2022.3.19美团笔试

原创已于 2022-03-19 12:21:00 修改 · 962 阅读

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于 2022-03-19 12:19:47 首次发布

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本文介绍了三道编程题目，涉及动态规划求解物品购买最优策略、字符串加密解密算法以及区间交集计算。第一题要求找到最优惠的购物方案，第二题考察字符串的加密与解密，第三题通过合并区间寻找重复点的数量。通过这些题目，可以锻炼程序员的区间处理和动态规划能力。

2个小时5道编程题，只做出了3道，感觉人无了

第一题

大意是有n个物品，可以按满减和折扣的方式购买，求前k个物品最划算的购买方式。直接模拟即可，细节没注意浪费了好久。

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<LL, LL> PII;

LL p1[50010], p2[50010];
vector<PII> rule;
LL t[50010];
LL idx;

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

    for (int i = 1; i <= n; i++)
        cin >> p1[i];
    for (int i =1;  i<= n; i++)
        cin >> p2[i];
    cin >> m;

    for (int i = 1; i <= m; i++) {
        cin >> t[i];
    }

    for (int i = 1; i <= m; i++) {
        LL x;
        cin >> x;
        rule.push_back({t[i], x});
    }
    sort(rule.begin(), rule.end());
    LL sum = 0, sum1 = 0;

    LL now = 0;
    for (int i = 1; i <= n; i++) {
        sum += p1[i];
        sum1 += p2[i];

        while(idx < rule.size() && rule[idx].first <= sum) {
            now = rule[idx].second;
            idx++;
        }
        // cout << sum << " " << sum1 << " " << now << endl;
        LL sum2 = sum - now;

        if (sum1 < sum2) cout << "Z";
        else if (sum1 > sum2) cout << "M";
        else cout << "B";

    }
    return 0;
}

第二题

给定一个字符串，求出其加密的结果或原先的字符串，加密方式为每次选取当前字符串第n/2个字符（下取整）加入空串中，直至取完。

#include <bits/stdc++.h>

using namespace std;

string code(string s) {
    string ans = "";
    int l, r;
    if (s.size() % 2 == 0) l = s.size() / 2 - 1, r = s.size() /2;
    else l = r = s.size() / 2;

    while(ans.size() < s.size()) {
        // cout << l << " " << r << endl;
        if (l >= 0) ans += s[l];
        if (r < s.size()) ans += s[r];
        l--, r++;
    }
    return ans;
}
string decode(string s) {
    string ans = "";
    int l = 0, r = 1;
    while(ans.size() < s.size())
    {
        if (l < s.size()) ans = s[l] + ans;
        if (r < s.size()) ans += s[r];
        l += 2, r += 2;
    }
    return ans;
}

int main()
{
    int n, t;
    cin >> n >> t;

    string str;
    cin >> str;

    if (t == 1) cout << code(str);
    else cout << decode(str);
    return 0;
}

第三题

题意提炼后大意为，A给出若干个区间，B也给出若干个区间，求A和B中重复的点的个数。分别区间合并后对A，B区间两两枚举求公共点，结果过了

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> PII;

vector<PII> block1, block2, temp;

int temp_l[30010], temp_r[30010];

long long get_common(PII x1, PII x2)
{
    if (x1.first > x2.second || x2.first > x1.second) return 0;
    int l = max(x1.first, x2.first);
    int r = min(x1.second, x2.second);
    return r - l + 1;
}

int main()
{
    int n, m1, m2;
    cin >> n >> m1 >> m2;

    for (int i = 1; i <= m1; i++)
        cin >> temp_l[i];
    for (int i = 1; i <= m1; i++)
    {
        cin >> temp_r[i];
        temp.push_back({temp_l[i], temp_r[i]});
    }

    sort(temp.begin(), temp.end());

    for (int i = 0; i < temp.size(); i++) {
        PII now = temp[i];
        while(i + 1 < temp.size() && temp[i + 1].first <= now.second) {
            now = {now.first, max(now.second, temp[i + 1].second)},
            i++;
        }
        block1.push_back(now);
    }

    //  -------------------------

    temp.clear();

    for (int i = 1; i <= m2; i++)
        cin >> temp_l[i];
    for (int i = 1; i <= m2; i++)
    {
        cin >> temp_r[i];
        temp.push_back({temp_l[i], temp_r[i]});
    }

    sort(temp.begin(), temp.end());

    for (int i = 0; i < temp.size(); i++) {
        PII now = temp[i];
        while(i + 1 < temp.size() && temp[i + 1].first <= now.second) {
            now = {now.first, max(now.second, temp[i + 1].second)},
            i++;
        }
        block2.push_back(now);
    }

    long long res = 0;
    for (int i = 0; i < block1.size(); i++)
        for (int j = 0; j < block2.size(); j++)
            res += get_common(block1[i], block2[j]);
    cout << res << endl;
    return 0;

}