题意：对于整数序列$A[1...n]$定义$f(l, r)$为区间$[l, r]$内等于区间最大值元素的个数，定义$z[i]$为所有满足$f(l, r)=i$的区间总数。对于所有的$1 \leq i \leq n$，计算$z[i]$。
分析：考虑由大往小枚举最大值，对于某一最大值为$M$的区间$[l, r)$，满足$a[p_i]=M$的元素将区间切割为若干子区间，那么这些子区间对长度的积对答案的某一项有等值的贡献，暴力枚举需要$O(n^2)$的时间，整体考虑那些对答案某一项有贡献的子区间对的积，它们满足卷积的性质。因此只需将长度序列与其反序列作类似多项式乘法的fft，即可将时间优化为$O(nlog(n))$。子区间递归处理即可。代码如下：
1 #include <algorithm>
2 #include <cstdio>
3 #include <cstring>
4 #include <string>
5 #include <queue>
6 #include <map>
7 #include <set>
8 #include <stack>
9 #include <ctime>
10 #include <cmath>
11 #include <iostream>
12 #include <assert.h>
13 #pragma comment(linker, "/STACK:102400000,102400000")
14 #define max(a, b) ((a) > (b) ? (a) : (b))
15 #define min(a, b) ((a) < (b) ? (a) : (b))
16 #define mp std :: make_pair
17 #define st first
18 #define nd second
19 #define keyn (root->ch[1]->ch[0])
20 #define lson (u << 1)
21 #define rson (u << 1 | 1)
22 #define pii std :: pair<int, int>
23 #define pll pair<ll, ll>
24 #define pb push_back
25 #define type(x) __typeof(x.begin())
26 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
27 #define FOR(i, s, t) for(int i = (s); i <= (t); i++)
28 #define ROF(i, t, s) for(int i = (t); i >= (s); i--)
29 #define dbg(x) std::cout << x << std::endl
30 #define dbg2(x, y) std::cout << x << " " << y << std::endl
31 #define clr(x, i) memset(x, (i), sizeof(x))
32 #define maximize(x, y) x = max((x), (y))
33 #define minimize(x, y) x = min((x), (y))
34 using namespace std;
35 typedef long long ll;
36 const int int_inf = 0x3f3f3f3f;
37 const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
38 const int INT_INF = (int)((1ll << 31) - 1);
39 const double double_inf = 1e30;
40 const double eps = 1e-14;
41 typedef unsigned long long ul;
42 typedef unsigned int ui;
43 inline int readint(){
44 int x;
45 scanf("%d", &x);
46 return x;
47 }
48 inline int readstr(char *s){
49 scanf("%s", s);
50 return strlen(s);
51 }
52
53 class cmpt{
54 public:
55 bool operator () (const int &x, const int &y) const{
56 return x > y;
57 }
58 };
59
60 int Rand(int x, int o){
61 //if o set, return [1, x], else return [0, x - 1]
62 if(!x) return 0;
63 int tem = (int)((double)rand() / RAND_MAX * x) % x;
64 return o ? tem + 1 : tem;
65 }
66 void data_gen(){
67 srand(time(0));
68 freopen("in.txt", "w", stdout);
69 int kases = 20;
70 printf("%d\n", kases);
71 while(kases--){
72 int sz = 6e4;
73 printf("%d\n", sz);
74 FOR(i, 1, sz) printf("%d ", Rand(sz, 1));
75 printf("\n");
76 }
77 }
78
79 struct cmpx{
80 bool operator () (int x, int y) { return x > y; }
81 };
82 const int maxn = 6e4 + 10;
83 int a[maxn];
84 int n;
85 struct Seg{
86 int l, r, v;
87 }seg[maxn << 2];
88 void build(int u, int l, int r){
89 seg[u].l = l, seg[u].r = r;
90 if(seg[u].r - seg[u].l < 2){
91 seg[u].v = l;
92 return;
93 }
94 int mid = (l + r) >> 1;
95 build(lson, l, mid), build(rson, mid, r);
96 if(a[seg[rson].v] > a[seg[lson].v]) seg[u].v = seg[rson].v;
97 else seg[u].v = seg[lson].v;
98 }
99 int query(int u, int l, int r){
100 if(seg[u].l == l && seg[u].r == r) return seg[u].v;
101 int mid = (seg[u].l + seg[u].r) >> 1;
102 if(r <= mid) return query(lson, l, r);
103 else if(l >= mid) return query(rson, l, r);
