uva 846 Steps（数学规律）

 Steps 

One steps through integer points of the straight line. The length of a step must be nonnegative and can be by one bigger than, equal to, or by one smaller than the length of the previous step.

What is the minimum number of steps in order to get from x to y? The length of the first and the last step must be 1.

Input and Output 

Input consists of a line containing n, the number of test cases. For each test case, a line follows with two integers: 0xy < 2³¹. For each test case, print a line giving the minimum number of steps to get from x to y.

Sample Input 

345 4845 4945 50

Sample Output 

334

题目大意：给出两个数字， 要求从a走到b, 第一步和最后一步只能走1， 每步的值可以是前面一步值-1， 不变和+1.求a到b最少走几步。

解题思路：1...k k ...1所需步为2 * k,

<1>2 * sum(1~k)  ->2 * k

<2>2 * sum(1~k)  ~2 *  sum(1~k) + 1 + k  -> 2 *k + 1

<3>2 * sum(1~k) + k + 1 ~2 * sum(1~k+1)  -> 2 *(k + 1)

#include<stdio.h>int main(){long long a, b, t, n, sum, k;scanf("%lld", &t);while (t--){scanf("%lld%lld", &a, &b);n = b - a;sum =  0;if (n == 0){printf("%lld\n", n);continue;}for (long long i = 0; ; i++){sum += 2 * i;k = sum - n;if (k >= 0 && k < i){printf("%lld\n", 2 * i);break;}else if(k >= i){printf("%lld\n", 2 * i - 1);break;}}}return 0;}