2018 Nowcoder Multi-University Training Contest 5

Posted 2022-02-28 dup4

Practice Link

A. gpa

题意：
有\(n\)门课程，每门课程的学分为\(s_i\)，绩点为\(c_i\)，要求最多删除\(k\)门课程，使得gpa最高。
gpa计算方式如下：
\[ \begineqnarray* gpa = \frac\sum s_ic_i\sum s_i \endeqnarray* \]

思路：
首先删去的课程越多，gpa肯定不会变得更差。
所以我们肯定是删去\(k\)门课程。
考虑二分答案，check的时候要满足：
\[ \begineqnarray* gpa &\leq& \frac\sum s_ic_i\sum s_i \gpa \cdot \sum s_i &\leq& \sum s_ic_i \\sum s_i \cdot gpa &\leq& \sum s_ic_i \\sum s_i \cdot (gpa - c_i) &\leq& 0 \endeqnarray* \]
那么check的时候贪心选取\(n - k\)个即可。

代码：

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define db double
#define N 100010
#define pii pair <int, int>
#define fi first
#define se second
const db eps = 1e-10;
int n, k; pii a[N];
 
bool ok(db x) 
    vector <db> vec;
    for (int i = 1; i <= n; ++i) 
        vec.push_back(a[i].fi * (x - a[i].se));
    
    sort(vec.begin(), vec.end());
    db tot = 0;
    for (int i = 0; i < n - k; ++i) 
        tot += vec[i];
    
    return tot <= 0 || fabs(tot - 0) < eps;

 
int main() 
    while (scanf("%d%d", &n, &k) != EOF) 
        for (int i = 1; i <= n; ++i) 
            scanf("%d", &a[i].fi);
        
        for (int i = 1; i <= n; ++i) 
            scanf("%d", &a[i].se);
        
        db l = 0, r = 1e3, res = 0;
        while (fabs(r - l) >= eps) 
            db mid = (l + r) / 2;
            if (ok(mid)) 
                l = mid;
                res = mid;
             else 
                r = mid;
            
        
        printf("%.10f\n", res);
    
    return 0;

G. max

题意：
给出\(c\)和\(n\)，要求找到一对\((a, b)\)满足\(1 \leq a, b \leq n\)使得\(gcd(a, b) = c\)并且最大化\(a \cdot b\)

思路：

• \(c > n\)时无解
• \(c = n\)时选择\((n, n)\)
• \(c < n\)时在\([1, \fracnc]\)中选取两个互质的数再分别乘上\(c\)

代码：

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
ll c, n;
 
int main() 
    while (scanf("%lld%lld", &c, &n) != EOF) 
        if (c > n) 
            puts("-1");
            continue;
        
        ll x = n / c;
        ll res = c * c;
        if (x > 1) 
            res *= x * (x - 1);
        
        printf("%lld\n", res);
    
    return 0;

J. plan

题意：
有\(n\)个人去住宿，双人房的价格为\(p_2\)， 三人房的价格为\(p_3\)，要求将\(n\)个人全都安排好住宿的最小代价是多少，不一定恰好住满。

思路：
大范围直接除2， 除3， 小范围暴力dp一下。

代码：

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

#define N 1000010
#define ll long long
#define ll long long
ll n, p2, p3;
ll f[N];

ll DFS(int x) 
    if (x <= 0) 
        return 0;
    
    if (f[x] != -1) 
        return f[x];
    
    return f[x] = min(p2 + DFS(x - 2), p3 + DFS(x - 3));


int main() 
    while (scanf("%lld%lld%lld", &n, &p2, &p3) != EOF) 
        memset(f, -1, sizeof f);
        if (n <= 1000000) 
            printf("%lld\n", DFS(n));
         else 
            ll res = 1e18;
            ll m;
            for (int i = 0; i < 1000000; ++i) 
                m = n - i;
                res = min(res, p2 * (m / 2) + DFS(i + m % 2));
                res = min(res, p3 * (m / 3) + DFS(i + m % 3));
            
            printf("%lld\n", res);
        
    
    return 0;