poj 2240
2015-08-20 19:41
288 查看
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
The input 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
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
Sample Output
Case 1: YesCase 2: No
弗洛伊德的变式题,还有一个问题,如何把字符数组变成节点
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
The input 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
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: YesCase 2: No
弗洛伊德的变式题,还有一个问题,如何把字符数组变成节点
#include <iostream> #include <string.h> #include <stdio.h> using namespace std; double e[50][50]; char a[50][50]; int n,m; int zhuanhuan(char s[]) { for(int i=1;i<=n;i++) { if(strcmp(a[i],s)==0) return i; } } int floyd() { for(int k=1;k<=n;k++) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(e[i][j]<e[i][k]*e[k][j]) e[i][j]=e[i][k]*e[k][j]; } } } for(int i=1;i<=n;i++) { if(e[i][i]>1) return 1; } return 0; } int main() { char start[50],end[50]; int tmp=1,x,y; double rate; while(cin>>n&&n!=0) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { e[i][j]=0; } } for(int i=1;i<=n;i++) { cin>>a[i]; } cin>>m; for(int i=1;i<=m;i++) { cin>>start; cin>>rate; cin>>end; x=zhuanhuan(start); y=zhuanhuan(end); e[x][y]=rate; } if(floyd()) printf("Case %d: Yes\n",tmp); else printf("Case %d: No\n",tmp); tmp++; } return 0; }
相关文章推荐
- 9个Java初始化和回收的面试题
- TCP 的那些事儿-1
- POJ-1442-Black Box-优先队列
- cocos2dx tolua传递参数分析
- 百度推广账户操作实践技巧
- poj 2602 大数相加(字符串输出)
- poj 3268
- linux中的块缓冲
- MVVM
- ural 1210. Kind Spirits
- JSP——>JSTL入门专用
- Hadoop它——跑start-all.sh时间namenode不启动
- 微分不等式
- Android性能优化典例(一)
- HDU 5416 CRB and Tree dfs
- Android性能优化典例(一)
- OGRE 简明扼要的分析(自己记录看看)
- 关于xutils中的BitmapUtil实现简单的缓存和下载
- 进程的创建以及进程间的通信
- Linux里UDP协议 广播和接收测试 C语言