刚刚交了一道题目，总结一下

# include 

long int recaman[500001];
int used[5000000] = {0};
	// if an integer is not used, the mark is 0 

int main(void)
{
	long int i, k;
	long int m;

	for (i = 0; i < 500001; i++)
	{
		if (i - 1 < 0)
		{
			recaman[i] = 0;
		}
		else
		{
			m = recaman[i-1] - i;
			if (m > 0 && used[m] == 0)
				recaman[i] = m;
			else
				recaman[i] = m + i + i;
		}
		used[recaman[i]] = 1;
	}
	while(scanf("%ld", &k) != EOF && k >= 0)
	{
			printf("%ld\n", recaman[k]);
	}

	return 0;
}

Recaman's SequenceTime Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB Total submit users: 1183, Accepted users: 1023 Problem 10010 : No special judgement

Problem description

The Recaman's sequence is defined by a0 = 0 ; for m > 0, a_m = a_m−1 − m if the rsulting a_m is positive and not already in the sequence, otherwise a_m = a_m−1 + m.
The first few numbers in the Recaman's Sequence is 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9 ...
Given k, your task is to calculate a_k.

The first few numbers in the Recaman's Sequence is 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9 ...
Given k, your task is to calculate a _k .

Input

The input consists of several test cases. Each line of the input contains an integer k where 0 <= k <= 500000. The last line contains an integer −1, which should not be processed.

Output

For each k given in the input, print one line containing a_k to the output.

Sample Input

7
10000
-1

Sample Output

20
18658

if (m > 0 && used[m] == 0)

注意一点，C语言中有一个特点，就是逻辑与运算，千万不要把上面这条语句&&两边的调换过来，否则数组访问会出错！！

while(scanf("%ld", &k) != EOF && k >= 0)

另外还有一点，写代码时候千万要认真，刚才就是在上面这个语句后面加了分号，，，，，我结果后来才发现。