标签：cli play get break 最小 lex ems can ras
注意到二叉查找树的一个性质：其中序遍历就是所有元素按权值排序的顺序。
所以我们可以离线地把这棵树的中序遍历求出来。然后我们在插入的时候就可以用一个set来维护前驱后继，这样就可以维护出整棵树的形态。
接着我们发现将最大、最小单旋到根后，一定会有一边儿子是空的，并且剩下的子树的深度+1。于是我们就只要支持单点修改、区间加、单点查询的数据结构即可。树状数组就好了。
然后树的形态维护的时候大力判断一下就好啦。
1 #include <cstdio> 2 #include <cmath> 3 #include <set> 4 #include <algorithm> 5 6 #define R register 7 #define P std::pair<int, int> 8 #define mkp std::make_pair 9 #define maxn 100010 10 #define fir first 11 #define sec second 12 int v[maxn], ch[maxn][2], fa[maxn], tot; 13 std::set<P> s; 14 int q[maxn]; 15 int dfn[maxn], inv[maxn], timer, pos[maxn], ltag; 16 int ql, qr; 17 int hash[maxn]; 18 int bt[maxn]; 19 void add(R int x, R int val) {for (; x <= tot; x += x & -x) bt[x] += val;} 20 int query(R int x) {R int ret = 0; for (; x; x -= x & -x) ret += bt[x]; return ret;} 21 void modify(R int l, R int r, R int v) 22 { 23 // printf("l = %d r = %d v = %d\n", l, r, v); 24 add(l, v); add(r + 1, -v); 25 } 26 int main() 27 { 28 R int n, root; scanf("%d", &n); 29 s.insert(mkp(0, 0)); 30 s.insert(mkp(1e9 + 7, 0)); 31 for (R int i = 1; i <= n; ++i) 32 { 33 R int opt; scanf("%d", &opt); q[i] = opt; 34 if (opt == 1) {R int key; scanf("%d", &key); v[++tot] = key; hash[tot] = key;} 35 } 36 std::sort(hash + 1, hash + tot + 1); 37 for (R int i = 1; i <= n; ++i) dfn[i] = std::lower_bound(hash + 1, hash + tot + 1, v[i]) - hash; 38 for (R int i = 1, nw = 0; i <= n; ++i) 39 { 40 R int opt = q[i]; 41 if (opt == 1) 42 { 43 R P now = mkp(v[++nw], nw); 44 R P prev = *(--s.upper_bound(now)), next = *s.upper_bound(now); 45 s.insert(now); 46 if (prev.sec && !ch[prev.sec][1]) 47 { 48 ch[prev.sec][1] = nw; 49 fa[nw] = prev.sec; 50 } 51 else if (next.sec && !ch[next.sec][0]) 52 { 53 ch[next.sec][0] = nw; 54 fa[nw] = next.sec; 55 } 56 !fa[nw] ? root = nw : 0; 57 58 R int dep; 59 printf("%d\n", dep = fa[nw] ? query(dfn[fa[nw]]) + 1 : 1); 60 modify(dfn[nw], dfn[nw], dep - query(dfn[nw])); 61 } 62 else if (opt == 2) 63 { 64 R P xmin = *(++s.begin()); R int xc = xmin.sec; 65 printf("%d\n", query(dfn[xc])); 66 // printf("fa %d\n", fa[xc]); 67 modify(!fa[xc] ? tot + 1 : dfn[fa[xc]], tot, 1); 68 modify(dfn[xc], dfn[xc], 1 - query(dfn[xc])); 69 70 if (root != xc) 71 { 72 ch[fa[xc]][0] = ch[xc][1]; 73 fa[ch[xc][1]] = fa[xc]; 74 fa[root] = xc; 75 ch[xc][1] = root; 76 root = xc; 77 fa[xc] = 0; 78 } 79 } 80 else if (opt == 3) 81 { 82 R P xmax = *(++s.rbegin()); R int xc = xmax.sec; 83 printf("%d\n", query(dfn[xc])); 84 modify(1, !fa[xc] ? 