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
实现代码：
#include <iostream>
using namespace std;
int main() {
int i, j,add1;
while (cin >> i >> j) {
cout << i << ' ' << j << ' ';
int add = 0;
if (j < i) {
int a;
a = j;j = i;i = a;
}//如果j小，调换位置。
for (int x = i; x <= j; x++) {
add1 = 1;//将每个数字的猜想初值设为1
int y = x;
//cout << y<<j;
while (y != 1) {
if (y % 2 == 1) y = 3 * y + 1;
else y = y / 2;
add1++;
}//进行3n+1猜想
if (add1 > add) add = add1;
}//查找最大数字
cout <<add<<endl;
}
return 0;
}