uva 104 Arbitrage

Description

Download as PDF

Background

The use of computers in the finance industry has been marked with controversy lately as programmed trading – designed to take advantage of extremely small fluctuations in prices – has been outlawed at many Wall Street firms. The ethics of computer programming is a fledgling field with many thorny issues.

The Problem

Arbitrage is the trading of one currency for another with the hopes of taking advantage of small differences in conversion rates among several currencies in order to achieve a profit. For example, if 1.00inU.S.currencybuys0.7Britishpoundscurrency,£1inBritishcurrencybuys9.5Frenchfrancs,and1Frenchfrancbuys0.16inU.S.dollars,thenanarbitragetradercanstartwith1.00 and earn tex2html_wrap_inline29 dollars thus earning a profit of 6.4 percent.

You will write a program that determines whether a sequence of currency exchanges can yield a profit as described above.

To result in successful arbitrage, a sequence of exchanges must begin and end with the same currency, but any starting currency may be considered.

The Input

The input file consists of one or more conversion tables. You must solve the arbitrage problem for each of the tables in the input file.

Each table is preceded by an integer n on a line by itself giving the dimensions of the table. The maximum dimension is 20; the minimum dimension is 2.

The table then follows in row major order but with the diagonal elements of the table missing (these are assumed to have value 1.0). Thus the first row of the table represents the conversion rates between country 1 and n-1 other countries, i.e., the amount of currency of country i ( tex2html_wrap_inline37 ) that can be purchased with one unit of the currency of country 1.

Thus each table consists of n+1 lines in the input file: 1 line containing n and n lines representing the conversion table.

The Output

For each table in the input file you must determine whether a sequence of exchanges exists that results in a profit of more than 1 percent (0.01). If a sequence exists you must print the sequence of exchanges that results in a profit. If there is more than one sequence that results in a profit of more than 1 percent you must print a sequence of minimal length, i.e., one of the sequences that uses the fewest exchanges of currencies to yield a profit.

Because the IRS (United States Internal Revenue Service) notices lengthy transaction sequences, all profiting sequences must consist of n or fewer transactions where n is the dimension of the table giving conversion rates. The sequence 1 2 1 represents two conversions.

If a profiting sequence exists you must print the sequence of exchanges that results in a profit. The sequence is printed as a sequence of integers with the integer i representing the tex2html_wrap_inline51 line of the conversion table (country i). The first integer in the sequence is the country from which the profiting sequence starts. This integer also ends the sequence.

If no profiting sequence of n or fewer transactions exists, then the line

no arbitrage sequence exists

should be printed.

Sample Input

3

1.2 .89

.88 5.1

1.1 0.15

4

3.1 0.0023 0.35

0.21 0.00353 8.13

200 180.559 10.339

2.11 0.089 0.06111

2

2.0

0.45

Sample Output

1 2 1

1 2 4 1

no arbitrage sequence exists

题目大意:套汇。给你n种货币相互之间的汇率。问你如何转换货币能在最少次数内获得利润。(利润 > 0.01)。

解题思路:.89就是0.89。他所求的不是最大利润,而是最少的获利交换次数。所以遍历交换次数,没个交换次数都算一次获利,当获利超过0.01,递归输出当前路径。用一个三维的dp数据dp[i][j][s],i, j代表由货币i转换到货币j。s代表转换次数,记录的是该情况下的汇率。然后用相似floyd的方式来更新这个数组。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int N = 25;
int n;
double rec[N][N][N], dp[N][N][N];
int temp[N];
void printfPath(int i, int j, int s) {
if (s == 0) {
printf("%d", i);
return;
}
printfPath(i, rec[i][j][s], s - 1);
printf(" %d", j);
return;
}
void DP() {
int cnt, flag = 0;
for (int s = 2; s <= n; s++) {
cnt = 0;
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (dp[i][j][s] < dp[i][k][s - 1] * dp[k][j][1]) {
dp[i][j][s] = dp[i][k][s - 1] * dp[k][j][1];
rec[i][j][s] = k;
}
}
}
}
for (int i = 1; i <= n; i++) {
if (dp[i][i][s] - 1.0 > 0.01) {
flag = 1;
printfPath(i, i, s);
printf("\n");
break;
}
}
if (flag) break;
}
if (!flag) printf("no arbitrage sequence exists\n"); }
void input() {
memset(dp, 0, sizeof(dp));
memset(rec, 0, sizeof(rec));
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (i == j) dp[i][j][1] = 1;
else scanf("%lf", &dp[i][j][1]);
}
}
}
int main() {
while (scanf("%d", &n) != EOF) {
input();
DP();
}
return 0;
}

uva 104 Arbitrage (DP + floyd)的更多相关文章

  1. UVa 104 - Arbitrage(Floyd动态规划)

    题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&pa ...

