P1428 小鱼比可爱

image-20240428213943899

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#include <iostream>
const int N = 105;
int a[N];
using namespace std;
int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	for (int i = 0; i < n; i++)
	{
		int count = 0;
		for (int j = 0; j < i; j++)
		{
			if (a[i] > a[j])
				count++;
		}
		cout << count << " ";
	}
}

P1427 小鱼的数字游戏

image-20240428230234332

看到的优秀题解，方便你快速过一遍知识：一个红题带你了解绿（黄）题知识点 - 洛谷专栏 (luogu.com.cn)

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#include <iostream>
#include <stack>
using namespace std;
const int N = 105;
int a[N];
stack<int> st;
int main()
{
	int tmp;
	while (1)
	{
		cin >> tmp;
		if (tmp == 0)
			break;
		st.push(tmp);
	}
	while (!st.empty())
	{
		cout << st.top() << " ";
		st.pop();
	}
}

P5727 【深基5.例3】冰雹猜想

image-20240514193755513

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#include <iostream>
#include <stack>
using namespace std;
const int N = 10005;
int a[N];
stack<int> st;
int main()
{
	int n;
	cin >> n;
	st.push(n);
	while (n != 1)
	{
		if (n % 2 == 0)
		{
			n = n / 2;
		}
		else
		{
			n = n * 3 + 1;
		}
		st.push(n);
	}
	while (!st.empty())
	{
		cout << st.top() << " ";
		st.pop();
	}
}

P1047 [NOIP2005 普及组] 校门外的树

image-20240429175906480

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#include <iostream>

using namespace std;
const int N = 100005;
int a[N];

int main()
{
	int n, m;
	cin >> n >> m;
	n += 1;
	int l, r;
	for (int i = 0; i < m; i++)
	{
		cin >> l >> r;
		for (int j = l; j <= r; j++)
			a[j] = 1;
	}
	int count=0;
	for (int i = 0; i < n; i++)
	{
		if (a[i] == 0)
			count++;

	}
	cout << count ;
}

P5728 【深基5.例5】旗鼓相当的对手

image-20240429182951046

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#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1005;
int a[N][3];

int main()
{
	int n, count = 0;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i][0] >> a[i][1] >> a[i][2];
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = i+1; j < n; j++)	//注意+1，不然比较的是自己，会多n次
		{
			if (abs(a[i][0] - a[j][0]) <= 5 && abs(a[i][1] - a[j][1]) <= 5 &&
				abs(a[i][2] - a[j][2]) <= 5 && abs(a[i][0] + a[i][1] + a[i][2] - a[j][0] - a[j][1] - a[j][2]) <= 10)			//绝对值小于
				count++;
		}
	}

	cout << count;

}

P5729 【深基5.例7】工艺品制作

image-20240429195454137

这题要注意x，y，z的范围。这里我采用边挖方块边计算的方法，可以省下一次三重循环遍历数组：

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#include <iostream>
#include <algorithm>
using namespace std;
const int N = 25;
int a[N][N][N];

int main()
{
	int w, x, h;
	cin >> w >> x >> h;
	int q;
	cin >> q;

	int x1, y1, z1;
	int x2, y2, z2;
	int count = 0;
	for (int i = 0; i < q; i++)
	{
		cin >> x1 >> y1 >> z1;
		cin >> x2 >> y2 >> z2;
		for (int j = x1; j <= x2; j++)
		{
			for (int k = y1; k <= y2; k++)
			{
				for (int l = z1; l <= z2; l++)
				{
				
					if (a[j][k][l] == 0)
					{
						count++;
						a[j][k][l] = -1;
					}
					
				}
			}
		}
	}
	/*
		若两坐标点的位置大小不确定，还需要
		for (int j = min(x1,x2); j <= max(x1,x2); j++)
			for (int k = min(y1,y2); k <= max(y1,y2); k++)
				for (int l = min(z1,z2); l <= (z1,z2); l++)
				
	*/

	cout << w*x*h-count;

}

P2550 [AHOI2001] 彩票摇奖

image-20240429202113371

我们可以将能数组中中奖的号码的数字在数组的位置设置为1，这样只需要对每个彩票号码取数组下标即可判断是否中奖。

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#include <iostream>
#include <algorithm>
using namespace std;
const int N = 35;
int a[N];
int b[10];		//中奖次数数组

int main()
{
	int n;
	cin >> n;
	for (int i = 1; i <= 7; i++)
	{	
		int tmp;
		cin >> tmp;
		a[tmp] = 1;
	}
	for (int i = 0; i < n; i++)
	{
		int count = 0;
		int tmp;
		for (int i = 1; i <= 7; i++)
		{
			cin >> tmp;
			if (a[tmp] == 1)
				count++;
		}
		b[count]++;
	}
	for (int i = 7; i >=1 ; i--)
	{
		cout << b[i] << " ";
	}

