数学uva846Steps

原创于 2015-08-18 15:10:21 发布 · 343 阅读

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本文介绍了一种求解从一点到另一点所需最短步数的算法，步长需为非负整数且每步可增减一或保持不变。输入为起点和终点坐标，输出是最少步数。文章通过实例解释了算法背后的数学原理，并提供了实现该算法的C语言代码。

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: 0 $\le$ x $\le$ y < 2³¹. For each test case, print a line giving the minimum number of steps to get from x to y.

Sample Input

Sample Output

3
3
4

  1 1
  2 1 1
  3 1 1 1
  4 1 2 1
  5 1 2 1 1
  6 1 2 2 1
  7 1 2 2 1 1
  8 1 2 2 2 1
  9 1 2 3 2 1
10 1 2 3 2 1 1
11 1 2 3 2 2 1
12 1 2 3 3 2 1
13 1 2 3 3 2 1 1
14 1 2 3 3 2 2 1
15 1 2 3 3 3 2 1
16 1 2 3 4 3 2 1


规律看了很久才看出来
找出有变化的是5 7 10 13
5 和 10都在2 × 2, 3 × 3之后
观察4 9 16都是2×sqrt（） - 1
56  10 11 12 是偶数
sqrt（5） sqrt（6） = 2    《=2×2 + 2
sqrt（10） 11 13 = 3         《=3×3 +3
都是2 × sqrt（）
789 13 14 15是奇数             
2×sqrt （）+ 1

代码如下

#include <stdio.h>
#include <math.h>
int main(){
	int n, step = 0, cha, s;
	long long x, y;
	scanf("%d", &n);
	while (n--) {
		scanf("%lld%lld", &x, &y);
		cha = y - x;
		if (cha <= 3)
			step = cha;
		else {
			s = sqrt(cha);
			if (s * s == cha)
				step = 2 * s - 1;
			else if (cha - s * s <= s)
				step = 2 * s;
			else
				step = 2 * s + 1;
		}
		printf("%d\n", step);
	}
	return 0;
}