题目原型

题

问题 1095: The 3n + 1 problem时间限制: 1Sec 内存限制: 64MB 提交: 6413 解决: 1766
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

答案错误33% 是的他没说a 与 b的大小关系，但是他说了，output i, j in the same order in which they appeared in the input
这不就是不能改变顺序嘛

代码

#include<iostream>
using namespace std;
int main()
{long long int a,b,c;while(cin>>a>>b){cout<<a<<' '<<b<<' ';//!!!!!!if(a>b){int aa=a;a=b;b=aa;}int maxx=0;for(int i=a;i<=b;i++){int ii=i;int ji=0;while(ii>1){if(ii%2==0){ii/=2;ji++;}else{ii=ii*3+1;ji++;}}if(maxx<ji)maxx=ji+1;}cout<<maxx<<endl;}return 0;
}

总结

1，比较a与b的大小因为题目中没有说明a一定小于b，所以必须比较，不然for循环没法写啊
2，比较a与b的大小但是你最后输出的时候不能是交换后的啊，所以提前输出！！！！！