}

P2615 [NOIP2015 提高组] 神奇的幻方

image-20240429221324882

做的有点麻，经典行列分不清，然后还自作聪明把条件合并。下次要细心，把行列分清QwQ。这题的话，看题写判断就OK了。

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#include <iostream>
#include <algorithm>
using namespace std;
const int N = 45;
int a[N][N];
int main()
{
	int n;
	cin >> n;
		
	int raw = 1, col = n/2 + 1;		//表示k-1数字的位置,初始化为1的位置·
	a[raw][col] = 1;
	for (int i = 2; i <=  n*n; i++)
	{
		if (raw == 1 && col != n )	
		{
			raw = n ;
			col = col + 1;
			a[raw][col] = i;
		
		}
		else if (raw != 1 && col==n )
		{		
			raw = raw - 1;
			col = 1;
			a[raw][col] = i;
			
		}
		else if(raw == 1 && col == n)
		{
			raw = raw + 1;
			col = col ;
			a[raw][col] = i;

		}
		else  
		{
			if (a[raw - 1][col + 1] == 0)		//右上
			{
				raw = raw - 1;
				col = col + 1;
				a[raw][col] = i;
				
			}
			else
			{
				raw = raw + 1;
				col = col ;
				a[raw][col] = i;
			}
		}
		
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			cout << a[i][j] << " ";
		}
		cout << endl;
	}

}

在题解中，有一种新的方案题解 P2615 【神奇的幻方】 - 洛谷专栏 (luogu.com.cn)，可以总结一下从第一行中点开始，把每一个下面的数放到“右上角”，若右上角有数，则放到正下方。当i=2时，1的右上方为2（取余），当i=3时 ,2的右上方为3（取余）,当i=4时，3的右上方已经被1占用，因此放入下方。

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8 1 6
3 5 7
4 9 2
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#include <iostream>
#include <algorithm>
using namespace std;
const int N = 45;
int a[N][N];

int main()
{
	int n;
	cin >> n;
	int raw = 1, col = n/2 + 1;		
	
	for (int i = 1; i <=  n*n; i++)
	{
		a[raw][col] = i;
		if (a[(raw - 2 + n) % n + 1][col % n + 1] == 0)		//要避免数组下标取0
			raw = (raw - 2 + n) % n + 1, col = col % n + 1;
		else
			raw = raw % n + 1;
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			cout << a[i][j] << " ";
		}
		cout << endl;
	}

}

P5730 【深基5.例10】显示屏

image-20240514193936527

打表仙人（bushi），我们每四列输出一个字母，先把.所在下标赋值，最后为未初始化的下标赋X，对于最后一列不能有.的问题，我们直接不输出即可：

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#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1035;
char a[5][N];


int main()
{
	int n;
	cin >> n;
	int col = 0;		//表示该数字开始的列
	for(int i = 0;i < n;i++)
	{
		char tmp;
		cin >> tmp;

		if (tmp == '0')
			a[1][col + 1] = '.', a[2][col + 1] = '.', a[3][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '1')
			a[0][col] = '.', a[1][col] = '.', a[2][col] = '.', a[3][col] = '.', a[4][col] = '.',
			a[0][col + 1] = '.', a[1][col + 1] = '.', a[2][col + 1] = '.', a[3][col + 1] = '.', a[4][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '2')
			a[1][col] = '.', a[1][col + 1] = '.', a[3][col + 1] = '.', a[3][col + 2] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '3')
			a[1][col] = '.', a[1][col + 1] = '.', a[3][col] = '.', a[3][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '4')
			a[3][col] = '.', a[3][col + 1] = '.', a[4][col] = '.', a[4][col + 1] = '.', a[0][col+1] = '.', a[1][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '5')
			a[1][col + 1] = '.', a[1][col + 2] = '.', a[3][col] = '.', a[3][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '6')
			a[1][col + 1] = '.', a[1][col + 2] = '.', a[3][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '7')
			a[1][col] = '.', a[2][col] = '.', a[3][col] = '.', a[4][col] = '.',
			a[1][col + 1] = '.', a[2][col + 1] = '.', a[3][col + 1] = '.', a[4][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '8')
			a[1][col + 1] = '.', a[3][col + 1] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		if (tmp == '9')
			a[1][col + 1] = '.', a[3][col + 1] = '.', a[3][col] = '.',
			a[0][col + 3] = '.', a[1][col + 3] = '.', a[2][col + 3] = '.', a[3][col + 3] = '.', a[4][col + 3] = '.';
		col += 4;
	}
	

	for (int i = 0; i < 5; i++)
	{
		for (int j = 0; j < n * 4 - 1; j++)
		{
			
			if (a[i][j] == 0)
				a[i][j] = 'X';
			cout << a[i][j];
		}
		cout << endl;
	}
}

