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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#211778#5051. Namomo SubsequencebigJTL 17ms18648kbC++203.5kb2023-10-12 21:09:352023-10-12 21:09:35

Judging History

你现在查看的是最新测评结果

  • [2023-10-12 21:09:35]
  • 评测
  • 测评结果:TL
  • 用时:17ms
  • 内存:18648kb
  • [2023-10-12 21:09:35]
  • 提交

answer

#include <bits/stdc++.h>
#define dug(x) cerr << "死了吧小丑" << x << endl
using namespace std;
using i64 = long long;
template <class T> bool chmax(T &a, T b)
{
    a > b ? (a = a) : (a = b);
    return a == b;
}
template <class T> bool chmin(T &a, T b)
{
    a > b ? (a = b) : (a = a);
    return a == b;
}
const int M = 998244353, N = 1e6 + 10;
struct Comb
{
#define ll long long
    ll n, p;
    vector<ll> fac, inv; // 阶乘 逆元

    long long qmi(long long k, long long p, long long M)
    {
        long long res = 1;
        while (p)
        {
            if (p & 1)
                res = res * k % M;
            k = k * k % M;
            p >>= 1;
        }
        return res;
    }
    Comb()
    {
    }
    Comb(int a, int b)
    {
        init(a, b);
    }
    void init(int a, int b)
    {
        n = a, p = b;
        fac.resize(n + 1), inv.resize(n + 1);
        fac[0] = 1;
        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;
        inv[n] = qmi(fac[n], p - 2, p);
        for (int i = n; i; i--)
            inv[i - 1] = inv[i] * i % p;
    }
    int C(int n, int m) // 在n里选m
    {
        if (n < m || m < 0)
            return 0;
        return (fac[n] * inv[m] % p) * inv[n - m] % p;
    }
    int lucas(int n, int m) // 要为质数
    {
        if (n < m || m < 0)
            return 0;
        if (m == 0)
            return 1;
        return 1ll * C(n % p, m % p) * lucas(n / p, m / p) % p;
    }
};
Comb co(N, M);
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(15) << fixed;
    int yi[128];
    vector<char> tm(62);
    for (int i = 0; i < 26; i++)
    {
        yi[i + 'A'] = i;
        tm[i] = i + 'A';
        yi[i + 'a'] = i + 26;
        tm[i + 26] = i + 'a';
    }
    for (int i = 0; i < 10; i++)
    {
        yi[i + '0'] = i + 52;
        tm[i + 52] = i + '0';
    }
    string a;
    cin >> a;
    int n = a.size();
    vector<array<i64, 62>> s(n + 1);
    fill(s[0].begin(), s[0].end(), 0);
    vector<vector<int>> mp(62);
    for (int i = 1; i <= n; i++)
    {
        s[i] = s[i - 1];
        s[i][yi[a[i - 1]]]++;
        mp[yi[a[i - 1]]].push_back(i);
    }
    vector<int> sa(n + 1, 0);
    for (int i = 1; i <= n; i++)
    {
        sa[i] = sa[i - 1];
        sa[i] += s[i - 1][yi[a[i - 1]]];
        sa[i] %= M;
    }
    i64 ans = 0;
    for (int i = 0; i < 62; i++)
    {
        if (mp[i].size() <= 1)
            continue;
        for (int j = 0; j < 62; j++)
        {
            if (i == j)
                continue;
            i64 f1 = 0, f2 = 0, dp1 = 0, dp2 = 0, res;
            f2 = s[n][j] - s[mp[i].back()][j];
            // if (tm[i] == 'f' && tm[j] == 'c')
            // {
            //     cout << f2 << endl;
            // }
            for (int z = mp[i].size() - 2; z >= 0; z--)
            {
                dp1 = (dp2 + (s[mp[i][z + 1]][j] - s[mp[i][z]][j]) * f2) % M;
                f1 = ((s[n][j] - s[mp[i][z]][j]) + f2) % M;
                i64 id = mp[i][z] - 1;
                if (id == 1)
                    continue;
                res = co.C(id, 2);
                i64 s1 = s[id][j], s2 = s[id][i];
                res = (res + M - sa[id] - s1 * s2 % M - (s1 + s2) * (id - s1 - s2) % M) % M;
                ans = (ans + res * dp1) % M;
                swap(f1, f2), swap(dp1, dp2);
            }
        }
    }
    cout << ans << endl;
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 13ms
memory: 18640kb

input:

wohaha

output:

1

result:

ok 1 number(s): "1"

Test #2:

score: 0
Accepted
time: 13ms
memory: 18632kb

input:

momomo

output:

0

result:

ok 1 number(s): "0"

Test #3:

score: 0
Accepted
time: 10ms
memory: 18548kb

input:

gshfd1jkhaRaadfglkjerVcvuy0gf

output:

73

result:

ok 1 number(s): "73"

Test #4:

score: 0
Accepted
time: 17ms
memory: 18648kb

input:

retiredMiFaFa0v0

output:

33

result:

ok 1 number(s): "33"

Test #5:

score: -100
Time Limit Exceeded

input:

bcdccccacccabdbdaddcabddbaccaaaaaabaccbbbcbbddabcbccabacdacbcbabccbcbddcdcbcaaadddddccdbabaabcbbcaaadadacdaadbdccbddddabcbaaddbcadadcbcbaaccacabdababaabdccadaddacdcacdaabbadadaddbbcccbcddaccaadbbcaaccccdcacbdbdddbaccaacbcaccaaabccdadddbaabdbcaaccacdcdcbcdddacbcacdbbbdccdddccccabdbacddacbaacbbcaccdcd...

output:


result: