zoukankan      html  css  js  c++  java
  • zoj 1092 Arbitrage (poj 2240)

    Arbitrage

    Time Limit: 2 Seconds      Memory Limit: 65536 KB

    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, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.

    Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

    Input Specification

    The input file will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
    Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.

    Output Specification

    
    For each test case, print one line telling whether arbitrage is possible or not in the
    format "Case case: Yes" respectively "Case case: No".

    Sample Input

    3
    USDollar
    BritishPound
    FrenchFranc
    3
    USDollar 0.5 BritishPound
    BritishPound 10.0 FrenchFranc
    FrenchFranc 0.21 USDollar
    
    3
    USDollar
    BritishPound
    FrenchFranc
    6
    USDollar 0.5 BritishPound
    USDollar 4.9 FrenchFranc
    BritishPound 10.0 FrenchFranc
    BritishPound 1.99 USDollar
    FrenchFranc 0.09 BritishPound
    FrenchFranc 0.19 USDollar
    
    0
    

    Sample Output

    Case 1: Yes
    Case 2: No
    
    //找是否存在大于1的回路
    //恶心点在于居然有钱汇同样钱汇率大于1、、不合常理呀
    #include <iostream> #include <stdio.h> #include <queue> #include <stack> #include <set> #include <vector> #include <math.h> #include <string.h> #include <algorithm> using namespace std; char rc[33][22]; double dis[33][33]; char a[22],b[22]; int n,m; int GetId(char s[]) { for(int i=1;i<=n;i++) if(strcmp(rc[i],s)==0) return i; } int main() { int i,j,k; int t=1; double d; while(scanf("%d",&n)==1&&n) { for(i=1;i<=n;dis[i][i]=1,i++) for(j=1;j<=n;j++) dis[i][j]=0; for(i=1;i<=n;i++) scanf("%s",rc[i]); scanf("%d",&m); while(m--) { scanf("%s %lf %s",a,&d,b); i=GetId(a); j=GetId(b); dis[i][j]=d; } for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(dis[i][j]<dis[i][k]*dis[k][j]) dis[i][j]=dis[i][k]*dis[k][j]; bool f=0; for(i=1;i<=n;i++) if(dis[i][i]>1) { f=1;break; } printf("Case %d: ",t++); if(f) printf("Yes\n");else printf("No\n"); } return 0; }
  • 相关阅读:
    Max关闭WPF
    InstallShield安装过程介绍
    InstallShield相关资料整理
    .net reflection的一点研究
    (转)VMware增加磁盘容量方法
    领域驱动设计《读书笔记》
    《领域驱动设计C#2008实现》读书笔记
    深入研究c++对象模型
    <转载>com之包容聚合
    基于插件开发的架构研究
  • 原文地址:https://www.cnblogs.com/372465774y/p/2777589.html
Copyright © 2011-2022 走看看