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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
#612504#7679. Master of Both IVzhangbojuWA 25ms12392kbC++172.4kb2024-10-05 11:38:072024-10-05 11:38:09

Judging History

你现在查看的是最新测评结果

  • [2024-10-05 11:38:09]
  • 评测
  • 测评结果:WA
  • 用时:25ms
  • 内存:12392kb
  • [2024-10-05 11:38:07]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
constexpr int N = 2e5 + 5, mod = 1e9 + 7;
struct Base{
    int a[20];
    void init() {
        for (int i = 0; i < 20; i++)
            a[i] = 0;
    }
    void insert(int x) {
        for (int i = 19; i >= 0; i--) {
            if (!((x >> i) & 1)) {
                continue;
            }
            if (!a[i]) {
                a[i] = x;
                break;
            }
            x ^= a[i];
        }
    }
    int count() {
        int cnt = 0;
        for (int i = 0; i < 20; i++)
            if (a[i])
                cnt++;
        return cnt;
    }
};
int n;
int pw[N];
int a[N];
int cnt[N];
vector<int> d[N];
int fac[N], inv[N];
int qpow(int a, int k) {
    int res = 1;
    while (k) {
        if (k & 1) res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        k >>= 1;
    }
    return res;
}
void init(int n) {
    pw[0] = 1;
    for (int i = 1; i <= n; i++)
        pw[i] = 2ll * pw[i - 1] % mod;
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = 1ll * fac[i - 1] * i % mod;
    inv[n] = qpow(fac[n], mod - 2);
    for (int i = n - 1; i >= 0; i--)
        inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
}
int C(int n, int m) {
    if (m < 0 || n < m) return 0;
    return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}
void solve() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        cnt[a[i]]++;
    }
    for (int i = 1; i <= n; i++)
        for (int j = 2 * i; j <= n; j += i)
            d[j].push_back(i);
    Base alls;
    alls.init();
    for (int i = 1; i <= n; i++)
        alls.insert(a[i]);
    int ans = pw[n - alls.count()] - 1;
    for (int i = 1; i <= n; i++) {
        if (!cnt[i]) continue;
        Base B;
        B.init();
        int sum = 0;
        for (int x : d[i]) {
            if (cnt[x]) {
                B.insert(x);
                sum += cnt[x];
            }
        }
        ans = (ans + 1ll * cnt[i] * pw[sum - B.count()]) % mod;
        for (int t = 3; t <= cnt[i]; t += 2)
            ans = (ans + C(cnt[i], t)) % mod;
    }
    printf("%d\n", ans);
}
void clear() {
    for (int i = 0; i <= n; i++)
        cnt[i] = 0, d[i].clear();
}
int main() {
    init(N - 1);
    int T;
    scanf("%d", &T);
    while (T--) {
        solve();
        clear();
    }
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 11716kb

input:

2
3
1 2 3
5
3 3 5 1 1

output:

4
11

result:

ok 2 number(s): "4 11"

Test #2:

score: -100
Wrong Answer
time: 25ms
memory: 12392kb

input:

40000
5
4 2 5 5 5
5
5 5 5 5 4
5
1 4 4 4 2
5
2 5 2 4 1
5
3 2 4 5 3
5
1 5 5 3 4
5
5 5 5 4 3
5
4 3 3 5 1
5
4 5 5 2 1
5
2 5 4 2 5
5
3 4 3 4 3
5
5 3 5 1 3
5
5 1 2 4 4
5
4 2 5 1 5
5
5 4 2 5 4
5
5 2 5 2 4
5
1 4 5 4 5
5
4 2 3 2 3
5
1 4 1 3 5
5
1 1 2 1 5
5
5 2 5 1 3
5
3 1 2 5 3
5
5 5 1 1 5
5
2 2 2 1 3
5
3 1 ...

output:

9
16
9
9
8
8
9
8
8
9
13
8
8
8
8
9
12
9
11
15
8
8
16
13
8
11
8
8
8
13
15
9
9
8
8
8
11
9
11
13
15
9
16
9
8
8
8
13
11
8
9
11
8
8
11
15
9
8
9
8
8
15
11
8
16
9
15
8
8
8
12
9
9
11
8
13
9
8
15
8
8
9
8
8
8
15
8
11
13
8
9
11
8
19
11
13
19
16
13
15
8
8
8
9
8
9
13
15
16
9
9
16
9
11
9
9
11
9
9
11
8
9
9
13
15
11...

result:

wrong answer 23rd numbers differ - expected: '17', found: '16'