2025-12-01

CF

Problem - 1455D - Codeforces

贪心，从小到大遍历
因为a[i]>x才能交换
所以把大的换小，使其满足单调不递减

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
int a[N];

void solve()
{
    int n, x;
    int ans = 0;
    cin >> n >> x;
    for (int i = 0; i < n;i++){
        cin >> a[i];
    }
    for (int i = 1; i < n;i++){
        if(a[i]>=a[i-1])
            continue;
        for (int j = 0; j < i;j++){
            if(a[j]>x){
                swap(a[j], x);
                ans++;
            }
        }
        if(a[i]<a[i-1]){
            cout << -1 << endl;
            return;
        }
    }
    cout << ans << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }
}

Problem - 1957C - Codeforces（dp好题）

计算方案数
通过

\[dp[i]=dp[i-1]+dp[i-2]*(i-1)*2 \]

推最后的\(dp[n]\)

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 1e9+7;
const int N=3e5+10;
LL dp[N];


void solve()
{
    int n, k;
    cin >> n >> k;
    int r, c;
    while(k--){
        cin >> r >> c;
        n -= (r == c ? 1 : 2);
    }
    dp[0] = dp[1] = 1, dp[2] = 3;
    for (int i = 3; i <= n;i++){
        dp[i] = dp[i - 1] + dp[i - 2] * (2 * i - 2);
        dp[i] %= mod;
    }
    cout << dp[n] << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }
}