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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
#814930#9870. ItemspropaneWA 7ms11284kbC++204.1kb2024-12-14 22:53:332024-12-14 22:53:38

Judging History

你现在查看的是最新测评结果

  • [2024-12-14 22:53:38]
  • 评测
  • 测评结果:WA
  • 用时:7ms
  • 内存:11284kb
  • [2024-12-14 22:53:33]
  • 提交

answer

#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
using LL = long long;
const int maxn = 2e6 + 5, mod = 998244353, G = 3, Gi = 332748118;

int qpow(int a, int b, int mod){
    int res = 1;
    while (b){
        if (b & 1) res = 1LL * res * a % mod;
        a = 1LL * a * a % mod;
        b >>= 1;
    }
    return res;
}

inline int mul(int a, int b){
    return 1LL * a * b % mod;
}

inline void add(int &a, int b){
    a += b;
    if (a >= mod) a -= mod;
}

inline void sub(int &a, int b){
    a -= b;
    if (a < 0) a += mod;
}

int inv[(1 << 21) + 5], fact[(1 << 21) + 5], invfact[(1 << 21) + 5];
void init(int n){
    inv[1] = 1;
    for(int i = 2; i <= n; i++)
        inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod; 
    fact[0] = invfact[0] = 1;
    for(int i = 1; i <= n; i++)
        fact[i] = mul(fact[i - 1], i);
    invfact[n] = qpow(fact[n], mod - 2, mod);
    for(int i = n - 1; i >= 1; i--)
        invfact[i] = mul(invfact[i + 1], i + 1);  
}

int C(int a, int b){
    if (a < 0 || b < 0 || a < b) return 0;
    return mul(mul(fact[a], invfact[b]), invfact[a - b]);
}

namespace NTT{
    vector<int> Omega(int L){
        int wn = qpow(G, mod / L, mod);
        vector<int> w(L); 
        w[L >> 1] = 1;
        for(int i = L / 2 + 1; i < L; i++) w[i] = mul(w[i - 1], wn);
        for(int i = L / 2 - 1; i >= 1; i--) w[i] = w[i << 1];
        return w;
    }
    auto W = Omega(1 << 21);

    void DIF(int *a, int n) {
        for(int k = n >> 1; k; k >>= 1)
            for(int i = 0, y; i < n; i += k << 1)
                for(int j = 0; j < k; j++){
                    y = a[i + j + k], a[i + j + k] = mul(a[i + j] - y + mod, W[k + j]), 
                    add(a[i + j], y);
                }
    }

    void IDIT(int *a, int n) {
        for (int k = 1; k < n; k <<= 1)
            for (int i = 0, x, y; i < n; i += k << 1)
                for(int j = 0; j < k; j++){
                    x = a[i + j], y = mul(a[i + j + k], W[k + j]),
                    a[i + j + k] = x - y < 0 ? x - y + mod : x - y;
                    add(a[i + j], y);
                }
        int inv = mod - (mod - 1) / n;
        for(int i = 0; i < n; i++) a[i] = mul(a[i], inv);
        reverse(a + 1, a + n);
    }
}

using Poly = std::vector<int>;

void DFT(Poly &a){
    NTT::DIF(a.data(), a.size());
}

void IDFT(Poly &a){
    NTT::IDIT(a.data(), a.size());
}

int normed(int n) { 
    return n == 1 ? 1 : (1 << (32 - __builtin_clz(n - 1))); 
}

void norm(Poly &a) { 
    if (!a.empty()) a.resize(normed((int)a.size()), 0); 
}

void dot(Poly &a, Poly &b) {
    for(int i = 0; i < a.size(); i++)
        a[i] = mul(a[i], b[i]);
}

Poly operator*(Poly a, Poly b) {
    int n = a.size() + b.size() - 1, L = normed(n);
    if (a.size() <= 8 || b.size() <= 8) {
        Poly c(n);
        for(int i = 0; i < a.size(); i++)
            for(int j = 0; j < b.size(); j++)
                c[i + j] = (c[i + j] + 1LL * a[i] * b[j]) % mod;
        return c;
    }
    a.resize(L), b.resize(L);
    DFT(a), DFT(b), dot(a, b), IDFT(a);
    return a.resize(n), a;
}

int main(){

#ifdef LOCAL
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
#endif

    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    int T;
    cin >> T;
    while(T--){
        int n; LL m;
        cin >> n >> m;
        int k = m / n;
        Poly a(2 * n + 1);
        for(int i = 0; i < n; i++){
            int x;
            cin >> x;
            a[n + x - k] = 1;
        }
        vector<int> ans(2 * n + 1);
        ans[n] = 1;
        int power = n;
        while(power){
            if (power & 1){
                auto c = ans * a;
                for(int i = -n; i <= n; i++){
                    ans[i + n] = (c[2 * n + i] != 0);
                }
            }
            power >>= 1;
            auto c = a * a;
            for(int i = -n; i <= n; i++){
                ans[i + n] = (c[2 * n + i] != 0);
            }
        }
        cout << (ans[n + m % n] ? "Yes" : "No") << '\n';
    }

}

詳細信息

Test #1:

score: 100
Accepted
time: 7ms
memory: 11176kb

input:

4
5 25
0 0 0 0 5
5 11
4 4 4 5 5
5 0
1 2 3 4 5
5 25
0 1 2 3 4

output:

Yes
No
No
No

result:

ok 4 token(s): yes count is 1, no count is 3

Test #2:

score: -100
Wrong Answer
time: 6ms
memory: 11284kb

input:

1314
1 0
0
1 0
1
1 1
0
1 1
1
2 0
0 0
2 0
0 1
2 0
0 2
2 0
1 0
2 0
1 1
2 0
1 2
2 0
2 0
2 0
2 1
2 0
2 2
2 1
0 0
2 1
0 1
2 1
0 2
2 1
1 0
2 1
1 1
2 1
1 2
2 1
2 0
2 1
2 1
2 1
2 2
2 2
0 0
2 2
0 1
2 2
0 2
2 2
1 0
2 2
1 1
2 2
1 2
2 2
2 0
2 2
2 1
2 2
2 2
2 3
0 0
2 3
0 1
2 3
0 2
2 3
1 0
2 3
1 1
2 3
1 2
2 3
2 0...

output:

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

result:

wrong answer expected NO, found YES [199th token]