104 int lhs = query(lson, l, mid), rhs = query(rson, mid, r);
105 if(a[rhs] > a[lhs]) return rhs;
106 return lhs;
107 }
108 int maxi;
109 vector<int> pos[maxn];
110 int idx[maxn];
111 void init(){
112 maxi = -1;
113 FOR(i, 1, n) maximize(maxi, a[i]);
114 FOR(i, 1, maxi) pos[i].clear();
115 FOR(i, 1, n){
116 int sz = pos[a[i]].size();
117 idx[i] = sz;
118 pos[a[i]].pb(i);
119 }
120 }
121 ll z[maxn];
122 ll c[maxn], d[maxn], e[maxn << 1];
123 int k;
124 const double PI = 2 * asin(1.);
125 struct Complex{
126 double x, y;
127 Complex(double x = 0, double y = 0) : x(x), y(y) {}
128 };
129 Complex operator + (const Complex &lhs, const Complex &rhs){
130 return Complex(lhs.x + rhs.x, lhs.y + rhs.y);
131 }
132 Complex operator - (const Complex &lhs, const Complex &rhs){
133 return Complex(lhs.x - rhs.x, lhs.y - rhs.y);
134 }
135 Complex operator * (const Complex &lhs, const Complex &rhs){
136 double tl = lhs.x * rhs.x, tr = lhs.y * rhs.y, tt = (lhs.x + lhs.y) * (rhs.x + rhs.y);
137 return Complex(lhs.x * rhs.x - lhs.y * rhs.y, lhs.x * rhs.y + lhs.y * rhs.x);
138 }
139 Complex w[2][maxn << 2], x[maxn << 2], y[maxn << 2];
140
141 void fft(Complex x[], int k, int v){
142 int i, j, l;
143 Complex tem;
144 for(i = j = 0; i < k; i++){
145 if(i > j) tem = x[i], x[i] = x[j], x[j] = tem;
146 for(l = k >> 1; (j ^= l) < l; l >>= 1) ;
147 }
148 for(i = 2; i <= k; i <<= 1) for(j = 0; j < k; j += i) for(l = 0; l < i >> 1; l++){
149 tem = x[j + l + (i >> 1)] * w[v][k / i * l];
150 x[j + l + (i >> 1)] = x[j + l] - tem;
151 x[j + l] = x[j + l] + tem;
152 }
153 }
154 int tot;
155 void solve(int l, int r){
156 if(l >= r) return;
157 if(r - l == 1){
158 ++z[1];
159 return;
160 }
161 int p = query(1, l, r);
162 int tem = idx[p];
163 int sz = pos[a[p]].size();
164 k = 0;
165 c[k++] = p - l + 1;
166 while(tem + 1 < sz && pos[a[p]][tem + 1] < r) c[k++] = pos[a[p]][tem + 1] - pos[a[p]][tem], ++tem;
167 c[k++] = r - pos[a[p]][tem];
168 ROF(i, k - 1, 0) d[i] = c[k - 1 - i];
169 int len;
170 for(len = 1; len < (k << 1); len <<= 1) ;
171 FOR(i, 0, len) w[1][len - i] = w[0][i] = Complex(cos(PI * 2 * i / len), sin(PI * 2 * i / len));
172 FOR(i, 0, k - 1) x[i] = Complex(c[i], 0);
173 FOR(i, k, len - 1) x[i] = Complex(0, 0);
174 fft(x, len, 0);
175 FOR(i, 0, k - 1) y[i] = Complex(d[i], 0);
176 FOR(i, k, len - 1) y[i] = Complex(0, 0);
177 fft(y, len, 0);
178 FOR(i, 0, len - 1) x[i] = x[i] * y[i];
179 fft(x, len, 1);
180 FOR(i, 0, 2 * k - 2) e[i] = (ll)(x[i].x / len + .5);
181 FOR(i, 0, k - 2) z[k - 1 - i] += e[i];
182 tem = idx[p];
183 solve(l, p);
184 while(tem + 1 < sz && pos[a[p]][tem + 1] < r) solve(pos[a[p]][tem] + 1, pos[a[p]][tem + 1]), ++tem;
185 solve(pos[a[p]][tem] + 1, r);
186 }
187 int main(){
188 //data_gen(); return 0;
189 //C(); return 0;
190 int debug = 0;
191 if(debug) freopen("in.txt", "r", stdin);
192 //freopen("out.txt", "w", stdout);
193 int T = readint();
194 while(T--){
195 n = readint();
196 FOR(i, 1, n) a[i] = readint();
197 build(1, 1, n + 1);
198 init();
199 clr(z, 0);
200 tot = 0;
201 solve(1, n + 1);
202 ll ans = 0;
203 FOR(i, 1, n) ans += z[i] ^ i;
204 printf("%lld\n", ans);
205 }
206 return 0;
207 }
code：
时间： 2024-08-06 07:50:23