也是看了一下题解题解 P5730 【【深基5.例10】显示屏】 - 洛谷专栏 (luogu.com.cn)，感觉这种方法是最优解，能够拥有最快的打表速度。

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string ans[10];
	if(a[i]=='1')
		{
		ans[1]+="..X.";
		ans[2]+="..X.";
		ans[3]+="..X.";
		ans[4]+="..X.";
		ans[5]+="..X.";
		}
//最后输出也是把最后的.截取掉

#

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#include <iostream>
using namespace std;
const int N = 15;
int a[N];

int main()
{
	int l, r;
	cin >> l >> r;
	for (int i = l; i <= r; i++)
	{
		int tmp = i;
		while (tmp  != 0)
		{
			a[tmp % 10]++;
			tmp /= 10;
		}
	}
	for (int i = 0; i < 10; i++)
	{
		cout << a[i] << " ";
	}
}

P5731 【深基5.习6】蛇形方阵

image-20240514194147792

把最后一次填空交给下一循环

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#include <iostream>
using namespace std;
const int N = 15;
int arr[N][N];

int main()
{
	int n;
	cin >> n;
	int x = 1, y = 0, z = 1;
	while (z <= n * n)
	{
		while (y < n && !arr[x][y+1]) {
			y++;
			arr[x][y] = z;
			z++;
		}
		while (x < n && !arr[x+1][y]) {
			x++;
			arr[x][y] = z;
			z++;
		}
		while (y >1  && !arr[x][y-1]) {
			y--;
			arr[x][y] = z;
			z++;
		}
		while (x > 1 && !arr[x-1][y]) {
			x--;
			arr[x][y] = z;
			z++;
		}
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++) printf("%3d", arr[i][j]);
		printf("\n");
	}
}

P2141 [NOIP2014 普及组] 珠心算测验

image-20240514203336359

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#include <iostream>
using namespace std;
const int N = 20005;
bool a[N];
int b[105];

int main()
{
	int n;
	cin >> n;
	int count = 0;
	for (int i = 0; i < n; i++)
	{
		cin >> b[i];
	}	
	for (int i = 0; i < n; i++)
	{
		for (int j = i + 1; j < n; j++)
		{
			a[b[i] + b[j]] = true;
		}
	}
	for (int i = 0; i < n; i++)
	{
		if (a[b[i]] == true)
		{
			count++;
		}
	}
	cout << count << endl;
}

P1161 开灯

image-20240514194446149

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#include <iostream>
#include<cmath>
using namespace std;


int main()
{
	int n;
	cin >> n;
	double a;
	int b;
	int ans = 0;
	while (n--)
	{
		cin >> a >> b;
		for (int i = 1; i <= b; i++)
		{
			ans ^= (int)floor(a * i);
		}

	}

	cout << ans;
}

P5732 【深基5.习7】杨辉三角

image-20240514202236754

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#include <iostream>

using namespace std;
const int N = 25;
int arr[N][N];

int main()
{
	int n;
	cin >> n;
	arr[0][0] = 1;
	for (int i = 1; i < n; i++)
	{
		 arr[i][i] =  1;
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= i; j++)
		{
			arr[i][j] = arr[i - 1 ][j] + arr[i - 1][j - 1];
		}

	}


	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= i; j++)
		{
			cout << arr[i][j] << " ";
		}
		cout << endl;
	}
}

P1789 【Mc生存】插火把

image-20240514202052617

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#include <iostream>
using namespace std;
const int N = 105;
int arr[N][N];

int main() {
  int Count = 0;
  int num;
  cin >> num;
  int a, b;
  cin >> a >> b;
  while (a--) {
    int x, y;
    cin >> x >> y;
    arr[x][y] = 1;
    for (int i = -1; i < 2; i++)
      for (int j = -1; j < 2; j++) {
        if (x + i >= 0 && y + j >= 0)
          arr[x + i][y + j] = 1;
      }
    arr[x + 2][y] = 1;
    arr[x][y + 2] = 1;

    if (x - 2 >= 0)
      arr[x - 2][y] = 1;
    if (y - 2 >= 0)
      arr[x][y - 2] = 1;
  }
  while (b--) {
    int x, y;
    cin >> x >> y;
    for (int i = -2; i < 3; i++) {
      for (int j = -2; j < 3; j++) {
        if (x + i >= 0 && y + j >= 0)
          arr[x + i][y + j] = 1;
      }
    }
  }

  for (int i = 1; i <= num; i++) {
    for (int j = 1; j <= num; j++) {
      if (arr[i][j] == 0)
        Count++;
    }
  }
  cout << Count;
}

