算法-数学-训练

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目前共计 5 道题。

质数距离

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<ll, ll>;

inline ll dis(PII p) { return p.second - p.first; }

bool is_comp[50010];
int primes[6000], tot = 0;
void prime(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!is_comp[i])
            primes[tot++] = i;
        for (int j = 0; j < tot && i * primes[j] <= n; j++)
        {
            is_comp[i * primes[j]] = true;
            if (i % primes[j] == 0)
                break;
        }
    }
}

bool st[1000010]; // [0~R-L] equals to [L~R]

int main()
{
    prime(50000);
    ll l, r;
    while (cin >> l >> r)
    {
        memset(st, 0, sizeof(st));
        if (l == 1)
            st[0] = true;
        for (int i = 0; i < tot; i++)
        {
            int p = primes[i];
            ll k = (l + p - 1) / p;
            if (k < 2)
                k = 2;
            for (ll j = k * p; j <= r; j += p)
                st[j - l] = true;
        }
        ll pre, cur, cnt = 0;
        PII ans1 = {0, 0x3f3f3f3f}, ans2 = {0, 0}; // 1 最小差距 2 最大差距
        for (ll i = l; i <= r; i++)
        {
            if (!st[i - l])
            {
                pre = cur, cur = i, cnt++;
                if (cnt >= 2)
                {
                    if (cur - pre < dis(ans1))
                        ans1 = make_pair(pre, cur);
                    if (cur - pre > dis(ans2))
                        ans2 = make_pair(pre, cur);
                }
            }
        }
        if (cnt < 2)
            printf("There are no adjacent primes.\n");
        else
            printf("%lld,%lld are closest, %lld,%lld are most distant.\n",
                   ans1.first, ans1.second, ans2.first, ans2.second);
    }
}

阶乘分解

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#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;

bool st[N];
int primes[N], tot = 0;
void prime(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!st[i])
            primes[tot++] = i;
        for (int j = 0; j < tot && i * primes[j] <= n; j++)
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0)
                break;
        }
    }
}

int main()
{
    prime(1e6);
    int n;
    cin >> n;
    for (int i = 0; i < tot && primes[i] <= n; i++)
    {
        int p = primes[i], c = 0, v = n;
        while (v)
        {
            v /= p;
            c += v;
        }
        cout << p << " " << c << endl;
    }
}

可见的点

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#include <bits/stdc++.h>
using namespace std;
const int N = 1010;

int phi[N];
int pre[N]; // 前缀和

int main()
{
    for (int i = 2; i <= 1000; i++)
        phi[i] = i;
    for (int i = 2; i <= 1000; i++)
        if (phi[i] == i) // 是一个质数
            for (int j = i; j <= 1000; j += i)
                phi[j] = phi[j] / i * (i - 1);
    for (int i = 2; i <= 1000; i++)
        pre[i] = pre[i - 1] + phi[i];
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++)
    {
        int n, ans = 0;
        cin >> n;
        cout << i << " " << n << " " << 2 * pre[n] + 3 << endl;
    }
}

最大公约数

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e7 + 10;

int n;
ll pri[N / 10], tot = 0;
ll phi[N];
ll pre[N];
bool st[N];

int main()
{
    cin >> n;
    for (int i = 2; i <= n; i++)
    {
        if (!st[i])
            pri[++tot] = i, phi[i] = i - 1;
        for (int j = 1; j <= tot && pri[j] * i <= n; j++)
        {
            int v = pri[j] * i;
            st[v] = true;
            if (i % pri[j] == 0)
            {
                phi[v] = pri[j] * phi[i];
                break;
            }
            else
                phi[v] = phi[i] * (pri[j] - 1);
        }
    }
    for (int i = 2; i <= n; i++)
        pre[i] = pre[i - 1] + phi[i];
    ll ans = 0;
    for (int i = 1; i <= tot; i++)
    {
        int m = n / pri[i];
        ans += pre[m] * 2 + 1;
    }
    cout << ans << endl;
}

台阶-Nim游戏

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

int main()
{
    int n;
    cin >> n;
    int res = 0;
    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        if (i & 1)
            res ^= x;
    }
    cout << (res != 0 ? "Yes" : "No") << endl;
}

集合-Nim游戏

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#include <bits/stdc++.h>
using namespace std;
const int N = 110, M = 10010;

int n, m;
int a[N]; // 石子对应的值
int b[N]; // 数字集合的值
int f[M]; // 某个数量的石子对应的mex值

int main()
{
    cin >> m;
    for (int i = 1; i <= m; i++)
        cin >> b[i];
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    f[0] = 0;
    for (int i = 1; i <= 1e4; i++)
    {
        f[i] = 0;
        set<int> st;
        for (int j = 1; j <= m; j++)
        {
            if (i - b[j] >= 0)
                st.insert(f[i - b[j]]);
        }
        for (int j = 0;; j++)
            if (!st.count(j))
            {
                f[i] = j;
                break;
            }
    }
    int res = 0;
    for (int i = 1; i <= n; i++)
        res ^= f[a[i]];
    cout << (res != 0 ? "Yes" : "No") << endl;
}