个点经过不超过x边权的边能到达的所有点中点权第k大的。
强制在线。
解题报告
Kruskal重构树!学了跟没学一样!想不到!
首先做一颗Kruskal重构树出来, 然后问题就转化成静态在线子树第k大了, 用dfn+主席树上整体二分解决。
代码
这里贴未进行强制在线解码的版本。
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cstdlib> using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 100050; const int MAXM = 500050; inline int getInt () { int ret; char ch; bool flg = false; while((ch = getchar()) < '0' || ch > '9') flg |= (ch == '0'); ret = ch-'0'; while((ch = getchar()) >= '0' && ch <= '9') ret = ret*10+(ch-'0'); return flg ? -ret : ret; } int len; namespace PST { const int MAXD = MAXN*35; int lc[MAXD], rc[MAXD], sum[MAXD], dtot; int roots[MAXN*2], tot; int insp, insv; void insert (int & u, int lb, int rb, int r) { if(lb > insp || rb < insp) {u = r; return ;} if(!u) u = ++dtot, sum[u] = sum[r]; sum[u] += insv; if(lb == rb) return ; int mid = (lb+rb)>>1; insert(lc[u], lb, mid, lc[r]); insert(rc[u], mid+1, rb, rc[r]); } void handleInsert (int val, int cnt) { ++tot, insp = val, insv = cnt; insert(roots[tot], 1, len, roots[tot-1]); } int queryKth (int l, int r, int k) { int x = roots[l-1], y = roots[r]; int lb = 1, rb = len; while(lb != rb) { int csum = sum[rc[y]]-sum[rc[x]]; int mid = (lb+rb)>>1; if(csum < k) k -= csum, x = lc[x], y = lc[y], rb = mid; else x = rc[x], y = rc[y], lb = mid+1; } if(k > sum[y]-sum[x]) return 0; // no solution return lb; } } struct Edge { int a, b, w; } edges[MAXM]; inline bool operator < (const Edge & a, const Edge & b) { return a.w < b.w; } int N, M, Q; int pfa[MAXN*2]; int root (int u) {return pfa[u]?(pfa[u]=root(pfa[u])):u;} int dtot; struct Node { int v, nxt; } d[MAXN*3]; int head[MAXN*2], etot; inline void addEdge (int a, int b) { etot ++; d[etot].v = b; d[etot].nxt = head[a]; head[a] = etot; } int dfn[MAXN*2], lst[MAXN*2], rev[MAXN*2], dfntot; int H[MAXN*2], hlinr[MAXN*2], fa[MAXN*2][20]; void dfs (int u, int f) { fa[u][0] = f; rev[dfn[u] = ++dfntot] = u; for(int e = head[u]; e; e = d[e].nxt) dfs(d[e].v, u); lst[u] = dfntot; } int maxv[MAXN*2][20]; void stSetup () { maxv[0][0] = INF; for(int i = 1; i<=dfntot; i++) maxv[i][0] = max(maxv[i][0], maxv[fa[i][0]][0]); for(int lvl = 1; lvl<20; lvl++) for(int i = 1; i<=dfntot; i++) { maxv[i][lvl] = max(maxv[fa[i][lvl-1]][lvl-1], maxv[i][lvl-1]); fa[i][lvl] = fa[fa[i][lvl-1]][lvl-1]; } } int stQuery (int u, int lim) { for(int i = 19; i>=0; i--) if(maxv[u][i] <= lim) u = fa[u][i]; return u; } int roots[MAXN*2], rtot; int main () { N = getInt(), M = getInt(), Q = getInt(); for(int i = 1; i<=N; i++) H[i] = getInt(); for(int i = 1; i<=M; i++) { edges[i].a = getInt(); edges[i].b = getInt(); edges[i].w = getInt(); } dtot = N; std :: sort(edges+1, edges+M+1); for(int i = 1; i<=M; i++) { int ra = root(edges[i].a), rb = root(edges[i].b); if(ra == rb) continue ; int u = ++dtot; pfa[ra] = u, pfa[rb] = u; maxv[u][0] = edges[i].w; addEdge(u, ra), addEdge(u, rb); } for(int