拓展欧几里得: 当 gcd ( a , b )= d 时,求绝对值和最小的 x , y 使得 x * a + y * b = d : d = gcd ( a , b ) = gcd ( b , a mod b ): 设: x1 * a + y1 * b = d : ① x2 * b + y2 * ( a mod b ) = d : ② 因为 a mod b = a - ( a / b )* b: ③(除法为整除) 将③代入①整理得: y2 * a + ( x2 - ( a / b ) * y2…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1576 A/B Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4020 Accepted Submission(s): 3091 Problem Description 要求(A/B)%9973,但由于A很大,我们只给出n(n=A%99…
Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; variable += C) statement; I.e., a loop which starts by setting variable to value A and <= x < 2k) modulo 2k. Input The input consists…