  2. POJ 2240 - Arbitrage(bellman_ford & floyd)

    题意: 给出一些货币和货币之间的兑换比率,问是否可以使某种货币经过一些列兑换之后,货币值增加. 举例说就是1美元经过一些兑换之后,超过1美元.可以输出Yes,否则输出No. 分析: 首先我们要把货币之 ...

  3. 洛谷P3502 [POI2010]CHO-Hamsters感想及题解(图论+字符串+矩阵加速$dp\&Floyd$)

    洛谷P3502 [POI2010]CHO-Hamsters感想及题解(图论+字符串+矩阵加速\(dp\&Floyd\)) 标签:题解 阅读体验:https://zybuluo.com/Junl ...

  4. uva 509 RAID!(磁盘数据)

    来自 https://blog.csdn.net/su_cicada/article/details/80085318 习题4-7 RAID技术(RAID!, ACM/ICPC World Final ...

  5. 取数字(dp优化)

    取数字(dp优化) 给定n个整数\(a_i\),你需要从中选取若干个数,使得它们的和是m的倍数.问有多少种方案.有多个询问,每次询问一个的m对应的答案. \(1\le n\le 200000,1\le ...

  6. 洛谷P2507 [SCOI2008]配对 题解(dp+贪心)

    洛谷P2507 [SCOI2008]配对 题解(dp+贪心) 标签:题解 阅读体验:https://zybuluo.com/Junlier/note/1299251 链接题目地址:洛谷P2507 [S ...

  7. [Codeforces722E] Research Rover (dp+组合数学)

    [Codeforces722E] Research Rover (dp+组合数学) 题面 给出一个N*M的方格阵,从(1,1)出发,到(N,M)结束,从(x,y)只能走到(x+1,y)或(x,y+1) ...

  8. 【wikioi】2800 送外卖(状压dp+floyd)

    http://www.wikioi.com/problem/2800/ 本题状压莫名其妙的tle了,(按照hzwer大神打的喂,他1000多ms,我就2000ms了?) (14.8.7更,将getnu ...

  9. UVA - 1347 Tour(DP + 双调旅行商问题)

    题意:给出按照x坐标排序的n个点,让我们求出从最左端点到最右短点然后再回来,并且经过所有点且只经过一次的最短路径. 分析:这个题目刘汝佳的算法书上也有详解(就在基础dp那一段),具体思路如下:按照题目 ...

随机推荐

  1. 路飞学城Python-Day15(模块二思维导图)

  2. 紫书 习题 8-16 UVa 1618 (中途相遇法)

    暴力n的四次方, 然而可以用中途相遇法的思想, 分左边两个数和右边两个数来判断, 最后合起来判断. 一边是n平方logn, 合起来是n平方logn(枚举n平方, 二分logn) (1)两种比较方式是相 ...

  3. iOS技术栈-Swift版

    在网上没找到基于Swift的技术栈介绍,或者技术已经不适用,所以自己画了一个,后面的博客中会重新整理基于Swfit的UI控件介绍和常用第三方库地图

  4. [terry笔记]11gR2_DataGuard搭建_拷贝数据文件

    11gR2搭建dataguard环境: 自己做的实验,后续按照rman模式搭建.主备切换.模式调整等实验会陆续发上来. primary: OS:oel 6.4 database:11.2.0.4.0 ...

  5. NYIST 1006 偷西瓜

    偷西瓜 时间限制:1000 ms  |  内存限制:65535 KB 难度:4   描述 对于农村的孩子来说最大的乐趣,莫过于和小伙伴们一块下地偷西瓜了,虽然孩子们条件不是很好,但是往往他们很聪明,他 ...

  6. Camera Calibration 相机标定:原理简介(二)

    2 针孔相机模型 常见的相机标定中,使用的相机多为针孔相机(Pinhole camera),也就是大家熟知的小孔成像理论.将其中涉及的坐标系之间的相互转换抽离出来,即为针孔相机模型的核心. 上图所示的 ...

  7. Qt之QTemporaryFile

    简述 QTemporaryFile类是操作临时文件的I/O设备. QTemporaryFile用于安全地创建一个独一无二的临时文件.临时文件通过调用open()来创建,并且名称是唯一的(即:保证不覆盖 ...

  8. Qt之表单布局(QFormLayout)

    简述 QFormLayout管理输入型控件和关联的标签组成的那些Form表单. QFormLayout是一个方便的布局类,其中的控件以两列的形式被布局在表单中.左列包括标签,右列包含输入控件,例如:Q ...

  9. Java读书笔记一(异常处理)

    1.介绍 在开发过程中.都会遇到程序执行时的异常,比如逻辑错误.硬件故障等,本篇博客将会具体的介绍一下Java中的异常处理机制. 2.Try和Catch捕获异常 我们在处理异常过程中经经常使用到的就是 ...

  10. angularjs 指令2

    <!DOCTYPE HTML> <html ng-app="myApp"> <head> <meta http-equiv="C ...