The 3n + 1 problemTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25347Accepted Submission(s): 9431
Problem Description
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 n2. print n3. if n = 1 then STOP4. if n is odd then n <- 3n + 15. else n <- n / 26. GOTO 2Given 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 1It 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).
Sample Input
1 10100 200201 210900 1000
Sample Output
1 10 20100 200 125201 210 89900 1000 174
Source
UVA
逼题 wa六次！服死了，啥时候我才能厉害啊。错误代码：
#include<iostream>using namespace std;int main(){int M,count=1,i,j,k;while(cin>>i>>j){if(i>j)//相当隐蔽的是题目并没有明确i，j的具体大小wa一次，接下来有更相当隐蔽的；swap(i,j);count=1; M=-1;//记得每个测试实例都要对某些变量重新初始化。for(int m=i;m<=j;m++){k=m;//必须用一个k代替循环变量，否则往下执行时被改变。while(k!=1){count++;if(k%2)k=3*k+1;elsek/=2;}if(count>M){M=count;}}cout<<i<<" "<<j<<" "<<M<<endl;//一开始加了个swap函数以为得了，操蛋的是如果交换了//两个变量的值，则对于原值的输出结果是非常可悲的，，解决方法是在未到达swap之前//就把i，j输出！}return 0;}
AC代码：
#include<iostream>using namespace std;int main(){int M,count=1,i,j,k;while(cin>>i>>j){cout<<i<<" "<<j<<" ";if(i>j)swap(i,j);M=-1;for(int m=i;m<=j;m++){k=m;while(k!=1){count++;if(k&1)k=3*k+1;elsek/=2;}if(count>M){M=count;}count=1;}cout<<M<<endl;}return 0;}
懂得倾听别人的忠告。
