Hailstone HOTPO
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 426 Accepted Submission(s): 258
Problem Description
The
hailstone sequence is formed in the following way:
(1) If n is even, divide it by 2 to get n'
(2) If n is odd, multiply it by 3 and add 1 to get n'
It is conjectured that for any positive integer number n, the sequence will always end in the repeating cycle: 4, 2, 1, 4, 2, 1,... Suffice to say , when n == 1, we will say the sequence has ended.
Write a program to determine the largest value in the sequence for a given n.
(1) If n is even, divide it by 2 to get n'
(2) If n is odd, multiply it by 3 and add 1 to get n'
It is conjectured that for any positive integer number n, the sequence will always end in the repeating cycle: 4, 2, 1, 4, 2, 1,... Suffice to say , when n == 1, we will say the sequence has ended.
Write a program to determine the largest value in the sequence for a given n.
Input
The first line of input contains a single integer P, (1<= P <= 100000), which is the number of data set s that follow. Each data set should be processed identically and independently.
Each data set consists of a single line of input consisting of two space separated decimal integers. The first integer is the data set number. The second integer is n, (1 <= n <= 100,000), which is the starting value.
Each data set consists of a single line of input consisting of two space separated decimal integers. The first integer is the data set number. The second integer is n, (1 <= n <= 100,000), which is the starting value.
Output
For each data set there is a single line of output consisting of the data set number, a single space, and the largest value in the sequence starting at and including n.
Sample Input
4 1 1 2 3 3 9999 4 100000
Sample Output
1 1 2 16 3 101248 4 100000
Source
Recommend
liuyiding
大水题。模拟过程。
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int cas,tag;
__int64 n,Max;
cin>>cas;
while(cas--)
{
scanf("%d %I64d",&tag,&n);
Max=n;
while(n!=1)
{
if(n%2)
n=3*n+1;
if(n>Max)Max=n;
n/=2;
}
printf("%d %I64d\n",tag,Max);
}
return 0;
}