SMU 2024 spring 天梯赛自主训练2

SMU 2024 spring 天梯赛自主训练2

7-1 I Love GPLT - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

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7-2 是不是太胖了 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int h;
    cin >> h;
    printf("%.1lf",(h-100)*0.9 * 2);

	return 0;
}

7-3 大笨钟 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int h,m;
    char op;
    cin >> h >> op >> m;
    int now = h * 60 + m;
    if(now >= 0 && now <= 12* 60){
        printf("Only %02d:%02d.  Too early to Dang.",h,m);
    }else{
        for(int i = 0;i < (now + 59) / 60 - 12 ;i ++)
            cout << "Dang";
    }

	return 0;
}

7-4 到底是不是太胖了 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int t;
    cin >> t;
    while(t -- ){
        double h,w;
        cin >> h >> w;
        double Per = (h-100)*0.9*2;
        double Nor = Per * 0.1;
        if(fabs(w - Per) < Nor)
        cout << "You are wan mei!\n";
        else if(w > Per)
            cout << "You are tai pang le!\n";
        else
            cout << "You are tai shou le!\n";

    }

	return 0;
}

7-5 情人节 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	vector<string> s;
    string str;
    cin >> str;
    while(str != "."){
        s.push_back(str);
        cin >> str;
    }

    if(s.size() < 2){
        cout << "Momo... No one is for you ...\n";
    }else if(s.size() < 14)
        cout << s[1] << " is the only one for you...\n";
    else cout<< s[1] << " and " << s[13] << " are inviting you to dinner...";

	return 0;
}

7-6 外星人的一天 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int t;
    cin >> t;
    while(t --){
        int d,h,m,now;
        char c;
        cin >> d >> h >> c >> m;
        now = h * 60 + m;
        if(!d)
            printf("%d %02d:%02d\n",d,h,m);
        else {
            if(d % 2 == 0) now += 24*60;
            printf("%d %02d:%02d\n",(d+1)/2,now / 60 / 2,(now/2)%60);
        }
    }

	return 0;
}

7-7 宇宙无敌加法器 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

模拟进制;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	string JIn;
    string NumA, NumB;
    cin >> JIn >> NumA >> NumB;
    vector<int> wei(30);
    reverse(JIn.begin(),JIn.end());
    reverse(NumA.begin(),NumA.end());
    reverse(NumB.begin(),NumB.end());
    string zero = "000000000000000000000";
    JIn += zero,NumA += zero, NumB += zero;
    vector<int> w(30);
    string ans = "000000000000000000000";
    for(int i = 0;i <= 20;i ++){
        int l = JIn[i] - '0';
        if(!l) l = 10;
        int num = (NumA[i] - '0') + (NumB[i] - '0') + w[i];
        w[i+1] += num / l;
        ans[i] = char('0' + num % l);
    }
    reverse(ans.begin(),ans.end());
    for(int i = 0;i < ans.size();i ++){
        if(ans[i] != '0'){
            cout << ans.substr(i) << '\n';
            return 0;
        }
    }
    cout << 0 << '\n';
	return 0;
}

7-8 古风排版 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

先补充到n的整数倍,然后倒序每n个输出;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
    cin >> n;
    string s;
    getline(cin,s);
    getline(cin,s);

    int m = s.size();
    if(m % n){
            for(int i = 0;i < n - m % n;i ++)
        s += ' ';
        m = s.size();
    }
    int dex = n;
    reverse(s.begin(),s.end());
    for(int i = 0;i < n;i ++){
        dex --;
        for(int j=dex;j < m;j += n)
            cout << s[j];
        cout << '\n';
    }

	return 0;
}

7-9 抢红包 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

排序;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

struct people{
	int id,num = 0;
	double val = 0;
};

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

 	int n;
 	cin >> n;
 	vector<people> Ke(n + 1);
 	for(int i = 1;i <= n;i ++){
 		int k,num = 0;
 		cin >> k;
 		Ke[i].id = i;
 		for(int j = 0;j < k;j ++){
 			int ni,pi;
 			cin >> ni >> pi;
 			num += pi;
 			Ke[ni].num ++;
 			Ke[ni].val += pi;
 		}
 		Ke[i].val -= num;
 	}