i = 1; i<=dtot; i++) if(root(i) == i) roots[++rtot] = i; for(int i = 1; i<=rtot; i++) dfs(roots[i], 0); stSetup(); memcpy(hlinr, H, sizeof H); sort(hlinr+1, hlinr+N+1); len = (unique(hlinr+1, hlinr+N+1)-hlinr)-1; for(int i = 1; i<=dfntot; i++) { int val = lower_bound(hlinr, hlinr+len+1, H[rev[i]])-hlinr; PST::handleInsert(val, rev[i]<=N); } hlinr[0] = -1; for(int q = 1; q<=Q; q++) { int u = getInt(), x = getInt(), k = getInt(); int pos = stQuery(u, x); printf("%d\n", hlinr[PST::queryKth(dfn[pos], lst[pos], k)]); } }
BZOJ 3439
题目大意: 给你N个串, 对每个串询问一次出现过的以这个串结尾的串中编号第k小的串编号。
解题报告
《论哈希之美与STLの胜利》
代码
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cstdlib> #include <map> #include <set> #include <queue> using namespace std; typedef unsigned long long ULL; const int MAXN = 100010; const int MAXL = 300030; const ULL SEED = 31; int N; char buf[MAXL]; ULL rec[MAXN]; map<ULL, priority_queue<int> > mp; map<ULL, priority_queue<pair<int, int> > > mp2; int ans[MAXN], K[MAXN]; int main () { scanf("%d", &N); for(int i = 1; i<=N; i++) { scanf("%s", buf); int len; reverse(buf, buf+(len=strlen(buf))); ULL hashc = 0; for(int j = 0; j<len; j++) { hashc = hashc*SEED+buf[j]; mp[hashc].push(i); } rec[i] = hashc; } for(int i = 1; i<=N; i++) scanf("%d", &K[i]), mp2[rec[i]].push(make_pair(K[i], i)); for(map<ULL, priority_queue<pair<int, int> > >::iterator it = mp2.begin(); it != mp2.end(); it++) { priority_queue<pair<int, int> > &que = it->second; while(!que.empty()) { pair<int, int> qry = que.top(); que.pop(); priority_queue<int>&st = mp[it->first]; while(st.size() > qry.first) st.pop(); if(st.size() < qry.first) ans[qry.second] = -1; else ans[qry.second] = st.top(); } } for(int i = 1; i<=N; i++) printf("%d\n", ans[i]); }
BZOJ 4105
题目大意: 给你一个数列A和数字P(P在数据范围中给出), 支持两种操作:
1.区间A[i]变成A[i]*A[i]%P。
2.区间求和。
解题报告
鬼畜题..可以发现对于每一个给定的P模数, 有一个神奇的性质。如果把x向x*x%P连边, 那么图上最大的环长度很小, 并且所有环长度的lcm最大为60, 并且每一个节点到环的距离不超过11。
然后就可以用线段树做, 每个节点保存如果这个区间进行操作1若干次(小于环lcm次)之后的区间和, 不在环上的stl暴力就可以了..
自己zz感觉不用写updata调了半个小时…rool(操作1)是需要updata的。
代码
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cstdlib> #include <cmath> #include <set> using namespace std; const int MAXN = 100010; int N, M, P; int vis[MAXN]; bool cvis[MAXN]; int cid[MAXN], clen[MAXN], ctot; bool dfs (int u, int d) { if(vis[u]) { if(cvis[u]) {clen[cid[u] = ++ctot] = d-vis[u]; return true;} return false; } vis[u] = d; cvis[u] = true; if(dfs(u*u%P, d+1)) { cvis[u] = false; if(cid[u] == ctot) return false; cid[u] = ctot; return true; } cvis[u] = false; return false; } int cirlcm; // 记住初始化! namespace IBT { const int MAXD = MAXN*5; const int BLEN = 80; int val[MAXD][BLEN]; // u节点经过i次操作的和/当前和 int rcnt[MAXD]; // 保存旋转次数 int buf[BLEN]; inline void rool (int u, int t) { rcnt[u] += t; memcpy(buf, val[u], sizeof buf); for(int i = 0; i<cirlcm; i++) val[u][i] = buf[(i+t)%cirlcm]; } inline