0 : dfn[fa[xc]], 1); 85 modify(dfn[xc], dfn[xc], 1 - query(dfn[xc])); 86 87 if (root != xc) 88 { 89 ch[fa[xc]][1] = ch[xc][0]; 90 fa[ch[xc][0]] = fa[xc]; 91 fa[root] = xc; 92 ch[xc][0] = root; 93 root = xc; 94 fa[xc] = 0; 95 } 96 } 97 else if (opt == 4) 98 { 99 R P xmin = *(++s.begin()); R int xc = xmin.sec; 100 printf("%d\n", query(dfn[xc])); 101 modify(dfn[xc] + 1 , !fa[xc] ? tot : dfn[fa[xc]] - 1, -1); 102 103 xc == root ? fa[root = ch[xc][1]] = 0 : 104 (ch[fa[xc]][0] = ch[xc][1], 105 fa[ch[xc][1]] = fa[xc]); 106 s.erase(xmin); 107 } 108 else 109 { 110 R P xmax = *(++s.rbegin()); R int xc = xmax.sec; 111 // printf("max %d %d\n", xc, fa[xc]); 112 printf("%d\n", query(dfn[xc])); 113 modify(!fa[xc] ? 1 : dfn[fa[xc]] + 1, dfn[xc] - 1, -1); 114 115 xc == root ? fa[root = ch[xc][0]] = 0 : 116 (ch[fa[xc]][1] = ch[xc][0], 117 fa[ch[xc][0]] = fa[xc]); 118 s.erase(xmax); 119 } 120 } 121 return 0; 122 } 123 /* 124 6 125 1 1 126 1 2 127 1 4 128 4 129 2 130 5 131 性质：二叉查找树的子树的权值对应的是一个区间。并且有且仅有所有子树内的点在这个区间内。 132 */
第1种的贡献可以用单调栈算。第2种的贡献没那么好算，所以我们来考虑1+2的贡献。1+2的贡献可以转为二维偏序，用排序+线段树来搞。
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 5 #define R register 6 #define maxn 200010 7 #define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b)) 8 inline int F() 9 { 10 R char ch; R int cnt; 11 while (ch = getchar(), ch < ‘0‘ || ch > ‘9‘) ; 12 cnt = ch - ‘0‘; 13 while (ch = getchar(), ch >= ‘0‘ && ch <= ‘9‘) cnt = cnt * 10 + ch - ‘0‘; 14 return cnt; 15 } 16 typedef long long ll; 17 struct Query { 18 int l, r; 19 } q[maxn]; 20 struct Chain { 21 Chain *next; 22 int id; 23 } *last[maxn], mem[maxn], *tot = mem; 24 struct Opt { 25 int x, id, type; 26 inline bool operator < (const Opt &that) const {return x < that.x;} 27 } op[maxn << 1]; 28 int ans1[maxn]; ll ans2[maxn]; 29 int st[maxn], top, a[maxn]; 30 int ql, qr; 31 ll tr[maxn << 2], tag[maxn << 2]; 32 inline void pushdown(R int o, R int l, R int r, R int mid) 33 { 34 if (tag[o]) 35 { 36 tag[o << 1] += tag[o]; 37 tag[o << 1 | 1] += tag[o]; 38 tr[o << 1] += 1ll * tag[o] * (mid - l + 1); 39 tr[o << 1 | 1] += 1ll * tag[o] * (r - mid); 40 tag[o] = 0; 41 } 42 } 43 inline void update(R int o) {tr[o] = tr[o << 1] + tr[o << 1 | 1];} 44 void modify(R int o, R int l, R int r) 45 { 46 if (ql <= l && r <= qr) 47 { 48 ++tag[o]; tr[o] += (r - l + 1); 49 return ; 50 } 51 R int mid = l + r >> 1; 52 pushdown(o, l, r, mid); 53 if (ql <= mid) modify(o << 1, l, mid); 54 if (mid < qr) modify(o << 1 | 1, mid + 1, r); 55 update(o); 56 } 57 ll query(R int o, R int l, R int r) 58 { 59 if (ql <= l && r <= qr) return tr[o]; 60 R int mid = l + r >> 1; 61 R ll ret = 0; 62 pushdown(o, l, r, mid); 63 if (ql <= mid) ret += query(o << 1, l, mid); 64 if (mid < qr) ret += query(o << 1 | 1, mid + 1, r); 65 return ret; 66 } 67 int bt[maxn], n, lef[maxn], rig[maxn], r[maxn], Fa[maxn]; 68 int Find(R int x) {return Fa[x] == x ? x : Fa[x] = Find(Fa[x]);} 69 inline