OI
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「BJ模拟」「CodeChef」「BZOJ-3509」COUNTARI 等差数列-暴力/FFT

给定 $n$ 个整数 $A_1,A_2,A_n$ ，求有多少个三元组 $(i,j,k)$ 满足 $1 \leq i < j < k \leq n$ 且 $A_j-A_i=A_k-A_j$ 。

链接

CodeChef-COUNTARI
BZOJ-3509

题解

此题可以暴力踩标程，用前后两个桶来统计，复杂度 $O(n max(a))$，这个暴力相信大家都会，这时我们只需要循环展开 $15$ 层，刺激 $CPU$ 并发，就拿下了 $BZOJ$ 和 $CC$ rank1….

然而被模拟赛的垃圾CPU坑T了….

暴力

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/*
 * created by xehoth on 18-03-2017
 * 不能并行?
 * CPU 并发!(好虚啊)
 * 假设不卡 CPU 并发...
 */
#include <bits/stdc++.h>

inline char read() {
    static const int IN_LEN = 1000000;
    static char buf[IN_LEN], *s, *t;
    if (s == t) {
        t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
        if (s == t) return -1;
    }
    return *s++;
}

template<class T>
inline bool read(T &x) {
    static bool iosig;
    static char c;
    for (iosig = false, c = read(); !isdigit(c); c = read()) {
        if (c == '-') iosig = true;
        if (c == -1) return false;
    }
    for (x = 0; isdigit(c); c = read())
        x = (x + (x << 2) << 1) + (c ^ '0');
    if (iosig) x = -x;
    return true;
}

const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *oh = obuf;

inline void print(char c) {
    if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
    *oh++ = c;
}

template<class T>
inline void print(T x) {
    static int buf[30], cnt;
    if (x == 0) {
        print('0');
    } else {
        if (x < 0) print('-'), x = -x;
        for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
        while (cnt) print((char)buf[cnt--]);
    }
}

inline void flush() {
    fwrite(obuf, 1, oh - obuf, stdout);
}

typedef unsigned long long ull;
typedef unsigned int uint;
#define long long long

const int MAXN = 100000;
const int MAX_VAL = 30000;
uint v[MAXN], bucL[MAX_VAL + 1], bucR[MAX_VAL + 1];

namespace Concurrent {

/*concurrent*/
inline void concurrentSolve() {
    /* real register %esp %ebp %eax */
    register int i, n, tmp;
    read(n);
    for (i = 0; i < n; i++)
        read(tmp), bucR[MAX_VAL - (v[i] = tmp)]++;

    bucR[MAX_VAL - v[0]]--;
    register int minL = v[0], maxL = v[0];

    register long ans = 0;

    n--;
    for (i = 1; i < n; i++) {
        register int last = v[i - 1], cur = v[i];
        if (last < minL) minL = last;
        else if (last > maxL) maxL = last;

        bucL[last]++;
        bucR[MAX_VAL - cur]--;

        register int bufx = cur << 1, low = std::max(minL, bufx - MAX_VAL),
                     high = std::min(maxL, bufx - 1);
        /*CPU 并发优化*/
        register uint tmp = 0, *p1 = bucL + low, *pr = bucL + high - 14,
                      *p2 = bucR + MAX_VAL - bufx + low;
        /*循环展开 + 刺激并发*/
        while (p1 <= pr) {
            tmp += (*p1) * (*p2) + (*(p1 + 1)) * (*(p2 + 1)) + (*(p1 + 2)) *
                   (*(p2 + 2)) + (*(p1 + 3)) * (*(p2 + 3)) + (*(p1 + 4)) * (*(p2 + 4))
                   + (*(p1 + 5)) * (*(p2 + 5)) + (*(p1 + 6)) * (*(p2 + 6)) + (*(p1 + 7))
                   * (*(p2 + 7)) + (*(p1 + 8)) * (*(p2 + 8)) + (*(p1 + 9)) * (*(p2 + 9))
                   + (*(p1 + 10)) * (*(p2 + 10)) + (*(p1 + 11)) * (*(p2 + 11))
                   + (*(p1 + 12)) * (*(p2 + 12)) + (*(p1 + 13)) * (*(p2 + 13))
                   + (*(p1 + 14)) * (*(p2 + 14));

            p1 += 15, p2 += 15;
        }
        while (p1 <= bucL + high) tmp += (*(p1++)) * (*(p2++));
        ans += tmp;
    }

    print(ans);
}

}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    Concurrent::concurrentSolve();
    flush();
    return 0;
}

FFT

这份是分块的，部分暴力部分正解

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/*
 * created by xehoth on 18-03-2017
 * 不能并行?
 * CPU 并发!(好虚啊)
 * 假设不卡 CPU 并发...
 ******************************
 * 垃圾CPU
 * 就这 OJ 被卡
 */
#include <bits/stdc++.h>

inline char read() {
    static const int IN_LEN = 1000000;
    static char buf[IN_LEN], *s, *t;
    if (s == t) {
        t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
        if (s == t) return -1;
    }
    return *s++;
}

template<class T>
inline bool read(T &x) {
    static bool iosig;
    static char c;
    for (iosig = false, c = read(); !isdigit(c); c = read()) {
        if (c == '-') iosig = true;
        if (c == -1) return false;
    }
    for (x = 0; isdigit(c); c = read())
        x = (x + (x << 2) << 1) + (c ^ '0');
    if (iosig) x = -x;
    return true;
}

const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *oh = obuf;

inline void print(char c) {
    if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
    *oh++ = c;
}

template<class T>
inline void print(T x) {
    static int buf[30], cnt;
    if (x == 0) {
        print('0');
    } else {
        if (x < 0) print('-'), x = -x;
        for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
        while (cnt) print((char)buf[cnt--]);
    }
}

inline void flush() {
    fwrite(obuf, 1, oh - obuf, stdout);
}

typedef unsigned long long ull;
typedef unsigned int uint;
#define long long long

const int MAXN = 100000;
const int MAX_VAL = 30000;
uint v[MAXN], bucL[MAX_VAL + 1], bucR[MAX_VAL + 1];

namespace Concurrent {

/*concurrent*/
inline void concurrentSolve(int n, int num1) {
    /* real register %esp %ebp %eax */
    register int i, tmp;
    bucR[MAX_VAL - (v[0] = num1)]++;
    for (i = 1; i < n; i++)
        read(tmp), bucR[MAX_VAL - (v[i] = tmp)]++;

    bucR[MAX_VAL - v[0]]--;
    register int minL = v[0], maxL = v[0];

    register long ans = 0;

    n--;
    for (i = 1; i < n; i++) {
        register int last = v[i - 1], cur = v[i];
        if (last < minL) minL = last;
        else if (last > maxL) maxL = last;

        bucL[last]++;
        bucR[MAX_VAL - cur]--;

        register int bufx = cur << 1, low = std::max(minL, bufx - MAX_VAL),
                     high = std::min(maxL, bufx - 1);
        /*CPU 并发优化*/
        register uint tmp = 0, *p1 = bucL + low, *pr = bucL + high - 14,
                      *p2 = bucR + MAX_VAL - bufx + low;
        /*循环展开 + 刺激并发*/
        while (p1 <= pr) {
            tmp += (*p1) * (*p2) + (*(p1 + 1)) * (*(p2 + 1)) + (*(p1 + 2)) *
                   (*(p2 + 2)) + (*(p1 + 3)) * (*(p2 + 3)) + (*(p1 + 4)) * (*(p2 + 4))
                   + (*(p1 + 5)) * (*(p2 + 5)) + (*(p1 + 6)) * (*(p2 + 6)) + (*(p1 + 7))
                   * (*(p2 + 7)) + (*(p1 + 8)) * (*(p2 + 8)) + (*(p1 + 9)) * (*(p2 + 9))
                   + (*(p1 + 10)) * (*(p2 + 10)) + (*(p1 + 11)) * (*(p2 + 11))
                   + (*(p1 + 12)) * (*(p2 + 12)) + (*(p1 + 13)) * (*(p2 + 13))
                   + (*(p1 + 14)) * (*(p2 + 14));

            p1 += 15, p2 += 15;
        }
        while (p1 <= bucL + high) tmp += (*(p1++)) * (*(p2++));
        ans += tmp;
    }

    print(ans);
}

}

namespace FastFourierTransform {

struct Complex {
    double r, i;

