2018 Nowcoder Multi-University Training Contest 5

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:
\[ \begin{eqnarray*} gpa = \frac{\sum s_ic_i}{\sum s_i} \end{eqnarray*} \]

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:
\[ \begin{eqnarray*} gpa &\leq& \frac{\sum s_ic_i}{\sum s_i} \\ gpa \cdot \sum s_i &\leq& \sum s_ic_i \\ \sum s_i \cdot gpa &\leq& \sum s_ic_i \\ \sum s_i \cdot (gpa - c_i) &\leq& 0 \end{eqnarray*} \]
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\([1, \frac{n}{c}]\)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;
}
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