题目分析

题目要求计算字符串所有前缀的神秘度，即统计满足条件的三元组 ((i,j,\text{len})) 的贡献和。核心在于高效识别字符串中的重复子串（重复运行），并计算每个重复结构对神秘度的贡献。

解题思路

1. 重复运行（Run）识别：利用后缀数组（SA）和Lyndon分解找到字符串中的所有重复运行（Run），每个Run表示一个具有最小周期 (p) 的最长重复区间 ([u, v))。
2. 贡献计算：对于每个Run，计算其内部所有合法三元组的贡献。具体来说，对于周期为 (p)、长度为 (L) 的Run，枚举子串长度 (\text{len})，统计满足条件的 ((i,j)) 对数并累加贡献。
3. 前缀和累加：将每个位置的贡献累加，得到所有前缀的神秘度。

代码实现（带注释）

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

// 输出重载，方便调试
template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) {
    return os << "(" << a.first << ", " << a.second << ")";
};
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) {
    const int sz = as.size();
    os << "[";
    for (int i = 0; i < sz; ++i) {
        if (i >= 256) { os << ", ..."; break; }
        if (i > 0) os << ", ";
        os << as[i];
    }
    return os << "]";
}
template <class T> void pv(T a, T b) {
    for (T i = a; i != b; ++i) cerr << *i << " ";
    cerr << endl;
}
template <class T> bool chmin(T &t, const T &f) {
    if (t > f) { t = f; return true; }
    return false;
}
template <class T> bool chmax(T &t, const T &f) {
    if (t < f) { t = f; return true; }
    return false;
}
#define COLOR(s) ("\x1b[" s "m")

// 模数类，支持模运算
template <unsigned M_> struct ModInt {
    static constexpr unsigned M = M_;
    unsigned x;
    constexpr ModInt() : x(0U) {}
    constexpr ModInt(unsigned x_) : x(x_ % M) {}
    constexpr ModInt(unsigned long long x_) : x(x_ % M) {}
    constexpr ModInt(int x_) : x(((x_ %= static_cast<int>(M)) < 0) ? (x_ + static_cast<int>(M)) : x_) {}
    constexpr ModInt(long long x_) : x(((x_ %= static_cast<long long>(M)) < 0) ? (x_ + static_cast<long long>(M)) : x_) {}
    ModInt &operator+=(const ModInt &a) {
        x = ((x += a.x) >= M) ? (x - M) : x;
        return *this;
    }
    ModInt &operator-=(const ModInt &a) {
        x = ((x -= a.x) >= M) ? (x + M) : x;
        return *this;
    }
    ModInt &operator*=(const ModInt &a) {
        x = (static_cast<unsigned long long>(x) * a.x) % M;
        return *this;
    }
    ModInt &operator/=(const ModInt &a) {
        return (*this *= a.inv());
    }
    ModInt pow(long long e) const {
        if (e < 0) return inv().pow(-e);
        ModInt a = *this, b = 1U;
        for (; e; e >>= 1) {
            if (e & 1) b *= a;
            a *= a;
        }
        return b;
    }
    ModInt inv() const {
        unsigned a = M, b = x;
        int y = 0, z = 1;
        for (; b;) {
            const unsigned q = a / b;
            const unsigned c = a - q * b;
            a = b; b = c;
            const int w = y - static_cast<int>(q) * z;
            y = z; z = w;
        }
        assert(a == 1U);
        return ModInt(y);
    }
    ModInt operator+() const { return *this; }
    ModInt operator-() const {
        ModInt a; a.x = x ? (M - x) : 0U;
        return a;
    }
    ModInt operator+(const ModInt &a) const { return (ModInt(*this) += a); }
    ModInt operator-(const ModInt &a) const { return (ModInt(*this) -= a); }
    ModInt operator*(const ModInt &a) const { return (ModInt(*this) *= a); }
    ModInt operator/(const ModInt &a) const { return (ModInt(*this) /= a); }
    template <class T> friend ModInt operator+(T a, const ModInt &b) { return (ModInt(a) += b); }
    template <class T> friend ModInt operator-(T a, const ModInt &b) { return (ModInt(a) -= b); }
    template <class T> friend ModInt operator*(T a, const ModInt &b) { return (ModInt(a) *= b); }
    template <class T> friend ModInt operator/(T a, const ModInt &b) { return (ModInt(a) /= b); }
    explicit operator bool() const { return x; }
    bool operator==(const ModInt &a) const { return (x == a.x); }
    bool operator!=(const ModInt &a) const { return (x != a.x); }
    friend std::ostream &operator<<(std::ostream &os, const ModInt &a) { return os << a.x; }
};

constexpr unsigned MO = 1000000007;
using Mint = ModInt<MO>;

