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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
#386526#7782. Ursa MinorA_programmerWA 130ms10048kbC++174.4kb2024-04-11 17:58:462024-04-11 17:58:47

Judging History

你现在查看的是最新测评结果

  • [2024-04-11 17:58:47]
  • 评测
  • 测评结果:WA
  • 用时:130ms
  • 内存:10048kb
  • [2024-04-11 17:58:46]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll mod = 1e9 + 7;
const ll Base = 10;
const int B = 455;
const int maxn = 2e5 + 5;

int a[maxn], b[maxn], f[maxn][18], lg[maxn];
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int query(int l, int r)
{
    int k = lg[r - l + 1];
    return gcd(f[l][k], f[r - (1 << k) + 1][k]);
}

ll fpow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

ll mi[maxn], inv[maxn], pre[maxn], iBase;
int bs, bn, bl[B], br[B], bel[maxn];

struct ds1 // O(1) query to O(sqrt(n)) update  a_i * x^i
{
    ll c[maxn], tag[maxn];
    void add(int pos, ll x)
    {
        for (int i = pos; i <= br[bel[pos]]; i++) (c[i] += x) %= mod;
        for (int i = bel[pos] + 1; i <= bn; i++) (tag[i] += x) %= mod;
    }
    ll sum(int pos) { return (c[pos] + tag[bel[pos]]) % mod; }
}ds1;

struct ds2 // O(sqrt(n)) query to O(1) update a_i * x^(i % k)
{
    ll c[maxn], sum[10005];
    ll d[maxn], sud[10005];
    void add(int pos, ll x) { (c[pos] += x) %= mod; (sum[bel[pos]] += x) %= mod; }
    void addD(int pos, ll x) { (d[pos] += x) %= mod; (sud[bel[pos]] += x) %= mod; }
    ll sumP(int pos)
    {
        ll ans = 0;
        for (int i = 1; i < bel[pos]; i++) (ans += sum[i]) %= mod;
        for (int i = bl[bel[pos]]; i <= pos; i++) (ans += c[i]) %= mod;
        return ans;
    }
    ll sumD(int pos)
    {
        ll ans = 0;
        for (int i = 1; i < bel[pos]; i++) (ans += sud[i]) %= mod;
        for (int i = bl[bel[pos]]; i <= pos; i++) (ans += d[i]) %= mod;
        return ans;
    }
}ds2[B];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n, m, q;
    cin >> n >> m >> q;
    mi[0] = inv[0] = pre[0] = 1; iBase = fpow(Base, mod - 2);
    for (int i = 1; i <= n; i++) mi[i] = mi[i - 1] * Base % mod, pre[i] = (pre[i - 1] + mi[i]) % mod;
    for (int i = 1; i <= n; i++) inv[i] = inv[i - 1] * iBase % mod;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1, k = 0; i <= m; i++)
    {
        cin >> b[i], f[i][0] = b[i];
        lg[i] = k;
        if (i == (1 << (k + 1))) lg[i] = ++k;
    }
    for (int j = 1; j <= 17; j++)
        for (int i = 1; i + (1 << j) - 1 <= m; i++)
            f[i][j] = gcd(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
    
    bs = sqrt(n);
    for (int i = 1; i <= n; i += bs)
        bl[++bn] = i, br[bn] = i + bs - 1;
    br[bn] = n;
    for (int i = 1; i <= bn; i++)
        for (int j = bl[i]; j <= br[i]; j++) bel[j] = i;
    
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= bs; j++)
        {
            ds2[j].add(i, a[i] * mi[i % j] % mod);
            if (i % j == 0) ds2[j].addD(i, a[i]);
        }
        ds1.add(i, a[i] * mi[i] % mod);
    }

    while (q--)
    {
        char s[4];
        cin >> s;
        if (s[0] == 'U')
        {
            int pos;
            ll v;
            cin >> pos >> v;
            v = (v + mod - a[pos]) % mod;
            for (int i = 1; i <= bs; i++)
            {
                ds2[i].add(pos, v * mi[pos % i] % mod);
                if (pos % i == 0) ds2[i].addD(pos, v);
            }
            ds1.add(pos, v * mi[pos] % mod);
            (a[pos] += v) %= mod;
        }
        else
        {
            int l, r, s, t;
            cin >> l >> r >> s >> t;
            int k = gcd(r - l + 1, query(s, t));
            ll hsh = 0, hs2 = 0;
            if (k <= bs)
            {
                hsh = ds2[k].sumP(r) - ds2[k].sumP(l - 1);
                hs2 = ds2[k].sumD(r) - ds2[k].sumD(l - 1);
                if (hsh < 0) hsh += mod;
                if (hs2 < 0) hs2 += mod;
            }
            else
            {
                for (int i = l; i <= r; i += k) (hs2 += a[i]) %= mod;
                for (int i = l / k; i <= r / k; i++)
                {
                    int L = i * k, R = L + k - 1;
                    L = max(L, l), R = min(R, r);
                    if (L > R) continue;
                    ll res = (ds1.sum(R) + mod - ds1.sum(L - 1)) * inv[i * k] % mod;
                    (hsh += res) %= mod;
                }
            }
            (hs2 *= pre[k - 1]) %= mod;
            cout << (hsh == hs2 ? "Yes\n" : "No\n");
        }
    }
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 9816kb

input:

6 4 5
1 1 4 5 1 4
3 3 2 4
Q 1 5 1 2
Q 2 5 3 4
U 5 2
Q 1 6 1 2
Q 2 5 3 4

output:

Yes
No
No
Yes

result:

ok 4 tokens

Test #2:

score: 0
Accepted
time: 1ms
memory: 7756kb

input:

1 1 1
0
1
Q 1 1 1 1

output:

Yes

result:

ok "Yes"

Test #3:

score: -100
Wrong Answer
time: 130ms
memory: 10048kb

input:

2000 2000 200000
1 1 2 0 0 2 0 2 0 0 0 0 0 2 2 1 2 0 0 2 2 2 1 0 1 2 1 2 0 0 1 1 1 2 0 0 2 2 2 2 0 2 0 0 2 1 2 0 0 1 2 2 1 0 2 0 0 0 1 2 2 1 2 2 0 0 1 1 1 0 0 2 0 0 1 1 0 2 2 2 1 0 0 1 0 1 2 2 2 1 1 2 2 1 2 1 0 2 2 3 1 3 2 3 1 0 1 2 0 1 1 1 0 2 2 3 2 0 3 2 3 3 1 2 3 1 2 0 1 0 3 1 0 0 2 0 1 2 1 3 2 2...

output:

Yes
Yes
No
Yes
Yes
No
No
No
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
Yes
Yes
Yes
No
Yes
Yes
No
No
No
No
No
Yes
No
No
No
Yes
Yes
No
Yes
No
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
Yes
Yes
No
No
Yes
No
Yes
No
No
No
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
Yes
Yes
Yes
No...

result:

wrong answer 1901st words differ - expected: 'No', found: 'Yes'