2020蓝桥杯模拟赛(一)

2020蓝桥杯校内模拟赛（一） -- 比较简单

前四题基础填空比较水为 计算器算就行

程序代码题

#include <string>
#include <iostream>
using namespace std;
char c[5] = { 'a','e','i','o','u' };
int check(char s) {
	int tag = 0;
	for (int i = 0; i < 5; i++) {
		if (s == c[i]) {
			tag = 1;
		}
	}
	if (tag == 1) return 1;
	return 0;
}
int main() {
	string s;
	cin >> s;
	int i = 0, j;
	int tag = 0, cnt = 0;//0为辅音
	
	while (i < s.size()) {
		while (check(s[i]) == tag && i < s.size()) {
			i++;
		}
		if (tag) tag = 0;//状态翻转
		else
			tag = 1;
		cnt++;

	}
	if (cnt == 4) {
		printf("Yes
");
	}
	else printf("No
");
	return 0;
}

#include <string>
#include <iostream>
using namespace std;
int Sum(int a) {
	int sum = 0;
	while (a) {
		sum += (a % 10) * (a % 10);
		a /= 10;
	}

	return sum;
}
int main() {
	int n, m;
	int ans = 0;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		if (Sum(i) == m) ans++;
	}
	cout << ans << endl;
	return 0;
}

#include <string>
#include <iostream>
using namespace std;
int main() {
	int n, a, b, c;
	cin >> n;
	cin >> a >> b >> c;
	int cnt = 0;
	for (int i = 1; i <= n; i++) {
		if (i % a != 0 && i % b != 0 && i % c != 0) {
			cnt++;
		}
	}
	cout << cnt << endl;
	return 0;
}

O(m * n)最坏时间复杂度

#include <string>
#include <iostream>
using namespace std;
int n, m;
int r, c;

bool in(int x, int y) {
	if (x + 1 == c && y + 1 == r) {
		return true;
	}
	else
		return false;
}
int main() {
	
	cin >> n >> m;
	cin >> c >> r;
	int x = 0, y = 0;//初始位置

	int a = 0 , b = 0;
	int s = 0;
	while (1) {//模拟过程
		
		for (int i = 0; i < m - 2 * a; i++) { // 右走
			s++;
			if (in(x, y)) {
				cout << s << endl;
				return 0;
			}
			y++;
			
		}
		y--;
		x++;
		for (int i = 0; i < n - 2 * a - 1; i++) {// 下走
			s++;
			if (in(x, y)) {
				cout << s << endl;
				return 0;
			}
			x++;
		}
		x--;

		y--;
		for (int i = 0; i < m - 2 * a - 1; i++) { // 左走
			s++;
			if (in(x, y)) {
				cout << s << endl;
				return 0;
			}
			y--;
		}
		y++;
		x--;
		for (int i = 0; i < n - 2 * a - 2; i++) { // 上走
			s++;
			if (in(x, y)) {
				cout << s << endl;
				return 0;
			}
			x--;
		}
		x++;
		y++;
		a++; //进入下一个模拟
	}
	return 0;
}

非DP做法

#include <string>
#include <iostream>
#include <vector>
#include <math.h>

using namespace std;
int num = 0;
void dfs( int a, int b);

void f( int i, int n)
{
	num++;
	num = num % 10000;
	
	if (abs(n - i) > 1)
		dfs( n, i);
}

void dfs( int a, int b)
{
	int c = abs(a - b);
	for (int i = 1; i < c; i++)
	{
		f( i, b);
	}
}

int main()
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++)
		f( i, n);

	cout << num;
	return 0;
}

//dfs
#include <bits/stdc++.h> 
using namespace std;
int n;double d;
struct s{
	double x,y,h;
};
s ss[1001];
double ma=0;
int c(s a,s b)
{
	return a.h>b.h;
} 
void f(int a,double sum,double hh)
{
	//cout<<a<<" "<<sum<<" "<<hh<<endl;
	if(sum>ma)ma=sum;
	for(int i=a+1;i<n;i++)
	{
		if(ss[i].h<ss[a].h)
		{
			double y=sqrt((ss[a].x-ss[i].x)*(ss[a].x-ss[i].x)+(ss[a].y-ss[i].y)*(ss[a].y-ss[i].y));
			double x=sqrt((ss[a].x-ss[i].x)*(ss[a].x-ss[i].x)+(ss[a].y-ss[i].y)*(ss[a].y-ss[i].y)+(ss[a].h-ss[i].h)*(ss[a].h-ss[i].h));
			if(x<=d)
			{
				f(i,sum+y,ss[i].h);
			}
		}
	}
}
int main() 
{

	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>ss[i].x>>ss[i].y>>ss[i].h;
	}
	sort(ss,ss+n,c);
	//cout<<ss[0].h<<endl;
	cin>>d;
	
	//for(int i=0;i<n;i++)
	//{
		f(0,0,ss[0].h);
	//}
	printf("%.2lf
",ma);
	return 0;
}

DP做法

• 数组dp[i][j]储存以ij为前缀，能衍生出的子序列的个数。
• 例如，以41为前缀，可以衍生出411、412两种，再加上41本身就是3种。
• 递推关系：dp[i][j] = dp[j][1] + … + dp[j][abs(i-j)-1] + 1（ij本身）

#include <iostream>
#include <vector>
using namespace std;
const int N = 1001;
vector<vector<int>> dp(N, vector<int>(N, -1));

int countSeq(int first, int second)
{
    if (dp[first][second] != -1)
        return dp[first][second];
    int t = 1;
    for (int i = 1; i < abs(first - second); i++)
    {
        t += countSeq(second, i);
    }
    dp[first][second] = t % 10000;
    return dp[first][second];
}
int main()
{
    int n, ans(0);
    cin >> n;
    dp[1][1] = 1;
    for (int i = 1; i <= n; i++)
    {
        ans = (ans + countSeq(n, i)) % 10000;
    }
    cout << ans << endl;
    return 0;
}