// SA-IS算法：构造后缀数组
template <class String> vector<int> suffixArrayRec(const String &as) {
    const int n = as.size();
    if (n == 0) return {};
    const auto minmaxA = minmax_element(as.begin(), as.end());
    const auto minA = *minmaxA.first, maxA = *minmaxA.second;
    if (static_cast<unsigned long long>(maxA) - minA >= static_cast<unsigned long long>(n)) {
        vector<int> us(n);
        for (int u = 0; u < n; ++u) us[u] = u;
        std::sort(us.begin(), us.end(), [&](int u, int v) -> bool { return (as[u] < as[v]); });
        int b = 0;
        vector<int> bs(n, 0);
        for (int i = 1; i < n; ++i) {
            if (as[us[i - 1]] != as[us[i]]) ++b;
            bs[us[i]] = b;
        }
        return suffixArrayRec(bs);
    }
    const int sigma = maxA - minA + 1;
    vector<int> pt(sigma + 1, 0), poss(sigma);
    for (int u = 0; u < n; ++u) ++pt[as[u] - minA + 1];
    for (int a = 0; a < sigma; ++a) pt[a + 1] += pt[a];
    vector<bool> cmp(n);
    cmp[n - 1] = false;
    for (int u = n - 1; --u >= 0;) {
        cmp[u] = (as[u] != as[u + 1]) ? (as[u] < as[u + 1]) : cmp[u + 1];
    }
    int nn = 0;
    vector<int> ids(n, 0);
    int last = n;
    vector<int> nxt(n, 0);
    vector<int> us(n, 0);
    memcpy(poss.data(), pt.data() + 1, sigma * sizeof(int));
    for (int u = n - 1; --u >= 1;)
        if (!cmp[u - 1] && cmp[u]) {
            ids[u] = ++nn;
            nxt[u] = last;
            last = u;
            us[--poss[as[u] - minA]] = u;
        }
    memcpy(poss.data(), pt.data(), sigma * sizeof(int));
    us[poss[as[n - 1] - minA]++] = n - 1;
    for (int i = 0; i < n; ++i) {
        const int u = us[i];
        if (u && !cmp[u - 1])
            us[poss[as[u - 1] - minA]++] = u - 1;
    }
    memcpy(poss.data(), pt.data() + 1, sigma * sizeof(int));
    for (int i = n; --i >= 0;) {
        const int u = us[i];
        if (u && cmp[u - 1])
            us[--poss[as[u - 1] - minA]] = u - 1;
    }
    if (nn) {
        int vsLen = 0;
        vector<int> vs(nn);
        for (const int u : us)
            if (ids[u])
                vs[vsLen++] = u;
        int b = 0;
        vector<int> bs(nn, 0);
        for (int j = 1; j < nn; ++j) {
            int v1 = vs[j - 1], v0 = vs[j];
            const int w1 = nxt[v1], w0 = nxt[v0];
            if (w1 - v1 != w0 - v0) {
                ++b;
            } else {
                for (; ; ++v1, ++v0) {
                    if (v1 == n) { ++b; break; }
                    if (as[v1] != as[v0]) { ++b; break; }
                    if (v1 == w1) break;
                }
            }
            bs[nn - ids[vs[j]]] = b;
        }
        for (int u = 0; u < n; ++u)
            if (ids[u])
                vs[nn - ids[u]] = u;
        const auto sub = suffixArrayRec(bs);
        memset(us.data(), 0, n * sizeof(int));
        memcpy(poss.data(), pt.data() + 1, sigma * sizeof(int));
        for (int j = nn; --j >= 0;) {
            const int u = vs[sub[j]];
            us[--poss[as[u] - minA]] = u;
        }
        memcpy(poss.data(), pt.data(), sigma * sizeof(int));
        us[poss[as[n - 1] - minA]++] = n - 1;
        for (int i = 0; i < n; ++i) {
            const int u = us[i];
            if (u && !cmp[u - 1])
                us[poss[as[u - 1] - minA]++] = u - 1;
        }
        memcpy(poss.data(), pt.data() + 1, sigma * sizeof(int));
        for (int i = n; --i >= 0;) {
            const int u = us[i];
            if (u && cmp[u - 1])
                us[--poss[as[u - 1] - minA]] = u - 1;
        }
    }
    return us;
}

