题目如下：

 

 

 

1.格式转换

#include <stdio.h>
int main(void)
{
    int h, p, m, d, y;
    while (~scanf("%d:%d,%d/%d/%d", &h, &p, &m, &d, &y))
    {
        printf("%04d%02d%02d,", y, m, d);
        if (h >= 12)
            printf("%02d:%02dPM\n", (h == 12) ? 0 : h-12, p);
        else
            printf("%02d:%02dAM\n", h, p);
    }
    return 0;
}

2.描绘闪电

#include <stdio.h>
int main(void)
{
    int n;
    while (~scanf("%d", &n))
    {
        for (int i = 0; i < n; i++)
        {
            for (int j = 1; j <= n - i; j++)
            {
                printf(" ");
            }
            printf("*");
            printf("\n");
        }//打印第一段

        for (int i = 1; i <= n + 1; i++)
            printf("*");
        printf("\n");//打印第二段

        for (int i = 0; i < n; i++)
        {
            for (int j = n - i-1; j >= 1; j--)
                printf(" ");
            printf("*\n");
        }//打印第三段

        printf("\n");
    }
    return 0;
}

拆分成三段来打印

3.密码设置

#include<stdio.h>
#include<string.h>
int main(void)
{
	char s[35];
	while (~scanf("%s", s))
	{
		int len = strlen(s), sz = 0, xx = 0, dx = 0, ts = 0, flag = 0,flag2=1;
		if (len < 8)
		{
			printf("no\n");
			continue;
		}
		else//首先需要长度大于等于8，才能继续判断具体字符
		{
			for (int i = 0; i < len; i++)
			{
				int e = 1;//e记录是否有题目范围外的其他字符
					if (s[i] >= '0' && s[i] <= '9')//判断数字
						sz = 1,e=0;
					if (s[i] >= 'a' && s[i] <= 'z')//判断小写字母
						xx = 1,e=0;
					if (s[i] >= 'A' && s[i] <= 'Z')//判断大写字母
						dx = 1,e = 0;
					if (s[i] == '!' || s[i] == '@' || s[i] == '#' || s[i] == '$' || s[i] == '*' || s[i] == '~')//判断特殊字符
						ts = 1, e = 0;
				if (e == 1)
				{
					flag2 = 0;
					printf("no\n");
					break;
				}
			}
			if (flag2)
			{
				flag = sz + xx + dx + ts;
				if (flag >= 3)//判断是否满足要求中至少三项
					printf("yes\n");
				else
					printf("no\n");
			}
		}
	}
	return 0;
}

4.伪素数列

#include<stdio.h>
int isprime(int x)
{
	if (x <= 1)
		return 0;
	else
	{
		for (int i = 2; i * i <= x; i++)
		{
			if (x % i == 0)
				return 0;
		}
		return 1;
	}
}
int main(void)
{
	int a[5005] = { 0 };
	int cnt = 0;
	for (int i = 2; cnt <= 5000; i++)
	{
		if (isprime(i))
			a[cnt++] = i;
		else
		{
			int flag = 0;
			int j = i;
			while (j>0&&isprime(j%10))
			{
				j = j / 10;
			}
			if (j == 0)
				a[cnt++] = i;
		}
	}
	
	int x;
	while (~scanf("%d", &x))
	{
		printf("%d\n", a[x - 1]);
	}
	return 0;
}

注意应尽量使代码简洁，不然会增加时间复杂度。

（我写这类题目经常时间超限...）

5.实数相加

本题思路：将两个实数拆分成整数部分和小数部分分别相加

这题有很多易错点（也可能是我的代码比较繁琐），具体的坑都在下面的代码中有标明。

#include <stdio.h>
#include<string.h>
int main(void)
{
    char x[405] = { 0 }, y[405] = { 0 };
    while (~scanf("%s %s", x, y))//用字符串的方式读入两个实数X和Y
    {
        int l1 = strlen(x), l2 = strlen(y);
        char x1[405] = { 0 }, x2[405] = { 0 };//x1是第一个数的整数部分，x2是第一个数的小数部分
        int i=l1, j=l2;//注意i与j的赋值不能为0，若找不到小数点那么整数部分长度就是l1与l2
        for (int k = 0; k < l1; k++)
        {
            if (x[k] == '.')
            {
                i = k;//找到小数点，就是整数和小数部分分割的位置
            }
        }
        for (int k = 0; k < i; k++)
        {
            x1[k] = x[k];
        }
        for (int k = i + 1; k < l1; k++)
        {
            x2[k-i-1] = x[k];
        }//分割x

        char y1[405] = { 0 }, y2[405] = { 0 };
        for (int k = 0; k < l2; k++)
        {
            if (y[k] == '.')
            {
                j = k;
            }
        }
        for (int k = 0; k < j; k++)
        {
            y1[k] = y[k];
        }
        for (int k = j + 1; k < l2; k++)
        {
            y2[k - j - 1] = y[k];
        }//分割y

        int r1[405] = { 0 }, t=0;//r1为储存整数最终结果的数组
        int p = i - 1, q = j - 1;
        

        int r2[405] = { 0 }, t2 = 0;//r2为储存小数最终结果的数组
//因为最终小数部分相加后可能会向整数部分进位，所以先计算小数部分
        int len;
        if (l1 - i > l2 - j)//len的长度应为两者中较长者
            len = l1 - 1 - i-1;
        else
            len = l2 - 1 - j-1;
        while (len >= 0)
        {
            int g=x2[len], r=y2[len];

