poj 2590 Steps

最新推荐文章于 2016-02-01 20:24:47 发布

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最新推荐文章于 2016-02-01 20:24:47 发布

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/22/2258598.html

本文介绍了一个关于在一维直线上从一点到另一点所需的最短步进次数的算法问题。该问题要求每一步的距离只能比前一步多一或少一，并且首尾两步长度必须为1。通过一个C++实现的示例程序，展示了如何计算从x到y的最少步数。

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Steps

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6566		Accepted: 3006

Description

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

Input consists of a line containing n, the number of test cases.

Output

For each test case, a line follows with two integers: 0 <= x <= y < 2^31. 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<iostream>
using namespace std;
int main()
{
long a,b,d;
int n;
int c;
int step;
    cin>>n;
while(n--)
    {
        cin>>a>>b;
        d=b-a;
        c=1;
        step=0;
while(1)
        {
if(d<2*c)
            {
break;
            }
else
            {
                d=d-2*c;
                step+=2;
                c++;
            }
        }
if(d>c)
            step+=2;
else if(d<=0)
            step+=0;
else
            step+=1; 
        cout<<step<<endl;
    }
return 0;
}