2013
12-04

Arbitrage

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.

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.

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

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

Case 1: Yes
Case 2: No

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define re(i,n) for(int i=0;i<n;i++)
#define re1(i,n) for(int i=1;i<=n;i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define inf (1<<29)
const int maxn =33;
int n , m;
double G[maxn][maxn] , diss;
char ch[maxn][maxn] , ch1[maxn] , ch2[maxn];
void floyd() {
re(k,n) re(i,n) re(j,n) if(G[i][j] < G[i][k]*G[k][j])
G[i][j] = G[i][k] * G[k][j];
}
int main() {
int cas = 1;
while(~scanf("%d",&n) && n) {
clr(G,0);
re(i,n) scanf("%s",ch[i]);
scanf("%d",&m);
while(m--) {
int u , v;
scanf("%s%lf%s",ch1,&diss,ch2);
re(i,n) if(strcmp(ch[i],ch1) == 0) { u = i; break; }
re(i,n) if(strcmp(ch[i],ch2) == 0) { v = i; break; }
G[u][v] = diss;
}
floyd();
int flag = 0;
re(i,n) if(G[i][i] > 1) { flag = 1; break; }
if(flag) printf("Case %d: Yes\n",cas++);
else printf("Case %d: No\n",cas++);
}
return 0;
}

1. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n

2. 问题3是不是应该为1/4 .因为截取的三段，无论是否能组成三角形， x， y-x ，1-y,都应大于0，所以 x<y,基础应该是一个大三角形。小三角是大三角的 1/4.