    Complex(double r = 0, double i = 0) : r(r), i(i) {}

    inline Complex operator+(const Complex &x) const {
        return Complex(r + x.r, i + x.i);
    }

    inline Complex operator-(const Complex &x) const {
        return Complex(r - x.r, i - x.i);
    }

    inline Complex operator*(const Complex &x) const {
        return Complex(r * x.r - i * x.i, r * x.i + i * x.r);
    }

    inline Complex conj() {
        return Complex(r, -i);
    }
};

const double PI = acos(-1);

inline void fft(Complex *a, const int n, const int f) {
    for (register int i = 0, j = 0; i < n; i++) {
        if (i > j) std::swap(a[i], a[j]);
        for (register int k = n >> 1; (j ^= k) < k; k >>= 1);
    }
    for (register int i = 1; i < n; i <<= 1) {
        Complex wn(cos(PI / i), f * sin(PI / i));
        for (register int j = 0; j < n; j += i << 1) {
            Complex w(1, 0);
            for (register int k = 0; k < i; k++, w = w * wn) {
                Complex x = a[j + k], y = w * a[i + j + k];
                a[j + k] = x + y, a[i + j + k] = x - y;
            }
        }
    }
    if (f == -1) for (register int i = 0; i < n; i++) a[i].r /= n;
}

const int MAXN = 700010;
Complex a[MAXN], b[MAXN];
int num[MAXN], l[MAXN], r[MAXN], st[MAXN], ed[MAXN];
int n;

inline void solve(int n, int num1) {
    num[1] = num1;
    register int max = num1;
    r[num1]++;
    for (register int i = 2; i <= n; i++)
        read(num[i]), max = std::max(max, num[i]), r[num[i]]++;
    max++;
    max = (max << 1) - 1;
    for (FastFourierTransform::n = 1; FastFourierTransform::n <= max;
            FastFourierTransform::n <<= 1);
    register int size = 2000;
    register int m = (n - 1) / size + 1;
    for (register int i = 1; i <= m; i++) {
        st[i] = ed[i - 1] + 1;
        ed[i] = i * size;
    }
    ed[m] = n;
    register long ans = 0;
    for (register int i = 1; i <= m; i++) {
        for (register int j = st[i]; j <= ed[i]; j++)
            r[num[j]]--;
        for (register int j = 0; j < FastFourierTransform::n; j++)
            b[j] = Complex(l[j], r[j]);
        fft(b, FastFourierTransform::n, 1);
        for (register int i = 0, j; i < FastFourierTransform::n; i++) {
            j = (FastFourierTransform::n - i) & 
            (FastFourierTransform::n - 1), 
            a[i] = (b[i] * b[i] - (b[j] * b[j]).conj()) 
            * Complex(0, -0.25);
        }
        fft(a, FastFourierTransform::n, -1);
        for (register int j = st[i]; j <= ed[i]; j++)
            ans += ((long)(a[2 * num[j]].r + 0.5));
        for (register int j = st[i]; j <= ed[i]; j++) {
            for (register int k = st[i]; k < j; k++)
                if (2 * num[j] - num[k] >= 0)
                    ans += r[2 * num[j] - num[k]];
            for (register int k = j + 1; k <= ed[i]; k++)
                if (2 * num[j] - num[k] >= 0)
                    ans += l[2 * num[j] - num[k]];
            l[num[j]]++;
        }
    }
    print(ans);
}
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    register int n, num1;
    read(n), read(num1);
    if ((n <= 100000 && num1 <= 50 && num1 != 1) || (n <= 5000)) {
        Concurrent::concurrentSolve(n, num1);
    } else {
        FastFourierTransform::solve(n, num1);
    }
    flush();
    return 0;
}
#
后缀自动机应用总结
「SPOJ-1811」LCS-后缀自动机

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