62. 醜數

class Solution {
public:
    int getUglyNumber(int n) {
        vector<int> q(1, 1);
        int i = 0, j = 0, k = 0;
        while( -- n) // 迴圈 n - 1 次 
        {
            int t = min(q[i] * 2, min(q[j] * 3, q[k] * 5));
            q.push_back(t);
            if(t == q[i] * 2) i ++ ;
            if(t == q[j] * 3) j ++ ;
            if(t == q[k] * 5) k ++ ;
        }
        return q.back();
    }
};
 

1262. 魚塘釣魚

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 110;

int n, T;
int a[N], d[N], l[N], spend[N];

int get(int k)
{
    return max(0, a[k] - d[k] * spend[k]);
}

int work(int n, int T)
{
    int res = 0;
    memset(spend, 0, sizeof spend);
    for(int i = 0; i < T; i ++ )
    {
        int t = 1;
        for(int j = 1; j <= n; j ++ )
            if(get(t) < get(j))
                t = j;

        res += get(t);
        spend[t] ++ ;
    }
    return res;
}

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
    for(int i = 1; i <= n; i ++ ) scanf("%d", &d[i]);
    for(int i = 2; i <= n; i ++ ) 
    {
        scanf("%d", &l[i]);
        l[i] += l[i - 1];
    }
    scanf("%d", &T);
    int res = 0;
    for(int i = 1; i <= n; i ++ )
        res = max(res, work(i, T - l[i]));

    printf("%d\n", res);
    return 0;
}

作者：NFYD
連結：https://www.acwing.com/activity/content/code/content/3475286/
來源：AcWing
著作權歸作者所有。商業轉載請聯絡作者獲得授權，非商業轉載請註明出處。
 

1541. 世界首富

#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int N = 210;

int n, m;

struct Person
{
    string name;
    int age, w;

    bool operator< (const Person& t) const
    {
        if(w != t.w) return w > t.w;
        if(age != t.age) return age < t.age;
        return name < t.name;
    }
};

vector<Person> ages[N];
int idx[N];

int main()
{
    scanf("%d%d", &n, &m);
    char name[10];
    for(int i = 0; i < n; i ++ )
    {
        int age, w;
        scanf("%s%d%d", name, &age, &w);
        ages[age].push_back({name, age, w});
    }

    for(auto& c : ages) sort(c.begin(), c.end());

    for(int T = 1; T <= m; T ++ )
    {
        printf("Case #%d:\n", T);
        int cnt, a, b;
        scanf("%d%d%d", &cnt, &a, &b);

        memset(idx, 0, sizeof idx);
        bool exists = false;
        while(cnt -- )
        {
            int t = -1;
            for(int i = a; i <= b; i ++ )
            {
                if(idx[i] < ages[i].size())
                {
                    if(t == -1 || ages[i][idx[i]] < ages[t][idx[t]])
                        t = i;
                }
            }
            if(t == -1) break;
            auto& p = ages[t][idx[t]];
            printf("%s %d %d\n", p.name.c_str(), p.age, p.w);
            idx[t] ++ ;
            exists = true;
        }
        if(!exists) puts("None");
    }
    return 0;
}

作者：NFYD
連結：https://www.acwing.com/activity/content/code/content/1790990/
來源：AcWing
著作權歸作者所有。商業轉載請聯絡作者獲得授權，非商業轉載請註明出處。
 

3280. 推薦系統

#include <iostream>
#include <algorithm>
#include <cstring>
#include <unordered_map>
#include <vector>
#include <set>

#define fi first
#define se second

using namespace std;

typedef pair<int, int> PII;

const int N = 55;

int m, n;
set<PII> g[N];
unordered_map<int, int> f[N];
int sum[N], cnt[N]; 
set<PII>::reverse_iterator it[N];
vector<int> ans[N];

int main()
{
    scanf("%d%d", &m, &n);
    for(int i = 1; i <= n; i ++ )
    {
        int id, score;
        scanf("%d%d", &id, &score);
        for(int j = 0; j < m; j ++ )
        {
            f[j][id] = score;
            g[j].insert({score, -id});
        }
    }
    int Q;
    scanf("%d", &Q);
    while(Q -- )
    {
        int t;
        scanf("%d", &t);
        if(t == 1)
        {
            int type, id, score;
            scanf("%d%d%d", &type, &id, &score);
            f[type][id] = score;
            g[type].insert({score, -id});
        }
        else if(t == 2)
        {
            int type, id;
            scanf("%d%d", &type, &id);
            g[type].erase({f[type][id], -id});
            f[type].erase(id);
        }
        else
        {
            int tot;
            scanf("%d", &tot);
            for(int i = 0; i < m; i ++ )
            {
                scanf("%d", &sum[i]);
                it[i] = g[i].rbegin();
                cnt[i] = 0;
                ans[i].clear();
            }
            while(tot -- )
            {
                int k = -1;
                for(int i = 0; i < m; i ++ )
                {
                    if(it[i] != g[i].rend() && cnt[i] < sum[i])
                    {
                        if(k == -1 || it[i]->fi > it[k]->fi)
                        {
                            k = i;
                        }
                    }
                }
                if(k == -1) break;
                ans[k].push_back(-it[k]->se);
                cnt[k] ++ ;
                it[k] ++ ;
            }
            for(int i = 0; i < m; i ++ )
            {
                if(ans[i].empty()) puts("-1");
                else
                {
                    for(auto x : ans[i])
                    {
                        printf("%d ", x);
                    }
                    puts("");
                }
            }
        }
    }
    return 0;
}

作者：NFYD
連結：https://www.acwing.com/activity/content/code/content/1742801/
來源：AcWing
著作權歸作者所有。商業轉載請聯絡作者獲得授權，非商業轉載請註明出處。

3874. 三元組的最小距離

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

int l, m, n;
int a[N], b[N], c[N];

int main()
{
    scanf("%d%d%d", &l, &m, &n);
    for(int i = 0; i < l; i ++ ) scanf("%d", &a[i]);
    for(int i = 0; i < m; i ++ ) scanf("%d", &b[i]);
    for(int i = 0; i < n; i ++ ) scanf("%d", &c[i]);

    LL res = 1e18;
    for(int i = 0, j = 0, k = 0; i < l && j < m && k < n; )
    {
        int x = a[i], y = b[j], z = c[k];
        res = min(res, (LL)max(max(x, y), z) - (LL)min(min(x, y), z));
        if(x <= y && x <= z) i ++ ;
        else if(y <= x && y <= z) j ++ ;
        else k ++ ;
    }
    printf("%lld\n", res * 2);
    return 0;
}

作者：NFYD
連結：https://www.acwing.com/activity/content/code/content/2883052/
來源：AcWing
著作權歸作者所有。商業轉載請聯絡作者獲得授權，非商業轉載請註明出處。