// 后缀数组结构体，包含SA、rank、LCP
struct SuffixArray {
    int n;
    bool rmq;
    vector<int> us, su, hs;
    SuffixArray() : n(0), rmq(false) {}
    SuffixArray(const string &as, bool rmq_) : rmq(rmq_) { build(as); }
    SuffixArray(const vector<int> &as, bool rmq_) : rmq(rmq_) { build(as); }
    SuffixArray(const vector<long long> &as, bool rmq_) : rmq(rmq_) { build(as); }
    template <class String> void build(const String &as) {
        n = as.size();
        us = suffixArrayRec(as);
        su.resize(n + 1);
        for (int i = 0; i < n; ++i) su[us[i]] = i;
        su[n] = -1;
        hs.assign(n, 0);
        for (int h = 0, u = 0; u < n; ++u)
            if (su[u]) {
                for (int v = us[su[u] - 1]; v + h < n && as[v + h] == as[u + h]; ++h) {}
                hs[su[u]] = h;
                if (h) --h;
            }
        if (rmq) {
            const int logN = n ? (31 - __builtin_clz(n)) : 0;
            hs.resize((logN + 1) * n - (1 << logN) + 1);
            for (int e = 0; e < logN; ++e) {
                int *hes = hs.data() + e * n;
                for (int i = 0; i <= n - (1 << (e + 1)); ++i) {
                    hes[n + i] = min(hes[i], hes[i + (1 << e)]);
                }
            }
        }
    }
    // 查询两个后缀的最长公共前缀
    inline int lcp(int u, int v) const {
        if (u == v) return n - u;
        int i = su[u], j = su[v];
        if (i > j) swap(i, j);
        const int e = 31 - __builtin_clz(j - i);
        return min(hs[e * n + i + 1], hs[e * n + j + 1 - (1 << e)]);
    }
};

// Lyndon分解：求最长Lyndon前缀
template <class String>
vector<pair<int, int>> lyndonSuffix(const String &as) {
    const int n = as.size();
    vector<pair<int, int>> pqs(n + 1);
    pqs[0] = make_pair(0, 0);
    for (int u = 0; u < n;) {
        for (int p = 1, q = 1, r = 0, v = u + 1; ; ++v) {
            pqs[v] = (r != 0) ? pqs[u + r] : make_pair(p, q);
            if (v == n || as[v - p] > as[v]) {
                u = v - r;
                break;
            } else if (as[v - p] < as[v]) {
                p = v + 1 - u;
                q = 1;
                r = 0;
            } else {
                if (++r == p) { ++q; r = 0; }
            }
        }
    }
    return pqs;
}

// 求最长Lyndon前缀
template <class String>
vector<int> lyndonPrefix(const String &as, const SuffixArray &sa) {
    const int n = as.size();
    int stackLen = 0;
    vector<int> stack(n + 1);
    vector<int> vs(n);
    for (int u = 0; u <= n; ++u) {
        for (; stackLen > 0 && sa.su[stack[stackLen - 1]] > sa.su[u]; --stackLen) {
            vs[stack[stackLen - 1]] = u;
        }
        stack[stackLen++] = u;
    }
    return vs;
}

// 反转字符串的最长Lyndon前缀
template <class String>
vector<int> lyndonPrefixInverted(const String &as, const SuffixArray &sa) {
    assert(sa.rmq);
    const int n = as.size();
    int stackLen = 0;
    vector<int> stack(n + 1);
    vector<int> vs(n);
    for (int u = 0; u <= n; ++u) {
        for (; stackLen > 0; --stackLen) {
            const int uu = stack[stackLen - 1];
            const int l = sa.lcp(uu, u);
            if (u + l < n && as[uu + l] > as[u + l]) break;
            vs[uu] = u;
        }
        stack[stackLen++] = u;
    }
    return vs;
}

