6.4 c++实战

超、凢脫俗 2023-06-12 11:06 15阅读 0赞

1 从键盘中任意输入5个整型数据存到数组a中，编程求出a中相邻两元素之和，并将这些和存到数组b中，按每行三个元素的形式输出。

#include <iostream>
    #include<stdio.h>
    
    using namespace std;
    
    int main()
    { 
        int a[5],b[5],i;
       printf("please input array a:\n");
       for(i=0;i<5;i++)
        scanf("%d",&a[i]);
        for(i=1;i<5;i++)
            b[i]=a[i]+a[i-1];
        for(i=1;i<5;i++)
            { printf("%5d",b[i]);
            if(i>=3)
             printf("\n");
    
            }
    
    }

结果  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDI2OTk1Mg_size_16_color_FFFFFF_t_70]  
**2 选票系统**  
有三个候选人,输入参加选举的人数及每个人选举的内容，输出三个候选人最终得票数以及无效的选票数。

#include <iostream>
    #include<stdio.h>
    
    using namespace std;
    
    int main()
    { 
        int v0=0,v1=0,v2=0,v3=0,i,n,a[50];
        printf("please input the number of electorate:\n");
        scanf("%d",&n);
        printf("please input 1or2or3\n");
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
            for(i=0;i<n;i++)
            { 
                if(a[i]==1)
                    v1++;
                else if(a[i]==2)
                    v2++;
                else if(a[i]==3)
                    v3++;
                else
                    v0++;
            }
            printf("the result:\n");
            printf("candidate1:%d\ncandidate2:%d\ncandidate3:%d\nonuser:%d\n",v1,v2,v3,v0);
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDI2OTk1Mg_size_16_color_FFFFFF_t_70 1]  
**6.4.3 字符串的加密和解密**  
实现字符串的加密与解密  
·加密方式：将字符串中每个字符加上它在字符串中的位置和一个偏移量3

#include <iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include<cstring>
    #define key 3 //偏移量
    
    //函数声明
    char * encrypt(char []);
    char * dencrypt(char []);
    
    int main()
    { 
        char password[50]="1234569";
        encrypt(password);
        printf("加密后字符串为：%s\n",password);
        dencrypt(password);
        printf("解密后字符串为：%s\n",password);
        return 0;
    }
    
    //函数定义
    char * encrypt(char password[])
    { 
        int i=0;
        int count=strlen(password);
        for(;i<count;i++)
        { 
            password[i]=password[i]+i+key;
        }
        return password;
    }
    
    char * dencrypt(char password[])
    { 
        int i=0;
        int count=strlen(password);
        for(;i<count;i++)
        { 
            password[i]=password[i]-i-key;
        }
        return password;
    }

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDI2OTk1Mg_size_16_color_FFFFFF_t_70]: https://img-blog.csdnimg.cn/20191112170458587.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDI2OTk1Mg==,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDI2OTk1Mg_size_16_color_FFFFFF_t_70 1]: https://img-blog.csdnimg.cn/20191113191442446.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDI2OTk1Mg==,size_16,color_FFFFFF,t_70