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Modulo Ruins the Legend题解 同余 - CSDN博客 Since we want to ruin the legend, please tell her the minimum sum of elements modulo m after the operation Note that you should minimize the sum after taking the modulo
2022ICPC杭州站A题解 - 洛谷专栏 - Luogu A Modulo Ruins the Legend 题目大意 : 给定一个长度为 n 的整数序列 a1 、 a2 、……、 an ，可以选择俩个整数个 s , d 将 ai 均加上 s+ id 并需要对操作后的新序列和对 m 取模后最小。 考察知识点 ： 扩展欧几里得算法 (exgcd) 解题过程 : ans mod m = ans−k1 ∗m = ∑i=0n ai
2022 杭州 A - ycllz - 博客园 A Modulo Ruins the Legend 因为是%m的意义下 我们尽量一直保持在%意义下 不然会很难注意一些不合法情况 原式可变形为（n*（n+1） 2 d + ns + sum ）%m 最小 我们设a=n*（n+1） 2 b=n （ad+bs+sum）%m 设g1=gcd (a,b) 里
2022 Hangzhou A. Modulo Ruins the Legend | GitSteve1025 A Modulo Ruins the Legend 求 求 (∑ i = 1 n a i + n s + n (n + 1) 2 d) 由 裴 蜀 定 理 知 由 裴 蜀 定 理 知 a x + b y = k × g c d (a, b) 令 原 式 令 s u m = ∑ i = 1 n a i ⇒ 原 式 = s u m + k × g c d (a, b) (s u m + k × g c d (a, b))
Problem - A - Codeforces Since we want to ruin the legend, please tell her the minimum sum of elements modulo m m after the operation Note that you should minimize the sum after taking the modulo
GYM104090A Modulo Ruins the Legend - exgcd - 博客园 GYM104090A Modulo Ruins the Legend - exgcd - 题目链接： https: codeforces com gym 104090 problem A 题解： 转化一下发现只需要求满足下式的解： ns+ n×(n+1) 2 d≡C(mod m) n s + n × (n + 1) 2 d ≡ C (mod m) 设 a=n,b= n(n+1) 2,p = gcd(a,b) a = n, b = n (n + 1) 2, p = g c d (a, b)