Uva 846 Steps-优快云博客

本文介绍了一种求解从起点到终点的最短步长路径算法，要求步长递增、不变或递减，且首尾步长必须为1。通过计算两点间距离并逐步逼近来确定最少步数。

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. Foreach 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 toget from x to y.

Sample Input

Sample Output

3
3
4

题意：有t组测试数据。要求为最少步得到给出的整数n。步长要求为：起步和停止的步长为一，每次只能在上次的基础上增加一，减少一或者不变。

先计算出两点间的长度d

因为是对称的从两边开始，依次往中间逼近 step++，用距离 d -= step 直到 d < 0

#include<stdio.h>
#include<string.h>

int main() {
	int t;
	while(scanf("%d",&t) != EOF) {
		int x,y;
		int d;
		while(t--) {
			scanf("%d%d",&x,&y);
			int step=1,count=0;
			d=y-x;
			int ans=0;
			while(d>0) {
				d -= step;
				ans++;
				if(count)
					step++;
				count = !count;
			}
			printf("%d\n",ans);
		}	
	}
	return 0;
}