C/C++ 作业笔记

7-4 查找整数

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#include <stdio.h>

int main() 
{
    int n, b;
    scanf("%d %d", &n, &b);
    int shu[n];
    int found = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", &shu[i]);
        if (shu[i] == b) {
            printf("%d", i);
            found = 1; 
            break;
        }
    }
    
    if (!found) 
    {
        printf("Not Found");
    }
    
    return 0;    
}

7-3 交换最小值和最大值

解题思路；

  • 将数组元素排序，找出最值。
  • 通过最值对原数组遍历，找到元素，并与[0],[n-]交换数值.
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    #include <stdio.h>

    int swap(int *x,int *y)//交换函数
    {
        int jh=0;
        jh = *x;
        *x = *y;
        *y = jh;
    }


    int main()
    {
        int shu[10];
        int shu1[10];
        int n;
        scanf("%d",&n);
        for (int i=0;i<n;i++)
            {
                scanf("%d",&shu[i]);
            }
       for (int u=0;u<n;u++)
           {
            shu1[u] = shu[u];      
           }// 复制原数组
       for (int j=0;j<n-1;j++) 
           {
            for (int k=0;k<n - 1 -j;k++)
                {
                    if (shu[k] > shu[k + 1])
                        {
                            swap(&shu[k],&shu[k+1]);
                        }
                }
           }// 排序

            for (int ch=0;ch<n;ch++)
                {
                  if (shu1[ch] == shu[0])
                  {
                    swap(&shu1[0],&shu1[ch]);
                    break;
                  }
                }// 查找min 并与[0] 交换
            for (int ch=0;ch<n;ch++)
                {
                  if (shu1[ch] == shu[n -1])
                  {
                    swap(&shu1[n - 1],&shu1[ch]);
                    break;
                  } // 查找max
                }
            for (int i=0;i<n;i++)
              {
                 printf("%d ",shu1[i]);
              }
    }

7-2 求矩阵各行元素之和

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#include <stdio.h>
int main()
{
    int m,n;
    scanf("%d %d",&m,&n);
    int shu[m][n];
    for (int i=0;i<m;i++)
        {
         for (int j=0;j<n;j++)
             {
                scanf("%d",&shu[i][j]);
             }
        }
     for (int i=0;i<m;i++)
        {
        int sum = 0;
         for (int j=0;j<n;j++)
             {
                sum += shu[i][j];
             }
             printf("%d\n",sum);
        }
            
}

7-4 字符串加密

解题思路;

  • 元素转换规律: str = ‘Z’ - (str - ‘A’)
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    #include <stdio.h>
    #include <string.h>

    void encrypt(char *str) {
        for (int i = 0; str[i] != '\0'; i++) {
            if (str[i] >= 'A' && str[i] <= 'Z') {
                str[i] = 'Z' - (str[i] - 'A');
            } else if (str[i] >= 'a' && str[i] <= 'z') {
                str[i] = 'z' - (str[i] - 'a');
            }
        }
    }

    int main()
    {
       char str[128];
       fgets(str,128,stdin);   
       encrypt(str);
       printf("%d\n",strlen(str));
       printf("%s",str);
       return 0;
    }

7-5 幸运数字

解题思路;

  • 使用一个数组(count[10])来保存0-9,出现的次数.下标表示lucky_number.
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    #include <stdio.h>
    #include <string.h>

    int main() {
        char input[101]; // 存储输入数字字符串，长度不超过100
        int count[10] = {0}; // 用于统计数字0-9出现的次数
        scanf("%s", input);

        // 遍历输入字符串，统计每个数字出现的次数
        for (int i = 0; i < strlen(input); i++) {
            if (input[i] >= '0' && input[i] <= '9') {
                count[input[i] - '0']++;
            }
        }

        // 找出出现次数最多的数字
        int max_count = 0, lucky_number = 0;
        for (int i = 0; i < 10; i++) {
            if (count[i] > max_count || (count[i] == max_count && i > lucky_number)) {
                max_count = count[i];
                lucky_number = i;
            }
        }

        printf("%d\n", lucky_number);
        return 0;
    }