A New Change Problem Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 533    Accepted Submission(s): 265 Problem Description Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Her…

A New Change Problem Problem Description Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Here you can assume that there are enough coins for both kinds.Please calculate the maximal value that you cannot pay and the total number tha…

/** 题目:Joseph's Problem 链接:https://vjudge.net/problem/UVA-1363 题意:给定n,k;求k%[1,n]的和. 思路: 没想出来,看了lrj的想法才明白. 我一开始往素数筛那种类似做法想. 想k%[1,n]的结果会有很多重复的,来想办法优化. 但没走通. 果然要往深处想. 通过观察数据发现有等差数列.直接观察很难确定具体规律:此处应该想到用式子往这个方向推导试一试. lrj想法: 设:p = k/i; 则:k%i = k-i*p; 容易想到…

Equations 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2086 ——每天在线,欢迎留言谈论. 题目大意: 有如下方程:Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, .... n). 若给出A0, An+1, 和 C1, C2, .....Cn. 求 A1 . 思路: 多写几个例子,找规律推导(抄的). 感想: 老啦,老啦,不行了. Java AC代码: import java.util.Scanner;…

https://leetcode.com/problems/water-and-jug-problem/description/ -- 365 There are two methods to solve this problem : GCD(+ elementary number theory) --> how to get GCF, HCD,  BFS Currently, I sove this by first method 1. how to compute GCD recursive…

Problem Description Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Here you can assume that there are enough coins for both kinds.Please calculate the maximal value that you cannot pay and the total number that you cannot pay. Inp…

Another Rock-Paper-Scissors Problem 题目连接: http://codeforces.com/gym/100015/attachments Description Sonny uses a very peculiar pattern when it comes to playing rock-paper-scissors. He likes to vary his moves so that his opponent can't beat him with hi…

题目链接:http://acm.fzu.edu.cn/problem.php?pid=2224 同hdu5869 //#pragma comment(linker, "/STACK:102400000, 102400000") #include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #inc…

题意:求互质的m和n的最大不能组合数和不能组合数的个数 思路:m和n的最大不能组合数为m*n-m-n,不能组合数的个数为(m-1)*(n-1)/2 推导: 先讨论最大不能组合数 因为gcd(m,n)=1,所以 0,n,2*n,3*n,...(m-1)*n(共m个数字)分别除以m,余数肯定不同,且为{0,1,2,3...m-1}中的某数 若存在非负数p,q使得pm+qn=x,x为可组合值,两边对m取余,则(q*n)%m==x%m,p*m>=0,所以只要x>q*n,x都能被组合出来.当q<m…

Hard problem 题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5858 Description cjj is fun with math problem. One day he found a Olympic Mathematics problem for primary school students. It is too difficult for cjj. Can you solve it? Give you the…