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10025 - The ? 1 ? 2 ? ... ? n = k problem
Time limit: 3.000 seconds
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=966
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
212-3646397
Sample Output
72701
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');elseflag = 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');elseflag = 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;
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