原题链接:http://www.dotcpp.com/oj/problem1095.html
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
解题思路
这道题求的是某个范围内的最大循环长度,何为最大循环长度,也就是
- 对于一个数n,是偶数就除以2
- 不是就乘以3再加1
- 再判断n是否为1
- 不为1继续执行第一步
- 循环长度就是这个循环执行了多少次(包括了 n= 1的时候)
所以直接定义函数用来求循环长度,再在i到j的区间判断一下最大循环长度。
在这里我一直有个bug,就是没有比较i与j,导致系统一直判我超时,注意一下就好了
参考代码
#include<iostream>
int caculation(int n);
using namespace std;
int main()
{
int i, j;
int max = 0, t;
while (cin >> i >> j)
{
cout << i << " " << j << " ";
if (i > j)
{
int temp;
temp = i;
i = j;
j = temp;
}
for (int k = i; k != j + 1; ++k)
{
t = caculation(k);
if (t > max)
{
max = t;
}
}
cout << max << endl;
max = 0;
}
return 0;
}
//用来计算循环长度
int caculation(int n)
{
int i = 0;
do
{
if (n % 2)
{
n = n * 3 + 1;
}
else
{
n /= 2;
}
++i;
} while (n != 1);
i = i + 1; //当n为1也算一次
return i;
}