A mountain consists of one or more hills, each of which consists of upwards, which we denote with `/‘, and downwards, which we denote with ‘\‘. We call upwards and downwards together as wards. Let /n be an upward of length n, and in the same way, let \n be a downward of length n. For example, the following upward is /2:
/ /
and the following downward is \3:
\
The set of all hills is given in the following:
Hills =def { /n\n | n is a natural number }
Now it comes to the definition of a mountain(in BNF):
Mountain ::= Hill | Mountain Mountain
That‘s to say, a hill is a mountain, and if M and N are mountains, then the concatenation of M and N is also a mountain.
For example,
//\/\/ \
is a mountain with 3 hills.
We define the width of a mountain as the number of upwards and downwards. We define the height of a mountain as the maximum length of upwards(or equivalently downwards).
We can give each mountain a unique number, called index, such that the following conditions are met:
For example, portion of the mountains and their indices are:
/\ /\ /\ //\ /\/\ / \ /\/\/\ /\/ \ ... /\/\/ \ /\/ \/\ ... 1 2 3 4 5 9 10
In this problem, you are print the mountain from the given index.
The input contains several lines, each of which contains an integer N<2^32. The last line contains 0, which you should not proceed.
For each N, your program should print the mountain, and you need make the mountain climbing.
The output of two consecutive instances should be separated by a blank line.
6 9 0
// \/
//\/\/
Author: fgjlwj
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
int num[50];
long long q_pow( long long a, long long b )
{
long long ans = 1;
while(b)
{
if(b & 1)
{
ans *= a;
}
b /= 2;
a *= a;
}
return ans;
}
int main()
{
long long n;
while(cin >> n&&n)
{
int w = log2( n )+1;
long long loc =n - (q_pow( 2, w - 1 ) - 1);
int h = 1;
int cnt = 1;
int digit = 1;
long long l = 1, r = q_pow( 2, w - 1 );
while(l < r)
{
long long mid = (l + r) / 2;
if(loc <= mid)
{
num[cnt++] = digit;
h = max( h, digit );
digit = 1;
r = mid ;
}
else
{
digit++;
l = mid + 1;
}
}
num[cnt] = digit;
h = max( h, digit );
for(int i = h; i >=1; i--)
{
for(int j = 1; j <= cnt; j++)
{
if(num[j] >= i)
{
int width = (num[j] - i+1) * 2;
cout << string( num[j] - width / 2, ' ' );
cout << "/" << string( width - 2 , ' ' ) << "\\";
cout << string( num[j] - width / 2, ' ' );
}
else
cout << string( 2 * num[j], ' ' );
}
cout << endl;
}
cout << endl;
}
return 0;
}原文地址：http://blog.csdn.net/maxichu/article/details/45593295
