uva 648 steps

本文探讨了一种通过等差数列解决从一个整数到另一个整数的最小步骤问题的方法,详细解释了算法逻辑并提供了C++代码实现。

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Steps
PC/UVa IDs: 110608/846, Popularity: A, Success rate: high Level: 2
Consider the process of stepping from integer x to integer y along integer points of
the straight line. The length of each step must be non-negative and can be one bigger
than, equal to, or 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
both the first and the last step must be 1.
Input
The input begins with a line containing n, the number of test cases. Each test case that
follows consists of a line with two integers: 0 ≤ x ≤ y < 231 .
Output
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



等差数列!!!


#include <cstdio>
#include <cmath>

int main () {
    int cs;
    scanf ("%d", &cs);
    while (cs--) {
        long long a, b;
        scanf ("%lld%lld\n", &a, &b);
        long long s = b - a;
        long long n = llrint (floor (sqrt (0.25 + s) - 0.5));
        if (n * (n + 1) == s) {
            printf ("%lld\n", n + n);
        } else if (n * (n + 1) + n + 1 >= s) {
            printf ("%lld\n", n + n + 1);
        } else {
            printf ("%lld\n", n + n + 2);
        }
    }
    return 0;
}




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