void add(R int x) {for (; x <= n; x += x & -x) ++bt[x];} 70 inline int query(R int x) {R int ret = 0; for (; x; x -= x & -x) ret += bt[x]; return ret;} 71 int main() 72 { 73 n = F(); R int m = F(), p1 = F(), p2 = F(); 74 for (R int i = 1; i <= n; ++i) a[i] = F(), r[a[i]] = i, Fa[i] = lef[i] = rig[i] = i; 75 R int opcnt = 0; 76 for (R int i = 1; i <= m; ++i) 77 { 78 q[i] = (Query) {F(), F()}; 79 *++tot = (Chain) {last[q[i].r], i}; last[q[i].r] = tot; 80 op[++opcnt] = (Opt) {q[i].l - 1, i, -1}; 81 op[++opcnt] = (Opt) {q[i].r, i, 1}; 82 } 83 84 memset(bt, 0, (n + 1) << 2); top = 0; 85 for (R int i = 1; i <= n; ++i) 86 { 87 R int rr = i - 2; 88 while (top && a[st[top]] < a[i]) add(st[top--]); 89 if (top) add(st[top]); 90 st[++top] = i; 91 for (R Chain *iter = last[i]; iter; iter = iter -> next) 92 ans1[iter -> id] += query(q[iter -> id].r) - query(q[iter -> id].l - 1); 93 } 94 95 std::sort(op + 1, op + opcnt + 1); 96 for (R int i = 1; i <= n; ++i) 97 { 98 R int ps = r[i], f1; 99 if (ps != 1 && a[ps - 1] < i) lef[ps] = lef[f1 = Find(ps - 1)], Fa[f1] = ps; 100 if (ps != n && a[ps + 1] < i) rig[ps] = rig[f1 = Find(ps + 1)], Fa[f1] = ps; 101 } 102 R int p = 1; 103 while (op[p].x == 0) ++p; 104 for (R int i = 1; i <= n; ++i) 105 { 106 // printf("%d %d\n", lef[i], rig[i]); 107 ql = lef[i]; qr = rig[i]; 108 modify(1, 1, n); 109 while (op[p].x == i) 110 { 111 ql = q[op[p].id].l; qr = q[op[p].id].r; 112 ans2[op[p].id] += op[p].type * query(1, 1, n); 113 ++p; 114 } 115 } 116 for (R int i = 1; i <= m; ++i) printf("%lld\n", 1ll * ans1[i] * p1 + (ans2[i] - (q[i].r - q[i].l + 1) - ans1[i]) * p2); 117 return 0; 118 } 119 /* 120 9 4 121 12 5 122 2 0 123 5 1 124 5 2 125 30 126 39 127 4 128 13 129 16 130 131 */
把式子拆开完就是一个卷积和二次函数的形式，二次函数用对称轴公式来算，卷积用FFT来算。
1 #include <cstdio> 2 #include <algorithm> 3 #include <cmath> 4 5 #define R register 6 #define maxn 262145 7 #define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0) 8 typedef double db; 9 const db pi = acos(-1); 10 struct Complex { 11 db x, y; 12 inline Complex operator - (const Complex &that) const {return (Complex) {x - that.x, y - that.y};} 13 inline Complex operator * (const Complex &that) const {return (Complex) {x * that.x - y * that.y, x * that.y + that.x * y};} 14 inline void operator += (const Complex &that) {x += that.x; y += that.y;} 15 } w[maxn << 1]; 16 int N; 17 void init() 18 { 19 R int h = N >> 1; 20 for (R int i = 0; i < N; ++i) w[i + h] = (Complex) {cos(2 * pi / N * i), sin(2 * pi / N * i)}; 21 for (R int i = h; i--; ) w[i] = w[i << 1]; 22 } 23 void bit_reverse(R Complex *a, R Complex *b) 24 { 25 for (R int i = 0; i < N; ++i) b[i] = a[i]; 26 for (R int i = 0, j = 0; i < N; ++i) 27 { 28 