UVA - 10025 The ? 1 ? 2 ? ... ? n = k problem

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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 题意：令S(n)=1 + 2 + 3 + ... + n. 其中一项为x.那么S(n)=1 + 2 + 3 + ...+ x + ... + n. 这样结果可能大于等于 | k | 
+ x 改成- x：S'(n)=1+2+...-x+...+n.
S(n)-S'(n)=2x;且k=S'(n).令y=2x.
利用 S(n)=n(n+1）/2#include<math.h>#include<stdio.h>#define lli long long intint main(){lli n;lli m;lli t;scanf("%lli",&n);while(n --){ scanf("%lli",&m);m = fabs(m);if(m){t = (sqrt(1 + 8*m) - 1) / 2;if(t*t + t != m *2 )t ++ ;while(((t*t+t)/2 - m)%2 == 1)t = t+ 1;printf("%lli\n",t);}elseprintf("3\n");if(n)printf("\n");}return 0;}

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