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:
    守望先锋-生涯数据信息抓取的实现
    Norm 数据库操作竟然可以如此简单
    java中关于转义字符的一个bug
    在Java中==的一个坑
    人机ai五子棋 ——五子棋AI算法之Java实现
    MySQL数据库罕见的BUG——Can't get hostname for your address
    [疑难杂症]__关于cmd命令正确而显示不是内部指令的错误(ps:已解决)
    [疑难杂症]__点击win10屏幕最上方的边界会莫名其妙打开Internet Explorer浏览器,不胜其烦(2次ps:已解决!!!).
    [Java初探外篇]__关于正则表达式
    [java初探外篇]__关于StringBuilder类与String类的区别
  • 原文地址:https://www.cnblogs.com/372465774y/p/2777589.html
Copyright © 2011-2022 走看看