ACM 846 Steps

可以先試著列出1步的可能情況、2步的可能情況 ......，看看是否有規律

#include <stdio.h>

int main(void)
{
    int i, x, y, n, steps, r, add, d;
   
    scanf("%d", &n);
    while (n--)
    {
        scanf("%d%d", &x, &y);
        d = y-x;
        if (d)
        {
            steps = 1;
            r = 0;
            add = 0;      
            do {
                if (steps%2)
                    add++;
                r = r + add;
                steps++;
            } while (r < d);
            printf("%d\n", steps-1);
        }
        else
            printf("0\n");
    }       
   
    return 0;
}