id=1465">http://poj.org/problem?id=1465 Multiple Time Limit: 1000MS Memory Limit: 32768K Total Submissions: 6164 Accepted: 1339 Description a program that, given a natural number N between 0 and 4999 (inclusively), and M distinct decimal digit…
题意:给出m个数字,要求组合成能够被n整除的最小十进制数. 分析:用到了余数判重,在这里我详细的解释了.其它就没有什么了. #include<cstdio> #include<cmath> #include<cstring> #include<queue> #include<algorithm> using namespace std; ; ; struct Node{ int pre; int r; int d; }; int vis[MAXN…
Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no mo…
The Embarrassed Cryptographer Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 11978 Accepted: 3194 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of…
The Embarrassed Cryptographer Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 15767 Accepted: 4337 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of…
题意:给一些n,求出最小的只包含0,1的n的倍数 设两数a, b满足: a < b 并且a % n = b % n. 如果 ( a * 10^x + c ) % n = z , 根据同余定理,( b * 10^x + c ) % n 也等于 z. b的情况其实与a相同,如果a不符合条件,那么b一定不符合条件. 因此我们在搜索时,从1开始,每次往后添加0或1,如果得到的数与之前得到的某数同余,则扔掉,否则放入队列继续搜索. #include <cstdio> #include <cs…
Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5930 Accepted Submission(s): 4146 Problem Description As we know, Big Number is always troublesome. But it's really important in our…
Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4594 Accepted Submission(s): 3175 Problem Description As we know, Big Number is always troublesome. But it's really important in ou…
D. Notepad time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other numb…
Description Edward 得到了一个长度为 N 的整数序列,他想找出这里面有多少个“幸运的”连续子序列.一个连续子序列被称为“幸运的”,当且仅当该子序列内的整数之和恰好是 K 的整数倍数.他请求你写一个程序来计算他喜欢的连续子序列个数. Input 输入第一行是一个整数 T,表示有 T 组数据.每组数据第一行是两个整数 N (1 <= N <= 10^6), K (1 <= K <= 10^9).接下来的一行包含 N 个整数 Ai (|Ai| &…
Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c. Input Input starts with an integer T (≤ 525), denot…
同余问题 基本定理: 若a,b,c,d是整数,m是正整数, a = b(mod m), c = d(mod m) a+c = b+c(mod m) ac = bc(mod m) ax+cy = bx+dy(mod m) -同余式可以相加 ac = bd(mod m) -同余式可以相乘 a^n = b^n(mod m) f(a) = f(b)(mod m) if a = b(mod m) and d|m then a = b(mod d) eg: 320 = 20(mod 100) and d =…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1212 Problem Description As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.To make the problem easier, I…
题目地址:http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1803 Knowledge Point: 同余定理:两个整数a.b,若它们除以整数m所得的余数相等,则称a与b对模m同余或a同余于b模m.记作 a≡b(mod m): 加法运用: (a + b) % m = (a % m + b % m) % m 乘法运用: (a * b) % m = ((a % m) * (b % m)) % m 高精度取模: 一个高精度数对一个数取余,可以把高精…
2018 China Collegiate Programming Contest Final (CCPC-Final 2018)-K - Mr. Panda and Kakin-中国剩余定理+同余定理 [Problem Description] \[ 求解x^{2^{30}+3}=c\pmod n \] 其中\(n=p\cdot q\),\(p\)为小于\(x\)的最大素数,\(q\)为大于\(x\)的最小素数,\(x\)为\([10^5,10^9]\)内随机选择的数.\(0< c<n\).…
