c/c++语言开发共享BZOJ3509: [CodeChef] COUNTARI(生成函数 分块)

题意

sol

这都能分块。。。。

首先移一下项，变为统计多少(i < j < k)，满足(2a[j] = a[i] + a[k])

发现(a[i] leqslant 30000)，那么有一种暴力思路是枚举(j)，对于之前出现过的数构造一个生成函数，对于之后出现过的数构造一个生成函数，求一下第(2a[j])项的值。复杂度(o(nvlogv))

每次枚举(j)暴力卷积显然太zz了，我们可以分一下块，对于每一块之前之后的数分别构造生成函数暴力卷积算，对于块内的直接暴力(这里的暴力不只是统计块内的((i, j, k))，还要考虑(j, k)在块内(i)在块外，以及(i, j)在块内，(k)在块外的情况，但都是可以暴力的)

如果分成(b)块的话，复杂度是(frac{n}{b} vlogv + b^2)，假设(n)与(v)同阶的话，(b)大概取(nlogn)是最优的。此时复杂度为(o(n sqrt{nlogn}))

下面的代码在原bzoj上可能会t

#include<bits/stdc++.h>  #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second #define ll long long  #define ull unsigned long long  #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 1e6 + 10, inf = 1e9 + 1; const double eps = 1e-9, pi = acos(-1); inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } namespace poly {     int rev[maxn], gpow[maxn], a[maxn], b[maxn], c[maxn], lim, inv2;     const int g = 3, mod = 1004535809, mod2 = 1004535808;     template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}     template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}     template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}     template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}     template <typename a, typename b> inline bool chmax(a &x, b y) {return x < y ? x = y, 1 : 0;}     template <typename a, typename b> inline bool chmin(a &x, b y) {return x > y ? x = y, 1 : 0;}     int fp(int a, int p, int p = mod) {         int base = 1;         for(; p > 0; p >>= 1, a = 1ll * a * a % p) if(p & 1) base = 1ll * base *  a % p;         return base;     }     int inv(int x) {         return fp(x, mod - 2);     }     int getlen(int x) {         int lim = 1;         while(lim < x) lim <<= 1;         return lim;     }     void init(/*int p,*/ int lim) {         inv2 = fp(2, mod - 2);         for(int i = 1; i <= lim; i++) gpow[i] = fp(g, (mod - 1) / i);     }     void ntt(int *a, int lim, int opt) {         int len = 0; for(int n = 1; n < lim; n <<= 1) ++len;          for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));         for(int i = 0; i <= lim; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);         for(int mid = 1; mid < lim; mid <<= 1) {             int wn = gpow[mid << 1];             for(int i = 0; i < lim; i += (mid << 1)) {                 for(int j = 0, w = 1; j < mid; j++, w = mul(w, wn)) {                     int x = a[i + j], y = mul(w, a[i + j + mid]);                     a[i + j] = add(x, y), a[i + j + mid] = add(x, -y);                 }             }         }         if(opt == -1) {             reverse(a + 1, a + lim);             int inv = fp(lim, mod - 2);             for(int i = 0; i <= lim; i++) mul2(a[i], inv);         }     }     void mul(int *a, int *b, int n, int m) {         memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));         int lim = 1, len = 0;          while(lim <= n + m) len++, lim <<= 1;         for(int i = 0; i <= n; i++) a[i] = a[i];          for(int i = 0; i <= m; i++) b[i] = b[i];         ntt(a, lim, 1); ntt(b, lim, 1);         for(int i = 0; i <= lim; i++) b[i] = mul(b[i], a[i]);         ntt(b, lim, -1);         for(int i = 0; i <= n + m; i++) b[i] = b[i];         memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));     } }; using namespace poly;  int n, a[maxn], mx, block, ll[maxn], rr[maxn], belong[maxn], mxblock, num[maxn]; ll solve1(int l, int r) {     for(int i = 1; i < l; i++) num[a[i]]++;      ll ret = 0;     for(int j = l; j <= r; j++)          for(int k = j + 1; k <= r; k++)              if(2 * a[j] - a[k] >= 0) ret += num[2 * a[j] - a[k]];     for(int i = 1; i < l; i++) num[a[i]]--;      for(int i = n; i > r; i--) num[a[i]]++;     for(int j = r; j >= l; j--)          for(int k = j - 1; k >= l; k--)              if(2 * a[j] - a[k] >= 0) ret += num[2 * a[j] - a[k]];     for(int i = n; i > r; i--) num[a[i]]--;          for(int j = l; j <= r; j++) {         for(int i = j - 1; i >= l; i--) num[a[i]]++;         for(int k = j + 1; k <= r; k++)              if(2 * a[j] - a[k] >= 0) ret += num[2 * a[j] - a[k]];         for(int i = j - 1; i >= l; i--) num[a[i]]--;     }     return ret; } int ta[maxn], tb[maxn], lim; ll solve2(int l, int r) {     memset(ta, 0, sizeof(ta));     memset(tb, 0, sizeof(tb));     for(int i = l - 1; i >= 1; i--) ta[a[i]]++;     for(int i = r + 1; i <= n; i++) tb[a[i]]++;     mul(ta, tb, mx, mx); ll ret = 0;     for(int i = l; i <= r; i++) ret += tb[2 * a[i]];     return ret; } signed main() {     //freopen("a.in", "r", stdin);       n = read(); block = sqrt(8 *  n * log2(n));      memset(ll, 0x3f, sizeof(ll));     for(int i = 1; i <= n; i++) {         belong[i] = (i - 1) / block + 1; chmax(mxblock, belong[i]);         chmin(ll[belong[i]], i);         chmax(rr[belong[i]], i);         a[i] = read(), chmax(mx, a[i]);     }     lim = getlen(mx); init(4 * lim);     ll ans = 0;     for(int i = 1; i <= mxblock; i++) {         ans += solve1(ll[i], rr[i]);         ans += solve2(ll[i], rr[i]);     }     cout << ans;     return 0; } /* 7 7 0 4 7 0 8 8  */