i > j ? std::swap(b[i], b[j]), 1 : 0; 29 for (R int l = N >> 1; (j ^= l) < l; l >>= 1) ; 30 } 31 } 32 void dft(R Complex *a) 33 { 34 for (R int l = 2, m = 1; m < N; l <<= 1, m <<= 1) 35 for (R int i = 0; i < N; i += l) 36 for (R int j = 0; j < m; ++j) 37 { 38 R Complex tmp = a[i + j + m] * w[j + m]; 39 a[i + j + m] = a[i + j] - tmp; 40 a[i + j] += tmp; 41 } 42 } 43 Complex a[maxn], b[maxn], c[maxn], ta[maxn], tb[maxn], tc[maxn]; 44 int main() 45 { 46 R int n, m, ans = 0x7fffffff, ret = 0, sum = 0; scanf("%d%d", &n, &m); 47 for (R int i = 0; i < n; ++i) scanf("%lf", &a[i].x), sum -= a[i].x; 48 for (R int i = 0; i < n; ++i) scanf("%lf", &b[i].x), sum += b[i].x; 49 std::reverse(b, b + n); 50 for (R int i = 0; i < n; ++i) b[i + n] = b[i]; 51 for (N = 1; N < (n << 1); N <<= 1); 52 init(); 53 R int ccc = round((db) sum / n); 54 55 for (R int i = 0; i < n; ++i) a[i].x += ccc, ret += a[i].x * a[i].x + b[i].x * b[i].x; 56 // printf("%d %d\n", ccc, ret); 57 bit_reverse(a, ta); 58 bit_reverse(b, tb); 59 dft(ta); dft(tb); 60 for (R int i = 0; i < N; ++i) c[i] = ta[i] * tb[i]; 61 std::reverse(c + 1, c + N); 62 bit_reverse(c, tc); dft(tc); 63 // for (R int i = 0; i < N; ++i) printf("%lf %lf\n", tc[i].x, tc[i].y); 64 for (R int i = 0; i < n; ++i) cmin(ans, (int) ret - (2 * tc[i + n].x / N)); 65 66 printf("%d\n", ans); 67 return 0; 68 }
玄学搜索题。搜索出来完用双指针扫一扫。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <map> 5 #include <bitset> 6 7 #define R register 8 #define maxn 110 9 int a[maxn], w[maxn]; 10 int f[maxn][maxn]; 11 const int oo = 1e8; 12 #define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0) 13 #define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b)) 14 #define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b)) 15 const int mod = 8260817; 16 typedef long long ll; 17 struct Queue { 18 int step, level, f; 19 } q[10000010]; 20 struct Hash { 21 Hash *next; 22 ll key; 23 } *last[mod], mem[mod], *tot = mem; 24 inline ll hash_key(R int a, R int b) 25 { 26 return 1ll * b * 233 + a; 27 } 28 inline bool insert(R int a, R int b) 29 { 30 R ll key = hash_key(a, b); 31 R int pos = key % mod; 32 for (R Hash *iter = last[pos]; iter; iter = iter -> next) 33 if (iter -> key == key) return 0; 34 *++tot = (Hash) {last[pos], key}; last[pos] = tot; 35 return 1; 36 } 37 struct Data { 38 int f, s; 39 inline bool operator < (const Data &that) const {return f < that.f;} 40 } st[maxn * maxn * maxn]; 41 int scnt, qu[30]; 42 int main() 43 { 44 R int n, m, mc; scanf("%d%d%d", &n, &m, &mc); 45 for (R int i = 1; i <= n; ++i) scanf("%d", &a[i]); 46 for (R int i = 1; i <= n; ++i) scanf("%d", &w[i]); 47 memset(f, -63, sizeof (f)); 48 f[0][mc] = 0; 49 for (R int i = 