压缩技术

image-20240514201704482

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#include <iostream>
#include <assert.h>
using namespace std;
const int N =205;
int arr[N][N];

int main()
{

	int num;
	cin >> num;
	int sum = num * num;
	int count = 0;
	int raw = 0, col = 0;
	while (count < sum)
	{
		int a, b;
		cin >> a >> b;
		count += a + b; 
		
		while(a--)
		{
			arr[raw][col] = 0;
			col++;
			if (col == num )
			{
				col = 0;
				raw++;
			}
		}
	
		while(b--)
		{
			arr[raw][col] = 1;
			col++;
			if (col == num)
			{
				col = 0;
				raw++;
			}
		}
	}
	for (int i = 0; i < num; i++)
	{
	
		for (int j = 0; j < num; j++)
			cout << arr[i][j] ;
		cout << endl;
	}

	return 0;
}

压缩技术[续集版]

image-20240514201732185

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#include <iostream>
#include<cmath>
using namespace std;
const int N =40005;
int arr[N];

int main()
{


	int total = 0, count = 0;

	char tmp;
	char flag = '0';		//注意赋值为字符而不是
	int i = 0;
	while (cin >> tmp)
	{
		total++;
		if (tmp == flag)
			count++;
		else
		{
			arr[i++] = count;
			flag = tmp;
			count = 1;
		}
	}

	cout << sqrt(total) << " ";
	
	for(int j =0; j < i; j++)
	{
		cout << arr[j] << " ";

	}
	cout << count;

	return 0;
}

P1205 [USACO1.2] 方块转换 Transformations

转180度即转两次90度，转270次即转三次90度，我们可以通过90度旋转函数调用来实现图片旋转。

旋转矩阵我们可以通过先转置矩阵，然后再将第i行与n-i行进行交换，得到顺时针旋转90度的矩阵。

image-20240514203408143

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#include<iostream>
using namespace std;

const int N = 15;
char matrix[N][N];
char matrixcopy[N][N];
char matrix2[N][N];

int n;

// 转置矩阵
void transpose() {
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            swap(matrix[i][j], matrix[j][i]);
        }
    }
}

// 水平翻转
void reverseRows() {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n / 2; j++) {
            swap(matrix[i][j], matrix[i][n - j - 1]);
        }
    }
}

// 顺时针旋转矩阵
void rotateClockwise() {
    transpose();
    reverseRows();
}

// 判断结果
bool JudgeRes() {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] != matrix2[i][j]) {
                return false;
            }
        }
    }
    return true;
}

// 初始化矩阵
void Init() {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            matrix[i][j] = matrixcopy[i][j];
        }
    }
}

// 检查旋转一次是否匹配
bool res1() {
    rotateClockwise();
    if (JudgeRes()) return true;
    return false;
}

// 检查旋转两次是否匹配
bool res2() {
    rotateClockwise();
    if (JudgeRes()) return true;
    return false;
}

// 检查旋转三次是否匹配
bool res3() {
    rotateClockwise();
    if (JudgeRes()) return true;
    return false;
}

// 检查水平翻转是否匹配
bool res4() {
    Init();
    reverseRows();
    if (JudgeRes()) return true;
    return false;
}

// 检查旋转后水平翻转是否匹配
bool res5() {
    bool flag;
    res1();
    if (JudgeRes()) return true;
    res2();
    if (JudgeRes()) return true;
    res3();
    if (JudgeRes()) return true;
    return false;
}

// 检查原始矩阵是否匹配
bool res6() {
    Init();
    if (JudgeRes()) return true;
    return false;
}

// 无操作，总是返回真
bool res7() {
    return true;
}

// 判断函数
void judge() {
    if (res1()) { cout << 1 << endl; return; }
    if (res2()) { cout << 2 << endl; return; }
    if (res3()) { cout << 3 << endl; return; }
    if (res4()) { cout << 4 << endl; return; }
    if (res5()) { cout << 5 << endl; return; }
    if (res6()) { cout << 6 << endl; return; }
    if (res7()) { cout << 7 << endl; return; }
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> matrix[i][j];
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            matrixcopy[i][j] = matrix[i][j];
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> matrix2[i][j];
        }
    }
    judge();
    return 0;
}