第四次集训—— Steps

最新推荐文章于 2022-07-26 22:43:16 发布

CODE_BALL

最新推荐文章于 2022-07-26 22:43:16 发布

阅读量647

点赞数

CC 4.0 BY-SA版权

分类专栏： VJ

本文链接：https://blog.youkuaiyun.com/codeball/article/details/17592347

VJ 专栏收录该内容

28 篇文章

订阅专栏

本文介绍了一种求解从坐标x到坐标y所需的最短步数算法，每一步长度必须是非负数，并且可以比前一步长一或短一。输入包含多个测试案例，输出每个案例从起点到终点所需的最少步数。

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 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

#include <stdio.h>
#include <math.h>

int main()
{
  int t,str,end,cha,zb,sh,ci;
  scanf("%d",&t);
  while(t--)
  {
      scanf("%d%d",&str,&end);
      if(str == end)
        printf("0\n");
      else
      {
          cha = end - str;
          zb = (int)sqrt(cha);
          sh = cha - zb * zb;
          ci = sh / zb + zb * 2 - 1;
          if(sh % zb !=0)
            ci++;
          printf("%d\n",ci);
      }
  }
    return 0;
}