milo
BAN USER#include <iostream>
using namespace std;
void reverse_by_word(char *str) {
if (*str == '\0') {
return;
}
char p[100];
int index = 0;
while (*str != ' ' && *str != '\0') {
p[index++] = *str;
++str;
}
p[index] = '\0';
reverse_by_word(++str);
cout << p << " ";
}
int main() {
char *str = "hello world";
reverse_by_word(str);
cout << endl;
}

milo
May 07, 2012 It' my solution:
#include <iostream>
using namespace std;
int main() {
int final_sum;
cin >> final_sum;
int array[10] = {11, 1, 3, 9, 10, 2, 5, 7, 4, 8};
int i = 0, j = 0, sum = array[0];
while (j < 10) {
while (sum < final_sum) {
++j;
if (j >= 10) break;
sum += array[j];
}
if (j >= 10) break;
if (sum == final_sum) {
for (int k = i; k <= j; ++k)
cout << array[k] << " ";
cout << endl;
sum = array[i];
++i;
} else {
sum = array[i];
++i;
sum = array[j];
j;
}
}
return 0;
}

milo
May 03, 2012 It' my solution:
#include <iostream>
using namespace std;
int main() {
int final_sum;
cin >> final_sum;
int array[10] = {11, 1, 3, 9, 10, 2, 5, 7, 4, 8};
int i = 0, j = 0, sum = array[0];
while (j < 10) {
while (sum < final_sum) {
++j;
if (j >= 10) break;
sum += array[j];
}
if (j >= 10) break;
if (sum == final_sum) {
for (int k = i; k <= j; ++k)
cout << array[k] << " ";
cout << endl;
sum = array[i];
++i;
} else {
sum = array[i];
++i;
sum = array[j];
j;
}
}
return 0;
}

milo
May 03, 2012 #include <iostream>
using namespace std;
void reverse_array(int array[], int index, int len) {
if (index >= len  1) return;
array[index] = array[index]^array[len  1];
array[len  1] = array[index]^array[len  1];
array[index] = array[index]^array[len  1];
reverse_array(array, index + 1, len  1);
}
int main() {
int array[10];
for (int i = 0; i < 10; ++i)
array[i] = i;
reverse_array(array, 0, 10);
for (int i = 0; i < 10; ++i)
cout << array[i] << endl;
}

milo
May 02, 2012 In the book 《programming pearls》 second edition, the 12th capter and the 10th question in the exercise. At last, the probabilty to choose any line is 1/n.
when there are n lines. the probabilty to choose n  1line is 1 / (n  1) * (n  1) / n = 1/ n
the probabilty to choose n  1 * the probability not choose n.
for a sorted array, 3 sum problem can be done in O(n):
1 # !/usr/bin/python
2 # * encoding: utf8 *
3
4 array = [1, 2, 3, 4, 5, 6, 7, 8, 9]
5
6 three_sum = 15
7
8 start = 0
9 end = 8
10 while start <= end:
11 if array[start] + array[end] == three_sum:
12 print '%d\t%d\t=\t%d' % (array[start], array[end], three_sum)
13 break
14 elif array[start] + array[end] > three_sum:
15 end = 1
16 elif array[start] + array[end] < three_sum:
17 start += 1

milo
April 16, 2012 1 # !/usr/bin/python
2 # * encoding: utf8 *
3
4 class DoubleStackQueue:
5
6 def __init__(self):
7 self.first_stack = []
8 self.second_stack = []
9
10 def en_queue(self, el = None):
11 self.first_stack.append(el)
12
13 def de_queue(self):
14 if len(self.second_stack) == 0:
15 while len(self.first_stack) != 0:
16 self.second_stack.append(self.first_stack.pop())
17 return self.second_stack.pop()
18
19 if __name__ == '__main__':
20 queue = DoubleStackQueue()
21 queue.en_queue(1)
22 queue.en_queue(2)
23
24 print queue.de_queue()
25 print queue.de_queue()
26
27 queue.en_queue(3)
28 print queue.de_queue()

milo
April 12, 2012 how about this method?
1 #!/usr/bin/python
2 #* encoding: utf8 *
3
4 a = [4, 7, 5, 1, 3, 8, 9, 6, 2]
5
6 def find_max_until(a, max_value, max_index):
7 tmp_max = 0
8 tmp_index = 0
9 for index in range(max_index):
10 if tmp_max < a[index] and a[index] < max_value:
11 tmp_max = a[index]
12 tmp_index = index
13 return tmp_max, tmp_index
14 max_value = 100
15 max_index = len(a)
16 for i in range(3):
17 max_value, max_index = find_max_until(a, max_value, max_index)
18 print max_value, max_index

milo
April 12, 2012
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 milo May 15, 2012