Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs,…

Arbitrage Problem Description Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British p…

题意:每两种货币之间都有不同的汇率  如果换回自己最后是赚的 输出Yes 否则是No 因为最多只有三十种货币 所以用Floyd是可行的 与一般的最短路板子不同的地方 汇率是要乘而不是加 如果乘上一个小于1的数就会比之前小 将每种货币看作点 汇率建边 如果这两种货币不能兑换 就设为0 最后与自己判断是否大于1 如果是 则存在套利 如果不是就不存在 代码如下: #include <iostream> #include <cstdio> #include <cstring>…

Arbitrage Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5679    Accepted Submission(s): 2630 Problem Description Arbitrage is the use of discrepancies in currency exchange rates to transform…

Problem Description Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1217 题目大意 在每种钱币间进行各种交换,最后换回自己如果能赚,那么就Yes,否则No 注意应为有负权所以dijsktra在这里行不通了可以用国产的spfa算法,可比bfs. 我的AC代码 #include<iostream>#include<cstdio>#include<cstring>#include<map>#include<queue>us…

正解:hash/二进制分解 解题报告: 传送门! umm首先提取下题意趴QAQ 大概是说给一棵树,每个点有一个权值,要求修改一些点的权值,使得同一个父亲的儿子权值相同,且父亲的权值必须是所有儿子权值之和 首先其实可以想到,只要树上一个点确定了,其实整棵树都确定了,太显然了懒得证QAQ 所以现在其实是只要确定任意一个点就能知道要修改的数量 显然对根进行考虑是最简单的,因为知道根就知道总量,然后就十分显然地出来ans了,甚至答案都能被直接表示出来,设当前节点的权值为a[x],有tot个儿子,那么它的…

hdu1217 Arbitrage Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4123    Accepted Submission(s): 1878 Problem Description Arbitrage is the use of discrepancies in currency exchange rates to tr…

题目传送门 /* 最短路:Floyd模板题 只要把+改为*就ok了,热闹后判断d[i][i]是否大于1 文件输入的ONLINE_JUDGE少写了个_,WA了N遍:) */ #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <string> #include <map> #include <cmath>…

题意  给你n种币种之间的汇率关系  推断是否能形成套汇现象  即某币种多次换为其他币种再换回来结果比原来多 基础的最短路  仅仅是加号换为了乘号 #include<cstdio> #include<cstring> #include<string> #include<map> using namespace std; map<string, int> na; const int N = 31; double d[N], rate[N][N],…