2020 11th Blue Bridge Cup C++B group first provincial real test

Question A: Running training

Answer: 65

#include <bits/stdc++.h>
using namespace std;

int main(void)
{
    
    
	int n = 10000, i;
	for (i = 0; n > 0; i++) {
    
    
		if (i % 2 == 0) {
    
    
			n -= 600;
		} else {
    
    
			n += 300;
		}
	}
	cout << i << endl;
	
	return 0;
}

Question B: Anniversary

Answer: 52038720

Can be calculated by excel, the difference between the two dates is 36138 days

So the answer is: 36138*24*60=52038720

Test Question C: Combined Test

Answer: 10

Assuming there are n people, the kits that need to be used are
$$n/k+0.01\ast n\ast k$$

Extract n, there are
$$n\ast (1/k+0.01\ast k)$$

So, when k=10, there is a minimum

Question D: REPEAT program

This is not QAQ yet

Question F: Dividing sequence

simulation

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(void)
{
    
    
	ll n;
	cin >> n;
	cout << n;
	n /= 2;
	
	while (n > 0) {
    
    
		cout << " " << n;
		n /= 2;
	}
	cout << endl;
	
	return 0;
}

Question G: Decoding

Simulation, expand the abbreviated string

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(void)
{
    
    
	string s;
	cin >> s;
	int n = s.size(), num;
	for (int i = 0; i < n; i++) {
    
    
		if (i != n - 1 && s[i + 1] >= '1' && s[i + 1] <= '9') {
    
    
			num = s[i + 1] - '0';
			while (num--) {
    
    
				cout << s[i];
			}
			i++;
		} else {
    
    
			cout << s[i];
		}
	}
	cout << endl;
	
	return 0;
}

Question H: Going the square

Simple dynamic programming

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int dp[35][35];

int main(void)
{
    
    
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		dp[i][1] = dp[1][i] = 1;
	for (int i = 2; i <= n; i++) {
    
    
		for (int j = 2; j <= m; j++) {
    
    
			if (i % 2 == 0 && j % 2 == 0)
				dp[i][j] = 0;
			else
				dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
		}
	}
	cout << dp[n][m] << endl;
	
	return 0;
}

Test Question I: Integer Concatenation

The complexity of splicing directly is $O\left(n^2\right)$, needs to be optimized
if the x and y splicing, splicing after the number of bits on the y-number bits
may first calculate the number of bits on k-0--9 times and recorded in the array
if The y-digit number exists in the array, indicating that the spliced ​​number may be a multiple of k, so that the splicing judgment is performed.
If the y-digit number is not in the array, it means that the spliced ​​array cannot be a multiple of k. Just skip

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;

ll a[N], num[15];
bool vis[15];

ll connect(ll a, ll b)
{
    
    
	ll bb = b;
	while (bb != 0) {
    
    
		a *= 10;
		bb /= 10;
	}
	a += b;
	
	return a;
} 

int main(void)
{
    
    
	int n, k, res = 0;
	cin >> n >> k;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	
	for (int i = 0; i <= 9; i++)
		vis[(i * k) % 10] = 1;
	
	for (int i = 0; i < n; i++) {
    
    
		for (int j = i + 1; j < n; j++) {
    
    
			ll x = a[i], y = a[j];
			if (vis[y % 10] && connect(x, y) % k == 0)
				res++;
			if (vis[x % 10] && connect(y, x) % k == 0)
				res++;
		}
	}
	cout << res << endl;
	
	return 0;
}

Test Question J: Network Analysis

For this question, I will only use the brute force method of union search, the complexity is O (nm) O(nm)
If$O$$($$n$$m$$)$ is 1, use union search set to merge. If it is 2, traverse all the elements, if in a set, add the corresponding value.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e4 + 5;

int n, m;
ll fa[N], sum[N];

void init()
{
    
    
	for (int i = 1; i <= n; i++)
		fa[i] = i;
}

int find(int x)
{
    
    
	if (fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}

void unite(int x, int y)
{
    
    
	x = find(x), y = find(y);
	fa[x] = y;
}

int main(void)
{
    
    
	int t, a, b;
	cin >> n >> m;
	init();
	while (m--) {
    
    
		cin >> t >> a >> b;
		if (t == 1) {
    
    
			unite(a, b);
		} else {
    
    
			for (int i = 1; i <= n; i++) {
    
    
				if (find(a) == find(i)) {
    
    
					sum[i] += b;
				}
			}
		}
	}
	for (int i = 1; i <= n - 1; i++)
		cout << sum[i] << " ";
	cout << sum[n] << endl;
	
	return 0;
}