CodeForces 1312

1312

• \(A题\)

测网速

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
int a[N];
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, m;
        cin >> n >> m;
        if (n % m == 0)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
    return 0;
}

• \(B题\)

把 \(a_i\) 从大到小排序便可，必然知足 \(j - i \not= a_j - a_i\) 由于 \(j - i\) 是正数，而 \(a_j - a_i\) 是负数

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
ll a[N];
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        sort(a + 1, a + n + 1);
        reverse(a + 1, a + n + 1);
        for (int i = 1; i <= n; i++)
            cout << a[i] << " ";
        cout << endl;
    }
    return 0;
}

• C题

没调出来，我 sb了

==出现 \(k^0,k^1,k^2...k^i\) 时能够考虑 \(k\) 进制== (重要)

那么就把全部的 \(a_i\) 转换成 \(k\) 进制，在相同位上出现基数的和若是超过 \(1\) 那么就不行，注意：\(k\) 进制所有的元素大小是 \(0\sim k - 1\)

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
ll a[N];
vector<ll> v[N];
ll k, n;
map<int, int> mp;
void solve(ll x, int pos)
{
    while (x)
    {
        v[pos].push_back(x % k);
        x /= k;
    }
}
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        mp.clear();
        cin >> n >> k;
        for (int i = 1; i <= n; i++)
        {
            cin >> a[i];
        }
        for (int i = 1; i <= n; i++)
            v[i].clear();
        int maxn = 0;
        for (int i = 1; i <= n; i++)
        {
            solve(a[i], i);
            maxn = max(maxn, (int)v[i].size());
        }
        for (int i = 1; i <= n; i++)
        {
            if ((int)v[i].size() <= maxn)
            {
                int cnt = maxn - (int)v[i].size();
                for (int j = 1; j <= cnt; j++)
                    v[i].push_back(0);
                reverse(v[i].begin(), v[i].end());
            }
        }
        int flag = 0;
        for (int j = 0; j < maxn; j++)
        {
            int cnt = 0;
            for (int i = 1; i <= n; i++)
            {
                cnt += v[i][j];
            }
            if (cnt > 1)
                flag = 1;
        }
        if (flag)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
    return 0;
}
/*
5 2
20 0 33 2 64
20:0001 0100
 0:0000 0000
33:0010 0001
 2:0000 0010
64:0100 0000
*/

• \(D题\)

考虑构造题目要求：

从 \(1\sim m\) 个 数中取 \(n - 1\) 个不一样的数出来，那么默认从小到大排序就是 \(m_1,m_2,m_3...m_{n-1}\)

从 \(m\) 个数中任取 \(n -1\) 个数是 \(C_{m}^{n - 1}\)

保持 \(n - 1\) 数中最大值 \(m_{n - 1}\) 不动，剩下的 \(n - 2\) 个数在最大值的左边知足上升的条件，任取一个在 \(n - 2\) 中的数做为相等的数放在右边，有 \(n - 2\) 种不一样的选法，那么剩下的 \(n - 3\) 个数自由选择放在左边或者右边后能够自动拍个序，天然就知足了条件，对于某一个数放在左边或者右边有两种选法，那么就是 \(2^{n - 3}\)

因此 \(C_{m}^{n - 1}\times (n - 2)\times2^{n - 3}\)

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int p = 998244353;
typedef long long ll;
typedef unsigned long long ull;
ll pow(ll a, ll b, ll m)
{
    ll ans = 1;
    a %= m;
    while (b)
    {
        if (b & 1)
            ans = (ans % m) * (a % m) % m;
        b /= 2;
        a = (a % m) * (a % m) % m;
    }
    ans %= m;
    return ans;
}
ll inv(ll x, ll p) //x关于p的逆元，p为素数
{
    return pow(x, p - 2, p);
}
ll C(ll n, ll m, ll p) //组合数C(n, m) % p
{
    if (m > n)
        return 0;
    ll up = 1, down = 1; //分子分母;
    for (int i = n - m + 1; i <= n; i++)
        up = up * i % p;
    for (int i = 1; i <= m; i++)
        down = down * i % p;
    return up * inv(down, p) % p;
}
ll Lucas(ll n, ll m, ll p)
{
    if (m == 0)
        return 1;
    return C(n % p, m % p, p) * Lucas(n / p, m / p, p) % p;
}
int main()
{
    ll n,mx;
    cin >> n >> mx;
    if(n <= 2)
    {
        cout << 0 << endl;
        return 0;
    }
    ll ans = Lucas(mx, n - 1,p);   
    ans = ans * (n - 2) % p;
    ans = ans * pow(2, n - 3,p) % p;
    cout << ans % p << endl;
    return 0;
}

• \(E题\)待补 \(DP\)