void updata (int u) { for(int i = 0; i<cirlcm; i++) val[u][i] = val[u<<1][i]+val[(u<<1)+1][i]; } inline void pushdown (int u) { if(rcnt[u]) { rool(u<<1, rcnt[u]); rool((u<<1)+1, rcnt[u]); rcnt[u] = 0; } } int insp, insv[BLEN], inslen; inline void mixin (int u) { for(int i = 0; i<cirlcm; i++) val[u][i] += insv[i%inslen]; } void insert (int u, int l, int r) { if(l > insp || r < insp) return ; mixin(u); if(l == r) return ; pushdown(u); int mid = (l+r)>>1; insert(u<<1, l, mid); insert((u<<1)+1, mid+1, r); } int argl, argr; void rooldown (int u, int l, int r) { if(l > argr || r < argl) return ; if(argl <= l && r <= argr) {rool(u, 1); return ;} pushdown(u); int mid = (l+r)>>1; rooldown(u<<1, l, mid); rooldown((u<<1)+1, mid+1, r); updata(u); } int query (int u, int l, int r) { if(l > argr || r < argl) return 0; if(argl <= l && r <= argr) return val[u][0]; pushdown(u); int mid = (l+r)>>1; return query(u<<1, l, mid)+query((u<<1)+1, mid+1, r); } } int A[MAXN]; void handleInsert (int pos, int num) { num %= P; int t = num*num%P, p = 0; IBT::insv[0] = num; while(t != num) { IBT::insv[++p] = t; t = t*t%P; } IBT::inslen = p+1; IBT::insp = pos; IBT::insert(1, 1, N); } set<int> st; inline void handleRool (int l, int r) { IBT::argl = l, IBT::argr = r; set<int>::iterator it = st.lower_bound(l), tmp; IBT::rooldown(1, 1, N); while(it != st.end() && *it <= r) { int p = *it; A[p] = A[p]*A[p]%P; tmp = it, ++ it; if(cid[A[p]]) { st.erase(tmp); handleInsert(p, A[p]); } } } inline int handleQuery (int l, int r) { IBT::argl = l, IBT::argr = r; set<int>::iterator it = st.lower_bound(l); int sum = 0; while(it != st.end() && *it <= r) sum += A[*it], ++ it; return sum+IBT::query(1, 1, N); } int gcd (int a, int b) {return b ? gcd(b, a%b) : a;} inline int lcm (int a, int b) {return a/gcd(a, b)*b;} int main () { scanf("%d%d%d", &N, &M, &P); cirlcm = 1; for(int i = 0; i<P; i++) if(!vis[i]) dfs(i, 1); for(int i = 1; i<=ctot; i++) cirlcm = lcm(cirlcm, clen[i]); for(int i = 1; i<=N; i++) { scanf("%d", &A[i]), A[i] %= P; if(cid[A[i]]) handleInsert(i, A[i]); else st.insert(i); } for(int i = 1; i<=M; i++) { int op, l, r; scanf("%d%d%d", &op, &l, &r); if(op == 0) handleRool(l, r); else printf("%d\n", handleQuery(l, r)); } } /* 6 100 11 1 7 10 4 10 4 0 2 4 1 4 5 15! 1 3 4 1 1 5 0 3 5 1 1 3 */
BZOJ 3744
题目大意: 给你一个串, 强制在线区间逆序对。
解题报告
%#$^%@!
分块+主席树乱搞!
真的那么简单么?
交上去, Time Limit Exceed!
生成大数据! 擀! 要跑18s!
怎么可能QAQ
Hineven: Ender我好像被常数附体了你的代码借我跑一下。
Ender: 好的。
Ender的代码只跑了3s+!
Hineven:%#$^%@!你怎么写的?!
Ender:树状数组加预处理啊。
Hineven:%#$^%@!
Hineven差点就转行写树状数组了!
JeremyGuo:我写的主席树没问题啊。
JermeyGuo跑了多久呢? 4s!
Hineven:[****哔****!]
JeremyGuo:我来帮你优化一下。
[30min之后…]
JeremyGuo:优化完了!PST每处理4k个询问查询次数下降了一半呢!
现在优化到13s了1
JeremyGuo:加油我先走了。
Hineven:你多少分一块?
JeremyGuo:128?
Hineven:$$%#$$$#@$!#!
然后Hineven就把块内元素从256个改到128个!现在只需要9s了!
Hineven:用暴力复杂度换PST果然正解!
然后开1k个块, 5s!交!过了!