// 寻找所有重复运行（Run）
template <class String>
vector<pair<int, pair<int, int>>> repetitions(const String &as, const SuffixArray &sa) {
    assert(sa.rmq);
    const int n = as.size();
    if (n == 0) return {};
    String asRev = as;
    std::reverse(asRev.begin(), asRev.end());
    const SuffixArray saRev(asRev, /*rmq=*/true);
    const vector<int> vs = lyndonPrefix(as, sa);
    const vector<int> vsInverted = lyndonPrefixInverted(as, sa);
    vector<pair<int, pair<int, int>>> runs;
    for (int u = 0; u < n; ++u) {
        {
            const int v = vs[u];
            const int p = v - u, uu = u - saRev.lcp(n - u, n - v), vv = v + sa.lcp(u, v);
            if (vv - uu >= 2 * p)
                runs.emplace_back(p, make_pair(uu, vv));
        }
        if (vs[u] != vsInverted[u]) {
            const int v = vsInverted[u];
            const int p = v - u, uu = u - saRev.lcp(n - u, n - v), vv = v + sa.lcp(u, v);
            if (vv - uu >= 2 * p)
                runs.emplace_back(p, make_pair(uu, vv));
        }
    }
    const int runsLen = runs.size();
    auto runsWork = runs;
    vector<int> pt(n + 1, 0);
    for (int i = 0; i < runsLen; ++i) ++pt[runs[i].second.first];
    for (int u = 0; u < n - 1; ++u) pt[u + 1] += pt[u];
    for (int i = runsLen; --i >= 0;) runsWork[--pt[runs[i].second.first]] = runs[i];
    memset(pt.data() + 1, 0, n * sizeof(int));
    for (int i = 0; i < runsLen; ++i) ++pt[runsWork[i].first];
    for (int p = 1; p < n; ++p) pt[p + 1] += pt[p];
    for (int i = runsLen; --i >= 0;) runs[--pt[runsWork[i].first]] = runsWork[i];
    runs.erase(std::unique(runs.begin(), runs.end()), runs.end());
    return runs;
}

int N;
char S[500'010];

int main() {
    for (; ~scanf("%s", S);) {
        N = strlen(S);
        vector<Mint> ans(N + 1, 0);
        const auto runs = repetitions(string(S)); // 找到所有重复运行

        // 遍历每个重复运行，计算贡献
        for (const auto &run : runs) {
            const int p = run.first;       // 最小周期
            const int i = run.second.first; // 起始位置
            const int j = run.second.second; // 结束位置
            Mint sum = 0;

            // 枚举子串长度l
            for (int l = 2 * p + 1; l <= j - i; ++l) {
                // 计算满足条件的k的数量及贡献
                const Int limK = (l - 1) / (2 * p);
                ans[i + l] += sum += (l * limK - p * (limK * (limK + 1) / 2));
            }
        }

        // 前缀和累加，得到每个前缀的神秘度
        for (int i = 1; i <= N; ++i)
            ans[i] += ans[i - 1];

        // 输出结果
        for (int i = 1; i <= N; ++i)
            printf("%u\n", ans[i].x);
    }
    return 0;
}

代码解释

1. 模数类：封装模运算操作，支持加法、乘法、逆元等，确保计算过程中数值不溢出。
2. 后缀数组：使用SA-IS算法构造后缀数组，高效计算后缀排序、rank值和LCP（最长公共前缀）。
3. Lyndon分解：找到字符串中的Lyndon前缀，用于识别重复运行。
4. 重复运行识别：通过后缀数组和Lyndon分解找到所有重复运行（Run），每个Run表示一个具有最小周期的最长重复区间。
5. 贡献计算：对每个Run，枚举子串长度，统计满足条件的三元组贡献，并累加到对应前缀位置。
6. 前缀和输出：将每个位置的贡献累加，输出所有前缀的神秘度。