UVA - 846-Steps

题目描述：
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 and Output
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

Sample Output

3
3
4

类型：数学题？。。考察思路的
思路： 因为步距由1到某值然后又降到1，要求最小步数，所以步距序列一定是由1升到n再降到1，中间插上平等步距。所以我的思路是求得一个数列：123…n…321前n项和刚好大于要求距离，然后降低n到n-1分析仍差距离（存在一个域）补上步数即可

#include<stdio.h>#include<stdlib.h>int main(){    int Case;    scanf("%d", &Case);    while(Case--)    {        int head, tail;        scanf("%d %d", &head, &tail);        int num = tail - head;        int n = 1;        while(((1+n)*n - n) < num)            n++;        n--;        if((num -((1+n)*n - n))>n)            n= 2*n - 1 + 2;        else            n= 2*n - 1 + 1;        printf("%d\n", n);    }}