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先不说题，我要吐槽！！！这题太坑了，你给了1-1,000,000的范围，然后你竟然不用任何精妙的算法，直接暴力循环就能AC，然而我还以为要用什么精妙的算法，写了大半天。。。（虽然最后还没写出来，手动黑脸），原因是忘了比较i和j的大小顺序，最后气哄哄的跑去看题解，f**k，竟然真的是水题，我还是太菜狗了。

正题：

没有废话，直接上代码：

//problems://Problems in Computer Science are often classified as belonging //to a certain class of problems (e.g., NP, Unsolvable, Recursive).// In this problem you will be analyzing a property of an algorithm //whose classification is not known for all possible inputs. //Consider the following algorithm: //    1.      input n //    2.      print n //    3.      if n = 1 then STOP //    4.           if n is odd then n <- 3n + 1 //    5.           else n <- n / 2 //    6.      GOTO 2 //Given the input 22, the following sequence of numbers will be printed //22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 ////It is conjectured that the algorithm above will terminate //(when a 1 is printed) for any integral input value. Despite the //simplicity of the algorithm, it is unknown whether this conjecture is true. //It has been verified, however, for all integers n such that 0 < n < 1,000,000 //(and, in fact, for many more numbers than this.) ////Given an input n, it is possible to determine the number of numbers printed //(including the 1). For a given n this is called the cycle-length of n. //In the example above, the cycle length of 22 is 16. //For any two numbers i and j you are to determine the maximum cycle //length over all numbers between i and j. //Input://The input will consist of a series of pairs of integers i and j, //one pair of integers per line. All integers will be less than 1,000,000 and greater than 0. ////You should process all pairs of integers and for each pair determine //the maximum cycle length over all integers between and including i and j. ////You can assume that no opperation overflows a 32-bit integer. //Output://For each pair of input integers i and j you should output i, j, //and the maximum cycle length for integers between and including i and j. //These three numbers should be separated by at least one space with all //three numbers on one line and with one line of output for each line of input. //The integers i and j must appear in the output in the same order in which //they appeared in the input and should be followed by the maximum cycle length (on the same line). #include 
   
    using namespace std;//求n得循环长度 int find(int n){	int cnt=0;	while(n!=1){		cnt++;		if(n%2)	n=3*n+1;		else	n/=2;	}	return cnt+1;}int main(){    int i,j,max;    while(cin>>i>>j){    	//你没有看错，我的那个拙略的算法，就是将i,j都放进一个数组中，然后遍历数组求出最大长度     	int vis[1000001]={0};     	max=0;    	cout<
    <<' '<
     
      <<' ';			//趁i,j位置没变，先偷偷输出他     	//将i,j按大小排列 		if(i>j){			int t=i;			i=j;			j=t;		}		//遍历i到j    		for(int k=i;k<=j;k++){    		if(!vis[k]){				//若k的长度未知     			int cnt=find(k);    			vis[k]=cnt;    			if(k%2 && k*3+1<=j)	vis[k*3+1]=cnt-1;//当k长度确定时，且k为奇数时，3k+1的长度确定     			else if(k*2<=j)	vis[2*k]=cnt+1;		//当k确定时，2k的长度确定 			}			if(vis[k]>max)	max=vis[k];	//求最大长度 		}   		cout<
      
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