2021.1.17高一模拟赛题解

点我下载题目

T1

绫濑的作业 | ayase.cpp

就根据题意慢慢模拟就行了
只用 if 就能AC的水题
不过还是有几个坑点的：
比如负号。
（mitruha逐渐CCF化）

#include <bits/stdc++.h>
using namespace std ;
 
double x, v1, v2, v3, t;
char ch1, ch2;
double level = 0;

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

    freopen("ayase.in", "r", stdin);
    freopen("ayase.out", "w", stdout);
	
    cin >> x >> v1;
    
    if (x < 0) v1 *= -1;
    level += v1;
    
    cin >> ch1 >> v2;
    
    if (ch1 == 'W') v2 *= -1;
    level += v2;
    
    cin >> ch2 ;
    if (ch2 == 'Y') 
    {
        cin >> ch2 >> v3;
        
        if (ch2 == 'W') v3 *= -1;	
        level += v3;
    }
    
    if (level == 0) 
    {
	puts("No solution!");
	exit(0);
    }
	
    t = x / level;
    
    if (t < 0)
	puts("No solution!");
    else
	cout << ceil(t);
    return 0;
}

T2

七海的烦恼 | nanami.cpp

不难想到用DFS做
先写出把每个灯更新的函数：

inline void update(int X, int Y)
{
	light[X][Y] = 1 - light[X][Y];
	light[X + 1][Y] = 1 - light[X + 1][Y];
	light[X - 1][Y] = 1 - light[X - 1][Y];
	light[X][Y + 1] = 1 - light[X][Y + 1];
	light[X][Y - 1] = 1 - light[X][Y - 1];
}

然后把每一个点都搜一遍
不过要注意剪枝，否则你会喜提TLE
AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 10;
int Min_ans = 10;
int light[N][N];
int ans = 0;

inline int mitru_read()
{
	int s = 0, w = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9')
	{
		if(ch == '-')
			w = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9')
		s = s * 10 + (ch ^ 48), ch = getchar();
	return s * w; 
}

inline void update(int X, int Y)
{
	light[X][Y] = 1 - light[X][Y];
	light[X + 1][Y] = 1 - light[X + 1][Y];
	light[X - 1][Y] = 1 - light[X - 1][Y];
	light[X][Y + 1] = 1 - light[X][Y + 1];
	light[X][Y - 1] = 1 - light[X][Y - 1];
}

inline void dfs(int cnt)
{
	if(cnt > Min_ans)
		return;
	int tot = 0;
	for(register int i = 1; i <= 3; i++)
	{
		for(register int j = 1; j <= 3; j++)
		{
			tot += light[i][j];
		}
	}
	if(tot == 9)
	{
		ans = cnt - 1;
		if(ans < Min_ans)
			Min_ans = ans;
	}
	for(register int i = 1; i <= 3; i++)
	{
		for(register int j = 1; j <= 3; j++)
		{
			update(i, j);
			dfs(cnt + 1);
			update(i, j);
		}
	}
	return;
}

int main()
{
	freopen("nanami.in", "r", stdin);
	freopen("nanami.out", "w", stdout);
	
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	for(register int i = 1; i <= 3; i++)
	{
		for(register int j = 1; j <= 3; j++)
		{
			light[i][j] = mitru_read();
		}
	}
	dfs(1);
	cout << ans;
	return 0;
}

T3

茉优的房间 | mayu.cpp

推出来是斐波那契数列
直接算，都可以算

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

inline int mitru_read()
{
	int s = 0, w = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9')
	{
		if(ch == '-')
			w = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9')
		s = s * 10 + (ch ^ 48), ch = getchar();
	return s * w; 
}

int m, n;
ll f[Maxn];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	freopen("mayu.in", "r", stdin);
	freopen("mayu.out", "w", stdout);
	
	m = mitru_read();
	n = mitru_read();
	f[1] = 1, f[2] = 2;
	
	for(int i = 3; i <= n - m; i++) 
		f[i] = f[i - 1] + f[i - 2];

	cout << f[n - m];
	return 0;
} 

TX
梦 | yami.cpp
az
我不会做