TOJ-1339 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 ≤ x ≤ y < 2³¹. For each test case, print a line giving the minimum number of steps to get from x to y.

Sample Input

3
45 48
45 49
45 50

Output for Sample Input

3
3
4

Source: Waterloo Local Contest Jan. 29, 2000

在前一个数的基础上依次加若干个数，求最少需要加几个数可得到后一个数。要求加的数的序列中，起始和末尾的数为1，其他的数

比前一个大1，小1，或相等。

规律总结：

　　对所给的两个数的差x，若有自然数i，i(i+1)<x<=(i+1)^2，需乘2i+1个数；若有i^2<x<=i(i+1)，需乘2i个数。

#include <iostream>  
#include <cmath>  
using namespace std;  
  
int main()  
{  
    int x,y,n;  
    cin >> n;  
    while(n--)  
    {  
        cin >> x >> y; 
        int n = y-x;
        int q = (int)(sqrt(y-x)-1e-9);
        if(n<=(q+1)*q) cout<<q*2<<endl;
        else cout<<q*2+1<<endl;
    }  
  
    return 0;  
}