买东西打折问题，整型提升练习，说谎问题，杨辉三角问题，菱形图案问题

买东西打折

int main()
{
    double p = 0.0;
    int m = 0;
    int d = 0;
    int flag = 0;
    scanf("%lf %d %d %d", &p, &m, &d, &flag);
    if (m == 12 && d == 12)
    {
        p = p * 0.8 - flag * 50;
    }
    else if (11 == m && 11 == d)
    {
        p = p * 0.7 - flag * 50;
    }
    if (p < 0)
        printf("0.00");
    else
        printf("%.2lf", p);

    return 0;
}

思路

1.先判断日期

2.根据日期进行相应计算

3.如果用券后小于0要赋值为0

整型提升

int main()
{
	//char -128~127
	//unsigned char - 0~255
	unsigned char a = 200;
	//00000000000000000000000011001000
	//11001000 - a截断
	//11111111111111111111111111001000
	unsigned char b = 100;
	//01100100 - b 截断
	//00000000000000000000000001100100
	unsigned char c = 0;

	c = a + b;
	//整型提升
	//00000000000000000000000011001000
	//00000000000000000000000001100100
	//00000000000000000000000100101100
	//00101100 - c截断
	//00000000000000000000000000101100
	//44
	printf("%d %d",a + b, c);

	return 0;
}

掌握数据的储存

半真半假问题

int main()
{
	int a = 0;
	int b = 0;
	int c = 0;
	int d = 0;
	int e = 0;
	for (a = 1; a <= 5; a++)
	{
		for (b = 1; b <= 5; b++)
		{
			for (c = 1; c <= 5; c++)
			{
				for (d = 1; d <= 5; d++)
				{
					for (e = 1; e <= 5; e++)
					{
						if (((b == 2) + (a == 3) == 1)
							&& ((b == 2) + (e == 4) == 1)
							&& ((c == 1) + (d == 2) == 1)
							&& ((c == 5) + (d == 3) == 1)
							&& ((d == 4) + (a == 1) == 1))
						{
							if(a*b*c*d*e==120)
							printf("a=%d,b=%d,c=%d,d=%d,e=%d\n", a, b, c, d, e);

						}
					}
				}
			}
		}
	}
	return 0;
} 

利用一真一假相加为1的原理，然后遍历每一种可能暴力求解

大小端存储

int main()
{
	unsigned int a = 0x1234;
	unsigned char b = *(unsigned char*)&a;
	printf("%x", b);
	return 0;
}

真假问题

int main()
{
	int killer = 0;
	for (killer = 'a'; killer < 'd'; killer++)
	{
		if ((killer != 'a') + (killer == 'c') + (killer == 'd') + (killer != 'd') == 3)
		{
			printf("%c\n", killer);
		}
	}

	return 0;
}

遍历四种凶手的可能性

判断每种组合是否满足三真一假

打印杨辉三角

int main()
{
	int arr[10][10] = { 0 };
	int i = 0;
	int j = 0;
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j < 10; j++)
		{
			if (j == 0)
			{
				arr[i][j] = 1;
			}
			if (i == j)
			{
				arr[i][j] = 1;
			}
			if (i>=2&&j>=1)
			{
				arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];
			}

		}
	}
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j + i < 10 - 1; j++)
		{
			printf(" ");
		}
		for (j = 0; j <= i; j++)
		{

			printf("%d ",arr[i][j]);
		}
		printf("\n");
	}
	return 0;
}

先初始化

然后打印

易错点：打印的时候要注意j的取值要与i联系

菱形图案打印

int main()
{
	int line = 0;
	scanf("%d", &line);
	//上
	int i = 0;
	for (i = 0; i < line; i++)
	{
		//打印一行
		//空格
		int j = 0;
		for (j = 0; j < line -1-i; j++)
		{
			printf(" ");
		}
		//*
		for (j = 0;j<2*i+1;j++)
		{
			printf("*");
		}
		printf("\n");
	}
	//下
	for (i = 0; i < line-1; i++)
	{
		//打印一行
		//空格
		int j = 0;
		for (j = 0; j <= i; j++)
		{
			printf(" ");
		}
		//*
		for (j = 0; j < 2*(line-1-i) - 1; j++)
		{
			printf("*");
		}
		printf("\n");
	}

笔记思路