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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
#594389#9302. Caesar Cipherzqx#WA 9ms22100kbC++233.8kb2024-09-27 22:55:092024-09-27 22:55:10

Judging History

你现在查看的是最新测评结果

  • [2024-11-04 17:10:09]
  • hack成功,自动添加数据
  • (/hack/1110)
  • [2024-10-21 09:47:48]
  • hack成功,自动添加数据
  • (/hack/1022)
  • [2024-10-21 09:39:50]
  • hack成功,自动添加数据
  • (/hack/1021)
  • [2024-10-21 09:31:34]
  • hack成功,自动添加数据
  • (/hack/1020)
  • [2024-10-03 10:14:59]
  • hack成功,自动添加数据
  • (/hack/928)
  • [2024-09-28 07:51:27]
  • hack成功,自动添加数据
  • (/hack/922)
  • [2024-09-28 07:42:39]
  • hack成功,自动添加数据
  • (/hack/921)
  • [2024-09-27 22:55:10]
  • 评测
  • 测评结果:WA
  • 用时:9ms
  • 内存:22100kb
  • [2024-09-27 22:55:09]
  • 提交

answer

#include <bits/stdc++.h>

#define int long long

using namespace std;

const int N = 5e5 + 10;
const int lim = 65536;
const int base = 13331;
const int mod1 = 1e9 + 7;
const int mod2 = 998244353;

int n;
int a[N];

struct Segment_Tree 
{
    static const int N = 5e5 + 10;

    int mod;
    int pw[N], ppw[N];
    void Init(int x)
    {
        mod = x;
        pw[0] = 1;
        ppw[0]=1;
        for (int i = 1; i < N; i++)
        {
            pw[i] = pw[i - 1] * base % mod;
            (ppw[i] = ppw[i - 1] + pw[i]) %= mod;
        }
    }

    struct node {
        int l, r;
        int hash, mx;
        int lazy;
    }t[N << 3];

    #define ls(x) (x << 1)
    #define rs(x) (x << 1 | 1)

    void Pushup(int p)
    {
        t[p].mx = max(t[ls(p)].mx, t[rs(p)].mx);
        t[p].hash = (t[ls(p)].hash * pw[t[rs(p)].r - t[rs(p)].l + 1] % mod + t[rs(p)].hash) % mod;
    }

    void Build(int p, int l, int r)
    {
        t[p].l = l, t[p].r = r;
        if (l == r) {
            t[p].hash = a[l] + 1;
            t[p].mx=a[l];
            return ;
        }
        int mid = (l + r) >> 1;
        Build(ls(p), l, mid);
        Build(rs(p), mid + 1, r);
        Pushup(p);
    }

    void Pushdown(int p)
    {
        int lz = t[p].lazy;
        if (lz == 0)
           return ;
        t[ls(p)].lazy += lz;
        t[rs(p)].lazy += lz;
        t[ls(p)].mx+=lz;
        t[rs(p)].mx+=lz;
        (t[ls(p)].hash += lz * ppw[t[ls(p)].r - t[ls(p)].l] % mod) %= mod;
        (t[rs(p)].hash += lz * ppw[t[rs(p)].r - t[rs(p)].l] % mod) %= mod;
        t[p].lazy = 0;
    }

    void Update(int p, int l, int r, int v)
    {
        if (t[p].l == t[p].r)
        {
            ((t[p].hash += v) %= lim)+=1;
            (t[p].mx += v) %= lim;
            return ;
        }

        if (l <= t[p].l && r >= t[p].r)
        {
            if (t[p].mx < lim - 1)
            {
                t[p].lazy += v;
                (t[p].hash += v * ppw[t[p].r - t[p].l] % mod) %= mod;
                return ;
            }
        }
        Pushdown(p);
        int mid = (t[p].l + t[p].r) >> 1;
        if (l <= mid)
            Update(ls(p), l, r, v);
        if (r > mid)
            Update(rs(p), l, r, v);
        Pushup(p);
    }

    int Query(int p, int l, int r)
    {
        if (l <= t[p].l && r >= t[p].r) 
        {
            return t[p].hash;
        }
        Pushdown(p);
        int mid = (t[p].l + t[p].r) >> 1;
        int res = 0;
        if (l > mid)
            res = Query(rs(p), l, r);
        else if(r <= mid)
            res = Query(ls(p), l, r);
        else 
            res = (Query(ls(p), l, mid) * pw[r -
             mid] % mod + Query(rs(p), mid + 1, r)) % mod;
        return res;
    }
}sgt1, sgt2;

void Solve()
{
    int q;
    cin >> n >> q;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    sgt1.Build(1, 1, n);
    sgt2.Build(1, 1, n);
    while (q--)
    {
        int opt;
        cin >> opt;
        if (opt == 1)
        {
            int l, r;
            cin >> l >> r;
            sgt1.Update(1, l, r, 1);
            sgt2.Update(1, l, r, 1);
        }
        else 
        {
            int l1, r1, l2, r2, len;
            cin >> l1 >> l2 >> len;
            r1 = l1 + len - 1; r2 = l2 + len - 1;
            bool ok1 = (sgt1.Query(1, l1, r1) == sgt1.Query(1, l2, r2));
            bool ok2 = (sgt2.Query(1, l1, r1) == sgt2.Query(1, l2, r2));
            cout << ((ok1 && ok2) ? "yes\n" : "no\n");
        }
    }
    return ;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int T = 1;
    sgt1.Init(mod1);
    sgt2.Init(mod2);
    while(T--)
    {
        Solve();
    }
}

详细

Test #1:

score: 0
Wrong Answer
time: 9ms
memory: 22100kb

input:

5 6
1 2 1 2 1
2 1 2 2
2 1 3 3
1 1 1
1 3 5
2 1 2 4
2 1 2 2

output:

no
yes
no
no

result:

wrong answer expected YES, found NO [4th token]