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 \(a_i\) 从大到小排序即可,必然满足 \(j - i \not= a_j - a_i\) 因为 \(j - i\) 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 \(k^0,k^1,k^2...k^i\) 时可以考虑 \(k\) 进制 (重要)
那么就把所有的 \(a_i\) 转换成 \(k\) 进制,在相同位上出现基数的和如果超过 \(1\) 那么就不行,注意:\(k\) 进制全部的元素大小是 \(0\sim k - 1\)
#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 \(n - 1\) 个不同的数出来,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满足上升的条件,任取一个在 \(n - 2\) 中的数作为相等的数放在右边,有 \(n - 2\) 种不同的选法,那么剩下的 \(n - 3\) 个数自由选择放在左边或者右边后可以自动拍个序,自然就满足了条件,对于某一个数放在左边或者右边有两种选法,那么就是 \(2^{n - 3}\)
所以 \(C_{m}^{n - 1}\times (n - 2)\times2^{n - 3}\)
#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 \(DP\)