AcWing 第60场周赛

AcWing 第60场周赛

https://www.acwing.com/activity/content/competition/problem_list/2048/

A. AcWing 4494. 吃饭

签到题

#include <bits/stdc++.h>

using namespace std;

int main () {
    int  n, m, k;
    cin >> n >> m >> k;
    if (n > m || n > k) cout << "No\n";
    else    cout << "Yes\n";
}

B. AcWing 4495. 数组操作

做法很多，一开始打算用堆做，但是没整出来。
后来写了一个二分查找第一个大于 x 的数，x是上一个最小的非零数（这样就避免了区间修改的的操作）

还是不太熟练啊...手速太慢了

#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 5;
int a[N];

int main () {
    int  n, k;
    cin >> n >> k;
    
    for (int i = 0; i < n; i ++)  {
        cin >> a[i];
    }
    sort (a, a + n);

    int st = 0;
    for (int i = 0; i < n; i ++)
        if (a[i]) {
            st = i;
            break;
        }

    int x = 0;
    a[n] = 0;
    bool zero = false;
    while (k --) {
        st = upper_bound (a, a + n, x) - a;
        //cout << "st=" << st << endl;
        if (st == n)    zero = true;
        if (zero)   cout << "0\n";
        else    cout << max(a[st] - x, 0) << endl;
        x = a[st];
    }
}

C. AcWing 4496. 吃水果

dp
定义f[i][j]，表示放到第i个，有j个不同

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 205, mod = 998244353;
int f[N][N]; //f[i][j] 放到第i个，有j个不同

signed main () {
    int  n, m, k;
    cin >> n >> m >> k;

    //init
    for (int i = 1; i <= n; i ++)   f[i][0] = m; //所有同学都一样

    for (int i = 2; i <= n; i ++) 
        for (int j = 1; j <= k; j ++) {
            f[i][j] = f[i-1][j]; //一样
            f[i][j] = (f[i][j] + f[i-1][j-1] * (m-1) % mod) % mod; //不一样
        }
            
    
    
    cout << f[n][k] << endl;
}
//dp