1; i <= n; ++i) 50 { 51 for (R int j = a[i]; j <= mc; ++j) 52 cmax(f[i][j - a[i]], f[i - 1][j] + 1), 53 cmax(f[i][dmin(j - a[i] + w[i], mc)], f[i - 1][j]); 54 } 55 R int Day = -1; 56 for (R int j = 0; j <= n; ++j) for (R int i = 0; i <= mc; ++i) cmax(Day, f[j][i]); 57 R int oo = 0; 58 for (R int i = 1; i <= m; ++i) scanf("%d", &qu[i]), cmax(oo, qu[i]); 59 60 R int head = 0, tail = 1; 61 q[1] = (Queue) {0, 0, 1}; 62 while (head < tail) 63 { 64 R Queue now = q[++head]; 65 if (insert(0, now.f)) st[++scnt] = (Data) {now.f, now.step + 1}; 66 if (now.step >= Day - 1) continue; 67 if (now.level > 0 && oo / now.level < now.f) continue; 68 if (insert(now.level + 1, now.f)) q[++tail] = (Queue) {now.step + 1, now.level + 1, now.f}; 69 70 if (1ll * now.f * now.level <= oo && now.f > 0) 71 { 72 if (insert(now.level, now.f * now.level)) 73 q[++tail] = (Queue) {now.step + 1, now.level, now.f * now.level}; 74 } 75 } 76 st[++scnt] = (Data) {0, 0}; 77 std::sort(st + 1, st + scnt + 1); 78 for (R int _ = 1; _ <= m; ++_) 79 { 80 R int c = qu[_], flag = 0; 81 if (Day <= 0) {puts("0"); continue;} 82 if (c <= Day) {puts("1"); continue;} 83 R int j = 0, mx = -oo * 2; 84 for (R int i = scnt; i; --i) 85 { 86 for (; j < scnt && st[j + 1].f + st[i].f <= c; ++j, cmax(mx, st[j].f - st[j].s)); 87 if (mx + st[i].f - st[i].s >= c - Day) 88 { 89 flag = 1; 90 break; 91 } 92 } 93 printf("%d\n", flag); 94 } 95 return 0; 96 } 97 /* 98 10 20 100 99 22 18 15 16 20 19 33 15 38 49 100 92 14 94 92 66 94 1 16 90 51 101 4 102 5 103 9 104 338 105 5222 106 549 107 7491 108 9 109 12 110 3288 111 3 112 1 113 2191 114 833 115 3 116 6991 117 2754 118 3231 119 360 120 6 121 122 1 123 1 124 1 125 0 126 0 127 0 128 0 129 1 130 1 131 0 132 1 133 1 134 0 135 0 136 1 137 0 138 0 139 0 140 0 141 1 142 */
看起来很nan，没去看。听说是一道不可做题。
广义Lucas定理。普通的Lucas一般指用于质数，广义的可以做质数的k次方的，然后用CRT（中国剩余定理）合并。
1 #include <cstdio> 2 3 #define R register 4 typedef long long ll; 5 int pw0[10], pw2[10], pw5[10]; 6 int counter; 7 inline int qpow(R int base, R ll power, R int mod) 8 { 9 R int ret = 1; 10 for (; power; power >>= 1, base = 1ll * base * base % mod) 11 power & 1 ? ret = 1ll * ret * base % mod : 0; 12 return ret; 13 } 14 int mod, k; 15 void exgcd(R int a, R int b, R int &x, R int &y) 16 { 17 if (!b) {x = 1, y = 0; return ;} 18 exgcd(b, a % b, y, x); 19 y -= a / b * x; 20 } 21 inline int inv(R int a, R int p) 22 { 23 R int x, y; 24 exgcd(a, p, x, y); 25 return (x % p + p) % p; 26 } 27 int f2[10][1024], f5[10][1953125]; 28 inline int fac(R ll n, R int fact, R int p, R int pk, R int *fk) 29 { 30 if (!n) return 1; 31 R ll y = n / pk; 32 R int ret = 1ll * qpow(fact, y, pk) * fk[n % pk] % pk; 33 return 1ll * fac(n / p, fact, p, pk, fk) * ret % pk; 34 } 35 int ft2[10], ft5[10]; 36 inline int Lucas(R ll n, R ll m, R int p, R int pk, R bool div) 37 { 38 R ll num = 0; 39 for (R ll tmp = p; tmp <= n; tmp *= p) num += n / tmp; 40 for (R ll tmp = p; tmp <= m; tmp *= p) num -= m / tmp; 41 for (R ll tmp = p; tmp <= n - m; tmp *= p) num -= (n - m) / tmp; 42 43 R int fact = p == 2 ? ft2[k] : ft5[k], 44 *fk = p == 2 ? f2[k] : f5[k], 45 fa = 1, fb, fc; 46 if (div) 47 { 48 if (p == 2) --num; 49 else fa = inv(2, pk); 50 } 51 if (num >= k) return 0; 52 53 fa = 1ll * fa * fac(n, fact, p, pk, fk) % pk; 54 fb = fac(m, fact, p, pk, fk); 55 fc = fac(n - m, fact, p, pk, fk); 56 fb = inv(fb, pk); fc = inv(fc, pk); 57 58 59 return 1ll * fa * fb % pk * fc % pk * qpow(p, num, pk) % pk; 60 } 61 inline int C(R ll n, R ll m, R bool div = 0) 62 { 63 if (m > n) return 0; 64 // printf("C(%lld, %lld) %d\n", n, m, div); 65 R int C2 = Lucas(n, m, 2, pw2[k], div); 66 R int C5 = Lucas(n, m, 5, pw5[k], div); 67 // printf("%d %d\n", C2, C5); 68 R int t2 = inv(pw5[k], pw2[k]); 69 R int t5 = inv(pw2[k], pw5[k]); 70 R int ans = (1ll * C2 * pw5[k] % mod * t2 + 1ll * C5 * pw2[k] % mod * t5) % mod; 71 // printf("%d\n", ans); 72 return ans; 73 } 74 int main() 75 { 76 R ll a, b; 77 pw0[0] = pw2[0] = pw5[0] = 1; 78 for (R int i = 1; i < 10; ++i) 79 pw0[i] = pw0[i - 1] * 10, 80 pw2[i] = pw2[i - 1] * 2, 81 pw5[i] = pw5[i - 1] * 5; 82 83 for (R int i = 1; i < 10; ++i) 84 { 85 f2[i][0] = 1; 86 for (R int j = 1; j < pw2[i]; ++j) 87 f2[i][j] = 1ll * f2[i][j - 1] * (j % 2 ? j : 1) % pw2[i]; 88 ft2[i] = f2[i][pw2[i] - 1]; 89 } 90 for (R int i = 1; i < 10; ++i) 91 { 92 f5[i][0] = 1; 93 for (R int j = 1; j < pw5[i]; ++j) 94 f5[i][j] = 1ll * f5[i][j - 1] * (j % 5 ? j : 1) % pw5[i]; 95 ft5[i] = f5[i][pw5[i] - 1]; 96 } 97 98 while (scanf("%lld%lld%d", &a, &b, &k) != EOF) 99 { 100 char str[10]; 101 R int ans = 0; 102 mod = pw0[k]; 103 ans = qpow(2, a + b - 1, mod); 104 if (a == b) (ans += mod - C(a + a, a, 1)) %= mod; 105 else 106 { 107 if (((a + b) & 1) == 0) (ans += C(a + b, (a + b) / 2, 1)) %= mod; 108 for (R ll j = (a + b) / 2 + 1; j < a; ++j) 109 (ans += C(a + b, j)) %= mod; 110 } 111 sprintf(str, "%%0%dd\n", k); 112 printf(str, ans); 113 } 114 return 0; 115 } 116 /* 117 488754688 118 1000000000000000 999999999990000 9 119 1000000000000000 999999999990000 8 120 1000000000000000 999999999990000 7 121 1000000000000000 999999999990000 6 122 1000000000000000 999999999990000 5 123 1000000000000000 999999999990000 4 124 1000000000000000 999999999990000 3 125 1000000000000000 999999999990000 2 126 1000000000000000 999999999990000 1 127 */
标签：cli play get break 最小 lex ems can ras
原文地址：http://www.cnblogs.com/cocottt/p/7073251.html
