UVa 846 - Steps

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本文介绍了一种计算从坐标x到y所需的最短步数的算法。每一步必须是非负整数，且长度只能比前一步长或短1单位或相等。输入包括测试案例数量及起点和终点坐标，输出为达到目标所需的最少步数。

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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 xto y.

Sample Input

Sample Output

3
3
4

Miguel Revilla 2002-06-15

#include <stdio.h>
#include <string.h>
int main()
{
    int i,j,n,m,s,t,k,step,k1;
    int x,y,d;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&x,&y);
        d=y-x;
        s=0;
        k=-1;
        step=1;
        while(d>0)
        {
            d-=step;
            s++;
            if(k>0)
            {
                step++;
            }
            k=-1*k;
        }
        printf("%d\n",s);
    }
    return 0;
}