//因为数组中原本储存的是数字0，所以此处应将其转换为字符0才能进行下面的转换
            if (x2[len] == 0)
                g = '0';
            if (y2[len] == 0)
                r = '0';

            r2[t2] = r2[t2]+g - '0' + r - '0';
            if (r2[t2] >= 10)
            {
                r2[t2] %= 10;
                r2[t2 + 1]++;
            }
            t2++,len--;
        }

        if (r2[t2] != 0)//向整数部分进位
        {
            r1[0] = r1[0] + 1;
        }
        
        while (p >= 0 && q >= 0)
        {
            r1[t] = r1[t] + x1[p] - '0' + y1[q] - '0';
            if (r1[t] >= 10)
            {
                r1[t] = r1[t] % 10;
                r1[t + 1]++;
            }
            t++, p--, q--;
        }

//计算多出来的部分
        while (p >= 0)
        {
            r1[t] = r1[t] + x1[p] - '0';
            t++, p--;
        }
        while (q >= 0)
        {
            r1[t] = r1[t] + y1[q] - '0';
            t++, q--;
        }//至此整数部分加好
        
        if (r1[t] == 0)//判断整数部分的长度
            t = t - 1;

        for (t; t >= 0; t--)
            printf("%d", r1[t]);//输出整数部分
        
        int fir = 0;
        while (r2[fir] == 0&&fir<t2)//找到最后一个不为0的小数
            fir++;
        if (fir != t2)
        {
            printf(".");
            int j = t2 - 1;
            for (j; j >= fir; j--)
                printf("%d", r2[j]);
        }
        printf("\n");
    }
    return 0;
}

6.谍影寻踪

本题思路：指针的基本操作（前插、后插）+结构体

当时写这题的时候没有注意到可能两个人都是曾经出现过的，所以后面的判断重合部分又加了che指针检查是否两者都是出现过的，幸好最后没有时间超限，不知道能不能有大佬提供更简单的解法。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(void)
{
	int n;
	while (~scanf("%d", &n))
	{
		struct node
		{
			int num;
			int cnt = 1;
			int n1;
			int n2;
			struct node* next;
		};

		struct node* head = NULL;
		struct node* A = (struct node*)malloc(sizeof(struct node));
		struct node* B = (struct node*)malloc(sizeof(struct node));
		scanf("%d,%d->%d,%d", &(A->num), &(A->n1), &(B->num), &(B->n1));
		B->next = head;
		A->next = B;
		head = A;
		for (int i = 1; i <= n - 1; i++)
		{
			A = (struct node*)malloc(sizeof(struct node));
			B = (struct node*)malloc(sizeof(struct node));
			scanf("%d,%d->%d,%d", &(A->num), &(A->n1), &(B->num), &(B->n1));
			B->next = NULL;
			A->next = B;
			struct node * cur = head,*pre=NULL;
			while ((cur->num) != A->num && (cur->num) != B->num)
				pre = cur, cur = cur->next;
			if (cur->num == A->num)
			{
				int flag = 1;
				struct node* che = head;
				for (; che->next != NULL; che = che->next)
				{
					
					if (che->num == B->num)
					{
						
						flag = 0;
						break;
					}
				}
				if (che->num == B->num)
				{

					flag = 0;
					
				}
				if (flag)
				{
					B->next = cur->next;
					cur->next = B;
				}
				else
				{
					if (che->n1 != B->n1)
					{
						che->cnt = 2;
						che->n2 = B->n1;
					}
				}
				if (cur->n1 != A->n1)
				{
					cur->cnt = 2;
					cur->n2 = A->n1;
				}
			}
			if (cur->num == B->num)
			{
				int flag = 1;
				struct node* che = head;
				for (; che->next != NULL; che = che->next)
				{
					if (che->num == A->num)
					{
						flag = 0;
						break;
					}
				}
				if (che->num == A->num)
				{
					flag = 0;
				}
				if (flag)
				{
					if (cur == head)
					{
						A->next = cur;
						head = A;
					}
					else
					{
						pre->next = A;
						A->next = cur;
					}
				}
				else
				{
					if (che->n1 != A->n1)
					{
						che->cnt = 2;
						che->n2 = A->n1;
					}
				}
				if (cur->n1 != B->n1)
				{
					cur->cnt = 2;
					cur->n2 = B->n1;
				}
			}
		}
		struct node* p = head;
		while (p->next != NULL)
		{
			if (p->cnt == 2)
			{
				printf("%d,%d#%d->", p->num, p->n1, p->n2);
			}
			else
			{
				printf("%d,%d->", p->num, p->n1);
			}
			p = p->next;
		}
		if (p->cnt == 2)
		{
			printf("%d,%d#%d", p->num, p->n1, p->n2);
		}
		else
		{
			printf("%d,%d", p->num, p->n1);
		}
		printf("\n");
	}
	return 0;
}

以上仅个人的一点浅薄的看法，这是一学期来第一次考试满分，希望能对大家有帮助。如果有更好的解法，欢迎批评指正。