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代码
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cstdlib> #include <cmath> #include <vector> #include <set> using namespace std; const int MAXN = 50050; int len; namespace PST { const int MAXD = MAXN*20; int lc[MAXD], rc[MAXD], sum[MAXD], dtot; int roots[MAXN], tot; int insp; /*void insert (int& u, int l, int r, int v) { if(l > insp || r < insp) {u = v; return ;} if(!u) u = ++dtot, sum[u] = sum[v]; sum[u] ++; if(l == r) return ; int mid = (l+r)>>1; insert(lc[u], l, mid, lc[v]); insert(rc[u], mid+1, r, rc[v]); }*/ void Insert(int& u, int l, int r, int pos){ ++dtot; lc[dtot] = lc[u]; rc[dtot] = rc[u]; sum[dtot] = sum[u]+1; u = dtot; if(l == r) return ; int mid=(l+r)>>1; if(pos <= mid) Insert(lc[u], l, mid, pos); else Insert(rc[u], mid+1, r, pos); } inline void handleInsert (int val) { insp = val, tot ++; //insert(roots[tot], 1, len, roots[tot-1]); roots[tot] = roots[tot-1]; Insert(roots[tot], 1, len, val); } int ql, qr; int query (int u, int u2, int l, int r) { if(u2 == u) return 0; if(ql <= l && r <= qr) return sum[u]-sum[u2]; int mid = (l+r)>>1; int ret=0; if(ql <= mid) ret += query(lc[u], lc[u2], l, mid); if(qr > mid) ret += query(rc[u], rc[u2], mid+1, r); return ret; } /*int Query(int lrt, int rrt, int l, int r, int L, int R){ if(L <= l && r <= R) return sum[rrt]-sum[lrt]; d_sum++; int mid=(l+r)>>1; int ret=0; if(L <= mid) ret += Query(lc[lrt], lc[rrt], l, mid, L, R); if(R > mid) ret += Query(rc[lrt], rc[rrt], mid+1, r, L, R); return ret; }*/ inline int handleQuery (int x1, int x2, int y1, int y2) { if(x1 > x2 || y1 > y2) return 0; x1 --, ql = y1, qr = y2; return query(roots[x2], roots[x1], 1, len); //return Query(roots[x1], roots[x2], 1, len, y1, y2); } } int N, qsize, qtot; int A[MAXN]; const int QCNT = 1000; vector<int> quads[1000]; int qconflict[1000]; int cconflict[1000][1000]; int cvalue[1000][1000]; int linr[MAXN]; inline int getConflict (int l, int r) { int ret = 0; for(int i = l; i<=r; i++) for(int j = i+1; j<=r; j++) if(A[j] < A[i]) ret ++; return ret; } void initialize () { memcpy(linr, A, sizeof linr); sort(linr+1, linr+N+1); len = (unique(linr+1, linr+N+1)-linr)-1; for(int i = 1; i<=N; i++) A[i] = lower_bound(linr+1, linr+len+1, A[i])-linr; qsize = 64; for(int i = 1; i<=N;) { ++qtot; int j; for(j = i; j-i<qsize && j<=N; j++) quads[qtot].push_back(A[j]); sort(quads[qtot].begin(), quads[qtot].end()); qconflict[qtot] = getConflict(i, j-1); i = j; } for(int i = 1; i<=qtot; i++) for(int j = i+1; j<=qtot; j++) { int&sum = cconflict[i][j]; vector<int>&v1 = quads[i], &v2 = quads[j]; unsigned q = 0; int csum = 0; for(unsigned p = 0; p<v1.size(); p++) { while(q<v2.size() && v2[q] < v1[p]) q++, csum++; sum += csum; } } for(int i = 1; i<=qtot; i++) for(int j = i+1; j<=qtot; j++) cconflict[i][j] += cconflict[i][j-1]; for(int r = qtot; r>=1; r--) for(int l = r; l>=1; l--) cvalue[l][r] = cvalue[l+1][r]+cconflict[l][r]-cconflict[l][l]; for(int i = 1; i<=N; i++) PST::handleInsert(A[i]); } int query (int l, int r) { if(l >= r) return 0; int bl = (l-1)/qsize+1, br = (r-1)/qsize+1; int ret = 0; if(bl == br) { for(int i = l; i<r; i++) ret += PST::handleQuery(i+1, r, 1, A[i]-1); return ret; } ret = cvalue[bl+1][br-1]; for(int i = bl+1; i<br; i++) ret += qconflict[i]; int edgl = bl*qsize, edgr = (br-1)*qsize+1; for(int i = l; i<=edgl; i++) ret += PST::handleQuery(i+1, r, 1, A[i]-1); for(int i = edgr; i<=r; i++) ret += PST::handleQuery(edgl+1, i-1, A[i]+1, len); // ¼ÇµÃ·µ»Ø0! return ret; } inline int getInt () { int ret; char ch; while((ch = getchar()) < '0' || ch > '9') ; ret = ch-'0'; while((ch = getchar()) >= '0' && ch <= '9') ret = ret*10+(ch-'0'); return ret; } int main () { N = getInt(); for(int i = 1; i<=N; i++) A[i] = getInt(); initialize(); int M = getInt(); int lastans = 0; for(int i = 1; i<=M; i++) { int a = getInt(), b = getInt(); a ^= lastans, b ^= lastans; printf("%d\n", lastans = query(a, b)); } }
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