10025 - The ? 1 ? 2 ? ... ? n = k problem
Time limit: 3.000 seconds
The problem
Given the following formula, one can set operators '+' or '-' instead of each '?', in order to obtain a given k
? 1 ? 2 ? ... ? n = k
For example: to obtain k = 12 , the expression to be used will be:
- 1 + 2 + 3 + 4 + 5 + 6 - 7 = 12
with n = 7
The Input
The first line is the number of test cases, followed by a blank line.
Each test case of the input contains integer k (0<=|k|<=1000000000).
Each test case will be separated by a single line.
The Output
For each test case, your program should print the minimal possible n (1<=n) to obtain k with the above formula.
Print a blank line between the outputs for two consecutive test cases.
Sample Input
2 12 -3646397
Sample Output
7 2701
首先，1~n的和sum一定要>=k。当sum >n时要减掉一个数， 比如在数字a前面加个-号， 相当于sum - 2 * a，也就是说每次减掉只能是偶数，那么就要求sum % 2 == k % 2.(负数同理）
O(√k)代码：
/*0.006s*/
#include<cstdio>
#include<algorithm>
using namespace std;
int main(void)
{
int T, k, sum, i;
bool flag = false;
scanf("%d", &T);
while (T--)
{
if (flag)
putchar('\n');
else
flag = true;
scanf("%d", &k);
k = abs(k), sum = 0;
for (i = 1;; i++)
{
sum += i;
if (sum >= k && ((sum - k) & 1) == 0)
{
printf("%d\n", i);
break;
}
}
}
return 0;
}
O(1)代码：
/*0.006s*/
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int main(void)
{
int T, k, n;
bool flag = false;
scanf("%d", &T);
while (T--)
{
if (flag)
putchar('\n');
else
flag = true;
scanf("%d", &k);
if (k == 0)
puts("3");
else
{
k = abs(k);
n = (int)ceil((sqrt(0.25 + (k << 1)) - 0.5));
if (k & 1)
{
if (n % 4 == 0 || n % 4 == 3)
n = ((n + 1) >> 2 << 2) + 1;
}
else
{
if (n % 4 == 1 || n % 4 == 2)
n = (n >> 2 << 2) + 3;
}
printf("%d\n", n);
}
}
return 0;
}
