Withnoidea

外观

PTA云巅码·520钻石争霸赛 2026

约 4526 字大约 15 分钟

c++算法

2026-05-21

提示

封榜后28.9% (0.2897125567322239)

520-1 吾眸唯卿

点击展开题目

思路

按照题目要求打印520 stars in the sky, but only one shines bright in my eye.即可。

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;


int main()
{
 	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	printf("520 stars in the sky, but only one shines bright in my eye.");
	return 0;
}

520-2 恋爱纪念日

点击展开题目

思路

考察分支结构，根据恋爱阶段判断暧昧、热恋、老夫老妻，再根据预算输入对应行为。

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;
int stage, money;

int main()
{
 	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> stage >> money;
	if(stage == 1)
	{
		if(money > 1000) cout << "zhu guang wan can";
		else cout << "he ka fei";
	}
	else if(stage == 2)
	{
		if(money > 10000 ) cout << "hao hua ying yuan VIP bao chang";
		else cout << "gong yuan ye can";
	}
	else
	{
		if(money > 200) cout << "chi huo guo";
		else cout << "shai tai yang";
	}
	return 0;
}

520-3 恋爱中的智商

点击展开题目

思路

记录0出现的所有位置进行输出，然后输出0出现的次数。用vector存储0出现的下标，vector的size就是出现的次数。

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;
int n;


int main()
{
 	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n;
	vector<int> res;
	for(int i = 1; i <= n; i ++)
	{
		int x;
		cin >> x;
		if(x == 0)
			res.push_back(i);
	}
	for(int i = 0; i < res.size(); i ++)
		cout << res[i] << " \n"[i == res.size() - 1];
	return 0;
}

520-4 天作之和

点击展开题目

思路

最大和res从最大b到最小a依次进行判断，若已知m, n = res - m 若((m+1)m+(n+1)n)%(mn)=0\big( (m+1)m + (n+1)n \big) \% (mn) = 0则当前res即为答案

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;
LL a, b;

int main()
{
 	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
 	cin >> a >> b;
 	bool flag = false;
 	for(LL res = b; res >= a; res --)
 	{
 		for(LL m = res - 1; m >= 1; m --)
 		{
 			LL n = res - m;
 			if((1LL * (m + 1) * m + 1LL * (n + 1) * n) % (1LL * m * n) == 0)
			{
				cout << res << endl;
			 	return 0;
			}
		}
	}
	cout << 0 << endl;
	return 0;
}

520-5 情侣号

点击展开题目

思路

判断是否有除了p位置以外，对应位置数字不同的情况，有则输出0,反之输出1。

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;
string p1, p2;
int p, n;

int main()
{
 	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> p >> n;
	for(int i = 0; i < n; i ++)
	{
		cin >> p1 >> p2;
		bool f = true;
		for(int i = 0; i < 11; i ++)
		{
			if(i == p - 1) continue;
			if(p1[i] != p2[i])
			{
				cout << 0 << endl;
				f = false;
				break;
			}
		}
		if(f)
			cout << 1 << endl;
	}
	return 0;
}

520-6 单身狗的派对

点击展开题目

思路

前缀和，寻找长度为ll的最小前缀和。

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;
int a[N];
int l, n;

int main()
{
 	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> l >> n;
	for(int i = 1; i <= n; i ++)
	{
		cin >> a[i];
		a[i] += a[i - 1];
	}
	int idx = 0;
	int res = 2e8;
	int s;
	for(int i = 1; i + l <= n; i ++)
	{
		int s = a[i + l - 1] - a[i - 1];
		if(s < res)
		{
			idx = i;
			res = s;
		}
	}
	cout << idx << ' ' << a[idx + l - 1] - a[idx - 1] << endl;
	return 0;
}

520-7 爱的标记

点击展开题目

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;
string s;
char c1, c2;

int main()
{
 	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> s;
	cin >> c1 >> c2;
	int idx = 0;
	string res;
	while(idx < s.size())
	{
		if(s[idx] == c1)
		{
			string tmp;
			int end = idx + 1;
			while(s[end] != c2 && end < s.size()) 
			{
				tmp += s[end];
				end ++;
			}
			if(end < s.size())
			{
				res += tmp;
				tmp = "";
				idx = end + 1;
			}
		}
		idx ++;
	}
	cout << res << endl;
	return 0;
}

520-8 丘比特之箭

点击展开题目

思路

unordermap

C++
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 1e5 + 10;


struct Couple {
    string u, v;
    int init_idx;
    int id; 
};


bool cmp_lucky(const pair<Couple, int>& a, const pair<Couple, int>& b) {
    if (a.y != b.y) return a.y > b.y;
    return a.x.u < a.x.u;
}


bool cmp_unlucky(const pair<Couple, int>& a, const pair<Couple, int>& b) {
    if (a.y != b.y) return a.y < b.y;
    return a.x.u < a.x.u;
}

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

    int n, k;
    if (!(cin >> n >> k)) return 0;

    vector<Couple> couples(n);
    for (int i = 0; i < n; i++) {
        cin >> couples[i].u >> couples[i].v >> couples[i].init_idx;
        couples[i].id = i;
    }


    unordered_map<string, vector<int>> arrow_cnt;
    
    for (int i = 0; i < k; i++) {
        string name;
        int type;
        cin >> name >> type;
        if (arrow_cnt.find(name) == arrow_cnt.end()) {
            arrow_cnt[name] = vector<int>(2, 0);
        }
        arrow_cnt[name][type]++;
    }

    vector<pair<Couple, int>> lucky_list;
    vector<pair<Couple, int>> unlucky_list;

    for (int i = 0; i < n; i++) {
        string u = couples[i].u;
        string v = couples[i].v;
        int init = couples[i].init_idx;


        int change_u = 0;
        int gold_u = 0, lead_u = 0;
        if (arrow_cnt.find(u) != arrow_cnt.end()) {
            gold_u = arrow_cnt[u][1];
            lead_u = arrow_cnt[u][0];
            change_u = gold_u * 520 - lead_u * 520;
        }


        int change_v = 0;
        int gold_v = 0, lead_v = 0;
        if (arrow_cnt.find(v) != arrow_cnt.end()) {
            gold_v = arrow_cnt[v][1];
            lead_v = arrow_cnt[v][0];
            change_v = gold_v * 520 - lead_v * 520;
        }


        int final_sum = (init + change_u) + (init + change_v);


        bool u_only_gold = (gold_u > 0 && lead_u == 0);
        bool v_only_gold = (gold_v > 0 && lead_v == 0);
        

        bool u_has_lead = (lead_u > 0);
        bool v_has_lead = (lead_v > 0);

        if (u_only_gold && v_only_gold) {
            lucky_list.push_back({couples[i], final_sum});
        }
        if (u_has_lead || v_has_lead) {
            unlucky_list.push_back({couples[i], final_sum});
        }
    }


    sort(lucky_list.begin(), lucky_list.end(), cmp_lucky);
    sort(unlucky_list.begin(), unlucky_list.end(), cmp_unlucky);

    int lucky_out = min(3, (int)lucky_list.size());
    for (int i = 0; i < lucky_out; i++) {
        cout << lucky_list[i].x.u << " " << lucky_list[i].x.v << " " << lucky_list[i].y << "\n";
    }


    int unlucky_out = min(3, (int)unlucky_list.size());
    for (int i = 0; i < unlucky_out; i++) {
        cout << unlucky_list[i].x.u << " " << unlucky_list[i].x.v << " " << unlucky_list[i].y << "\n";
    }

    return 0;
}