ACM练习3*n+1问题
2014-10-24 16:38
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题目描述
Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example,the following sequence of numbers will be generated for n = 22: 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for every integer n. Still, the conjecture holds for all integers
up to at least 1, 000, 000. For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length
over all numbers between i and j, including both endpoints.
输入
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.输出
For each pair of input integers i and j, output i, j in the same order in which they appeared in the input and then the maximum cycle length for integers between and including i and j. These three numbers should be separated by one space,with all three numbers on one line and with one line of output for each line of input.
样例输入
1 10 100 200 201 210 900 1000
样例输出
1 10 20 100 200 125 201 210 89 900 1000 174
提示
杭电1032这是题目,大概意思是对于一个数,如果这个数的是偶数,就除以2,如果是奇数就乘以3然后加上1,直到最终结果为1为止,这个规律对于任何数都成立,记录从开始到1的步骤,要求输入一个数,i和j,求出这两个数之间步骤最少的一个。没有告诉一共输入多少组。
我应该注意的是:不是把所有的输入都输入完成,然后做个总的输出,而是每个输入对应一组输出。
#include<stdio.h> int main() { int i,j,k,k2,max=0,count=0,tem,j1,j2; while(scanf("%d %d",&i,&j)!=EOF)//这里不能改成while(1),改后会提示超出限制 { max=0; j1=i; j2=j; if(i>j)//这里必须判断大小,不能看例子都是i小于j的所以就自以为是i<j { tem=i; i=j; j=tem; } for(k=i;k<=j;k++) { for(k2=k,count=1;count==0||k2!=1;count++) { if(k2%2==0) { k2=k2/2; } else { k2=k2*3+1; } } if(count>max) { max=count; } } printf("%d %d %d\n",j1,j2,max); } return 0; }
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