蓝桥杯 普及题 The 3n + 1 problem

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

一开始找了半天规律……，然后发现数据量很小根本不会超时……，注意如果你们交换了两个数的值的话输出时也要交换，否则答案错误33%，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    int m,n;
   while(~scanf("%d%d",&m,&n)){
        int ans=0;
        int s=m,e=n;
        if(s>e) swap(s,e);
        for(int i=s;i<=e;i++){
            int k=i,cnt=0;
            while(1){
                cnt++;
                if(k%2==0) k/=2;
                else k=k*3+1;
                if(k==1) {cnt++;break;}
            }
            ans=max(ans,cnt);
        }
        printf("%d %d %d\n",m,n,ans);
   }
}