371 - Ackermann Functions
Time limit: 3.000 seconds
An Ackermann function has the characteristic that the length of the sequence of numbers generated by the function cannot be computed directly from the input value. One particular integer Ackermann function is the following:
This Ackermann has the characteristic that it eventually converges on 1. A few examples follow in which the starting value is shown in square brackets followed by the sequence of values that are generated, followed by the length of the sequence in curly braces:
[10] 5 16 8 4 2 1 {6}
[13] 40 20 10 5 16 8 4 2 1 {9}
[14] 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 {17}
[19] 58 29 88 44 22 ... 2 1 {20}
[32] 16 8 4 2 1 {5}
[1] 4 2 1 {3}
Input and Output
Your program is to read in a series of pairs of values that represent the first and last numbers in a closed sequence. For each closed sequence pair determine which value generates the longest series of values before it converges to 1. The largest value in the sequence will not be larger than can be accomodated in a 32-bit Pascal LongInt or C long. The last pair of values will be 0, 0. The output from your program should be as follows:
Between L and H, V generates the longest sequence of S values.
Where:
L = the lower boundary value in the sequence
H = the upper boundary value in the sequence
V = the first value that generates the longest sequence, (if two or more values generate the longest sequence then only show the lower value) S = the length of the generated sequence.
In the event that two numbers in the interval should both produce equally long sequences, report the first.
Sample Input
1 20 35 55 0 0
Sample Output
Between 1 and 20, 18 generates the longest sequence of 20 values. Between 35 and 55, 54 generates the longest sequence of 112 values.
同这一题。
完整代码:
/*0.126s*/
#include<bits/stdc++.h>
using namespace std;
int main()
{
///坑1
long long i, j, n, m, count, maxi, max;
while (scanf("%lld%lld", &n, &m), n || m)
{
if (n > m) swap(n, m);///坑2
max = 0;
for (i = n; i <= m; ++i)
{
j = i;
count = 0;
if (j == 1) count = 3;
while (j != 1)
{
if (j & 1) j += (j << 1) + 1;
else j >>= 1;
++count;
}
if (count > max)
{
max = count;
maxi = i;
}
}
printf("Between %lld and %lld, %lld generates the longest sequence of %lld values.\n", n, m, maxi, max);
}
return 0;
}
Ackermann函数解析
本文介绍了一种特定的整数Ackermann函数,该函数生成的序列最终收敛于1,并提供了求解最长序列及其长度的方法。通过示例展示了如何计算从特定起始值到1的序列长度,并给出了一段实现该功能的C++代码。
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