 	Ke.erase(Ke.begin());
 	sort(Ke.begin(),Ke.end(),[&](people a,people b){
 		if(a.val == b.val && a.num == b.num) return a.id < b.id;
 		if(a.val == b.val) return a.num > b.num;
 		return a.val > b.val;
 	});

 	for(auto [id,num,val] : Ke){
 		printf("%d %.2lf\n",id, val / 100);
 	}

	return 0;
}

7-10 家庭房产 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

并查集处理家庭关系,然后遍历一遍统计家庭所有信息,最后根据题目要求排序;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

struct people {
    int id = -1, num, fang, s;
    double avgs, avgfang;
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<int> fa(10010);
    iota(fa.begin(), fa.end(), 0);

    auto find = [&](auto self, int x)-> int{
        return fa[x] == x ? x : fa[x] = self(self, fa[x]);
    };

    bitset<10010> vis;
    vector<people> info(10010);
    for (int i = 0; i < n; i ++) {
        int id, x, y, k, f, s;
        cin >> id >> x >> y >> k;
        int u = find(find, id);
        vis[id] = 1;
        if (x != -1) {
            x = find(find, x);
            vis[x] = 1;
            fa[x] = u;
        }
        if (y != -1) {
            y = find(find, y);
            vis[y] = 1;
            fa[y] = u;
        }
        for (int j = 0; j < k; j ++) {
            int z;
            cin >> z;
            vis[z] = 1;
            z = find(find, z);
            fa[z] = u;
        }
        cin >> f >> s;
        info[id] = {id, 1, f, s, 0, 0};
    }

    vector<people> res(10010), ans;
    set<int> exit;
    for (int i = 0; i < 10000; i ++) {
        if (vis[i]) {
            int u = find(find, i);
            if (res[u].id == -1) res[u].id = i;
            res[u].num ++;
            res[u].fang += info[i].fang;
            res[u].s += info[i].s;
            exit.insert(u);
        }
    }

    for (auto v : exit) {
        ans.push_back(res[v]);
    }
    for (auto &i : ans) {
        i.avgs = i.s * 1.0 / i.num;
        i.avgfang = i.fang * 1.0 / i.num;
    }

    sort(ans.begin(), ans.end(), [&](people a, people b) {
        if (a.avgs != b.avgs) return a.avgs > b.avgs;
        return a.id < b.id;
    });

    printf("%d\n", ans.size());
    for (auto i : ans) {
        printf("%04d %d %.3lf %.3lf\n", i.id, i.num, i.avgfang, i.avgs);
    }

    return 0;
}

7-11 功夫传人 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

dfs弟子递归,遇到传道者累加答案即可;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    double r;
    cin >> n;
    vector<double> z(n + 1);
    vector<int> DeDao(n + 1);
    cin >> z[0] >> r;
    vector g(n + 1, vector<int>());
    for (int i = 0; i < n; i ++) {
        int k, x;
        cin >> k;
        if (!k) cin >> x, DeDao[i] = x;
        else {
            for (int j = 0; j < k; j ++) {
                cin >> x;
                g[i].push_back(x);
            }
        }
    }

    double ans = 0;
    auto dfs = [&](auto self, int now) {
        if (DeDao[now]) {
            ans += z[now] * DeDao[now];
            return;
        }

        for (auto v : g[now]) {
            z[v] = z[now] * (100 - r) * 0.01;
            self(self, v);
        }

    };

    dfs(dfs, 0);

    cout << (int) ans << '\n';

    return 0;
}

7-12 链表去重 - SMU 2024 spring 天梯赛自主训练2 (pintia.cn)

开1e6的数组用对应地址记录结点信息以及下一个地址,按初始地址构建两条链,碰到重复的就丢到另一条链;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int st, n;
	cin >> st >> n;
	vector<PII> L1(n), L2(n);
	vector<PII> Num(100100);
	for (int i = 0; i < n; i ++) {
		int x, a, addr;
		cin >> addr >> a >> x;
		Num[addr] = {a, x};
	}

	int cnt1 = 0, cnt2 = 0;
	set<int> vis;
	while (st != -1) {
		if (vis.count(abs(Num[st].first))) {
			L2[cnt2 ++] = {Num[st].first, st};
		} else {
			vis.insert(abs(Num[st].first));
			L1[cnt1 ++] = {Num[st].first, st};
		}
		st = Num[st].second;
	}

	auto print = [](int cnt, vector<PII> &L) -> void{
		for (int i = 0; i < cnt; i ++) {
			printf("%05lld %lld ", L[i].second, L[i].first);
			if (i == cnt - 1) printf("-1\n");
			else printf("%05lld\n", L[i + 1].second);
		}
	};

	print(cnt1, L1);
	print(cnt2, L2);

	return 0;
}