PAT Level B 1005 Continue (3n+1) Conjecture (25 points)

PAT Level B exercise summary

PAT Level B 1005 Continue (3n+1) Conjecture (25 points) I hope my ideas can help you.

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1. Question 1005

输入样例：
6
3 5 6 7 8 11

输出样例：
7 6

2. Analysis

Enter up to 100 digits according to the title, and the input number is 1<n<=100; think of using index, take the input number as the array subscript, cut this number n and compare it with the index [n]. If in the index [ If n] exists, then this number is not a key number. Change the position of index [n] to 0; finally remove the number 0 in the index array, fill in the subscripts in turn, and finally output it backwards.

(The demonstration is for reference only, the value of index [n] is changed to 0)

输入后----->
索引：3  5  6  7  8  11
--------->
算3：5 8 4 2 1；
索引：3  0  6  7  0  11
算5：8 4 2 1；
索引：3  0  6  7  0  11
算6：3 5 8 4 2 1；
索引：0  0  6  7  0  11
算7：11 17 26 13 20 10 5 8 4 2 1；
索引：0  0  6  7  0  0
算8：4 2 1；
索引：0  0  6  7  0  0
算11：17 26 13 20 10 5 8 4 2 1；
索引：0  0  6  7  0  0

Shown below 成功代码

// 1005代码
#include<stdio.h>
#define MAXS 101
int main(){
    
    
	int n,i,j;
	scanf("%d",&n);
	int num,suo[MAXS]={
    
    0};//索引中初始为0 
	for(i=0;i<n;i++){
    
    
		scanf("%d",&num);
		suo[num]=1;//以输入值为下标，标记为1 
	}
	for(i=1;i<100;i++){
    
    
		if(suo[i]){
    
    //如果索引【i】存在 
			for(j=i;j>1;){
    
    
				if(j%2==0){
    
    
					j/=2;
				}
				else{
    
    
					j=(3*j+1)/2;
				}
				if(j<100&&suo[j]){
    
    //索引中有值存在并且j不越界 
					suo[j]=0;
				}
			}
		}
	}
	for(i=1,j=1;i<=100;i++){
    
    //从下标1开始 
		if(suo[i]!=0){
    
    //如果索引【i】不等于0则i为关键字 ，从1开始存 
			suo[j]=i;
			j++;
		}//如果索引【i】==0则i++，j不变等着存下一个 
	}
	suo[j]='\0';
	printf("%d",suo[j-1]);//按要求输出 
	for(j=j-2;j>=1;j--){
    
    
		printf(" %d",suo[j]);
	}
	return 0;
}