第十一届蓝桥杯C++B组省赛第一场

最新推荐文章于 2022-04-08 10:04:36 发布

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#算法 #c++ #蓝桥杯 #蓝桥杯省赛

蓝桥杯刷题专栏收录该内容

23 篇文章

订阅专栏

本文集包含多种算法题目的解决方案，涉及跑步训练模拟、日期计算、矩阵遍历等问题，通过C++实现并提供详细代码示例。

跑步训练

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

using namespace std;

int main()
{
    int a = 600 / 60, b = 300 / 60;
    
    int t = 0, m = 10000;
    int ans = 0;
    while (m >= a * 60)
    {
        m -= a * 60;
        m += b * 60;
        ans += 120;
    }
    
    if (m)
    {
        ans += m / a;
        cout << ans << endl;
    }
    return 0;
}

纪念日

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

using namespace std;

bool is_leap(int y)
{
    return y % 4 == 0 && y % 100 != 0 || y % 400 == 0;
}

int main()
{
    int cnt = 0;
    for (int i = 1922; i <= 2020; i ++ )
    {
        cnt += 365 + is_leap(i);
    }
    cout << (cnt - 22) * 60 * 24 << endl;
    return 0;
}

合并检测

这题概率论

在这里插入代码片

REPEAT程序

#include<iostream>

using namespace std;

int main()
{
	int ans = 0;
	for(int i = 0; i < 2; i++)
	{
		ans += 4;
		for(int j = 0; j < 5; j++)
		{
			for(int k = 0; k < 6; k++)
			{
				ans += 5;
			}
			ans += 7;
		}
		ans += 8;
	}
	ans += 9;
	cout << ans <<endl;
	return 0;
}

整除序列

#include <iostream>

using namespace std;

int main()
{
    long long int n; 
    cin >> n;
    while(n > 0)
    {
        cout << n << ' ';
        n = n / 2;
    }
    return 0;
}

解码

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

using namespace std;

bool check(char c)
{
    return c >= '0' && c <= '9';
}

int main()
{
    string str;
    cin >> str;
    
    string res;
    for (int i = 0; i < str.size(); i ++ )
    {
        if (check(str[i]))
        {
            int j = i;
            while (j < str.size() && check(str[j])) j ++ ;
            int cnt = stoi(str.substr(i, j - i)) - 1;
            while (cnt -- ) res += str[i - 1];
            i = j - 1;
        }
        else res += str[i];
    }
    cout << res << endl;
    return 0;
}

走方格

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

using namespace std;
const int N = 35;
int n, m;
int f[N][N];

int main()
{
    scanf("%d%d", &n, &m);
    f[1][1] = 1;
    
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
        {
            if (i == 1 && j == 1) continue;
            if (i % 2 || j % 2)
            {
                f[i][j] = f[i - 1][j] + f[i][j - 1];
            }
        }
    
    cout << f[n][m] << endl;
    
    return 0;
}

整数拼接

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

using namespace std;
const int N = 100010;
typedef long long LL;

int n, m;
int a[N], s[11][N];

int main()
{
    scanf("%d%d", &n, &m);
    
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
    
    for (int i = 0; i < n; i ++ )
    {
        LL t = a[i] % m;
        for (int j = 0; j < 11; j ++ )
        {
            s[j][t] ++ ;
            t = t * 10 % m;
        }
    }
    
    LL res = 0;
    for (int i = 0; i < n; i ++ )
    {
        LL t = a[i] % m;
        int len = to_string(a[i]).size();
        res += s[len][(m - t) % m];
        
        LL r = t;
        while (len -- ) r = r * 10 % m;
        if (r == (m - t) % m) res -- ;
    }
    printf("%lld\n", res);
    return 0;
}

超级胶水

这题其实很坑，本来时想着用DP做，发现数据范围不对。模拟了一下数据发现交换顺序不影响答案，好家伙，玩因的是叭…

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

using namespace std;
const int N = 100010;
typedef long long LL;
LL res = 0;
int n;

int main()
{
    scanf("%d", &n);
    LL last = 0;
    for (int i = 1; i <= n; i ++ )
    {
        int x;
        cin >> x;
        if (i > 1)
        {
            res += last * x;
            last = last + x;
        }
        else last = x;
    }
    printf("%lld\n", res);
    return 0;
}

网络分析

不创建新节点写法

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

using namespace std;
const int N = 110010;

int n, m;
int p[N], d[N];

int find(int x)
{
    if (p[x] == x || p[p[x]] == p[x]) return p[x];
    int r = find(p[x]);
    d[x] += d[p[x]];
    p[x] = r;
    return r;
}

int main()
{
    scanf("%d%d", &n, &m);
    
    for (int i = 1; i < N; i ++ ) p[i] = i;
    
    int op, a, b;
    int k = n;
    while (m -- )
    {
        scanf("%d%d%d", &op, &a, &b);
        if (op == 1)
        {   
            a = find(a), b = find(b);
            if (a != b)
            {
                d[a] -= d[b];
                p[a] = b;
            }
        }
        else
        {
            a = find(a);
            d[a] += b;
        }
    }
    
    for (int i = 1; i <= n; i ++ )
        if (find(i) == i) printf("%d ", d[i]);
        else printf("%d ", d[i] + d[find(i)]);
    puts("");
    
    return 0;
}

创建新结点写法

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

using namespace std;
const int N = 110010;

int n, m;
int p[N], d[N];

int find(int x)
{
    if (p[x] == x || p[p[x]] == p[x]) return p[x];
    int r = find(p[x]);
    d[x] += d[p[x]];
    p[x] = r;
    return r;
}

int main()
{
    scanf("%d%d", &n, &m);
    
    for (int i = 1; i < N; i ++ ) p[i] = i;
    
    int op, a, b;
    int k = n;
    while (m -- )
    {
        scanf("%d%d%d", &op, &a, &b);
        if (op == 1)
        {   
            a = find(a), b = find(b);
            if (a != b)
            {
                k ++ ;
                p[a] = p[b] = k;
            }
        }
        else
        {
            a = find(a);
            d[a] += b;
        }
    }
    
    for (int i = 1; i <= n; i ++ )
        if (find(i) == i) printf("%d ", d[i]);
        else printf("%d ", d[i] + d[find(i)]);
    puts("");
    
    return 0;
}