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HDU 1032
The 3n + 1 problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23729 Accepted Submission(s): 8853
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 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.
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.
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 10 100 200 201 210 900 1000
Sample Output
1 10 20 100 200 125201 210 89 900 1000 174
Source
UVA
题意:输入两个数。假如是 1 10.。 对1到10之间的所有整数,进行操作。假如取6。。因为6是偶数,所以除以2得到3。因为3是奇数,所以乘三+1得到10。10是偶数,除以2.
得到5,乘三加一。得到16.除以2,得到8,除以2,得到4,除以二,得到2,除以2,得到1.停止操作。一共操作了8次。问 的是给定区间内所有的整数,进行如上操作,得到一为止,求最大的操作次数是多少?
上代码吧。 WA了好多遍,原因是没有考虑到n,m的输出位置固定,而错误的交换了他们的位置。。orz。。
#include <stdio.h> int main() { int n,m,t,max,i,sum,n1,n2; while(scanf("%d%d",&n,&m)!=EOF) { if(n>m) //n,m大小不一定,所以保证让n1为小的。 { n1=m,n2=n; } else n1=n,n2=m; printf("%d %d ",n,m); //输出n,m的位置不能变。 max=0; for(i=n1;i<=n2;i++) //枚举n1到n2的所有整数。 { t=i;sum=1; while(t!=1) //如果不等于一的话,一直循环以下操作。 { if(t%2==1) t=3*t+1; else t=t/2; sum++; //每操作一次,加一个。 } if(sum>max) max=sum; //求出最大操作次数。 } printf("%d\n",max); //输出。 } return 0; }
HDU 1032
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