1、将一个数倒转输出:

  (1)限定使用递归,第一反应想到的方法 

 1 #include<stdio.h>
 2  int array[10];
 3  int i=0;
 4  void convert(int n){
 5      if(n==0)
 6          return;
 7      else{
 8          array[i++]=n%10;
 9          convert(n/10);
10      }
11  }
12  
13  int main(){
14      int number=12345;
15      convert(number);
16      for(int j=0;j<i;j++){
17          printf("%d",array[j]);
18      } 
19  }

  (2)使用数组指针:

 1  #include<stdio.h>
 2  #include<stdlib.h>
 3  
 4  void convert(int * result,int num){
 5      if(num>10)
 6          convert(result +1,num/10);
 7      *result=num%10;
 8  }
 9  
10  int main(){
11      int number=12468;
12      int result[10]={-1};
13      convert(result,number);
14      printf("%d\n",number);
15      for(int i=0;i<10;i++)
16          if(result[i]>0)
17              printf("%d",result[i]);
18  }

 

2、输出学生的成绩大于平均成绩的学生:

  (1)C 语言版用递归:

  

 1 /*
 2      这里有一个疑问:在加与不加return average, average的值在每次返回的时候是一样的 
 3      按理说,如果不加return average的值每次average的值应该不一样 ? 除非 average每次递归后保存一次值 
 4  */
 5   
 6  #include<stdio.h>
 7      double find(int total,int n){
 8      int number,score,average;
 9      scanf("%d",&number);
10      if(number!=0){
11          scanf("%d",&score);
12          average=find(total+score,n+1);
13          if(score>=average){
14              printf(" number=%d: score=%d\n",number,score);
15          }
16          printf(" average=%d ",average);
17         // return average;
18      }else{
19          printf("Average=%d\n",total/n);
20          return total/n;
21      }
22  }
23  
24  int main(){
25      find(0,0);
26  }

  (2)Java用的for循环:

    

 1 package employeer_algorithm;
 2 
 3 import java.util.Scanner;
 4 
 5 class Student {
 6     private String name="";
 7     private double store=0;
 8     public String getName() {
 9         return name;
10     }
11     public void setName(String name) {
12         this.name = name;
13     }
14     public double getStore() {
15         return store;
16     }
17     public void setStore(double store) {
18         this.store = store;
19     }
20 }
21 public class Test{
22     public static void main(String args[]){
23         double sum=0;
24         Student students[]= new Student [5];
25         System.out.println("Please input informations of students:");
26         Scanner sc = new Scanner(System.in);
27         for(int i=0;i<students.length;i++){
28             System.out.println("The informations of"+i+"th student are format as name,store ");
29             students[i]= new Student();//刚开始的时候没有进行new 对象,以为在开辟数组的时候已经new好
30             students[i].setName(sc.next());
31             students[i].setStore(sc.nextDouble());
32             sum+=students[i].getStore();
33         }
34         double average = sum/students.length;
35         for(int i=0;i<students.length;i++){
36             if(average<students[i].getStore())
37                 System.out.println(students[i].getName()+"'s score above average store");
38         }
39         
40     }
41 }

 

3、组合问题:从M个不同字符中选择N个字符的不同组合:

  

 1 #include<stdio.h>
 2  #include<stdlib.h>
 3  #include<cmath>
 4  void find(char *source, char *result, int n){
 5      if(n==1){
 6          while(*source){
 7              printf("%s %c\n",result,*source++);
 8          }
 9      }else{
10          int i,j;
11          for(i=0;source[i]!=0;i++);
12          for(j=0;result[j]!=0;j++);
13          for(;i>=n;i--){
14              result[j]=*source++;
15              result[j+1]='\0';
16              find(source,result,n-1);
17          }
18      }
19  }
20  
21  int main(){
22      int const n = 3;
23      char * source="ABCDEF";
24     char result[n+1]={0};
25      if(n>0)
26      find(source,result,3); 
27  }