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BZOJ3509: [CodeChef] COUNTARI(生成函数 分块)

2019年03月15日  | 移动技术网IT编程  | 我要评论

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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 
*/

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