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Java做ACM-ICPC的特点：(1) 在一般比赛中，Java程序会有额外的时间和空间，而实际上经过实验，在执行计算密集任务的时候Java并不比C/C++慢多少，只是IO操作较慢而已。(2) Java 简单而功能强大，有些东西用Java实现起来更为方便，比如高精度。(3) 用Java不易犯细微的错误，比如C/C++中的指针， “if (n = m) ... ” 等(4) 目前来看Eclipse已成基本配置，写Java程序反而比C/C++更方便调试。在具体竞赛时也算多一种选择。(5) 学会Java对以后工作有好处。现在国外很多地方会Java的人比会C/C++的人多。下面说一下ACM-ICPC队 阅读全文
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Problem DescriptionThe three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. You are to calculate ho 阅读全文
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关于ACM的输入输出（一）一般来说ACM的现场赛会规定输入输出或者是文件输入标准输出也可能是文件输入文件输出如果没有规定的话那么一般就是标准的输入输出了那说一下输入输出的重定向一般用下面两种方法c++常用:#include ifstream filein("data.in"); // 定义一个文件输入流ofstream fileout("data.out"); //cout fileout>都要改成filein>>, cout> str) { } getline读字符串时遇到换行符结束,用于读一整行 char str[BUFFER 阅读全文
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所谓的快速幂，实际上是快速幂取模的缩写，简单的说，就是快速的求一个幂式的模(余)。在程序设计过程中，经常要去求一些大数对于某个数的余数，为了得到更快、计算范围更大的算法，产生了快速幂取模算法。我们先从简单的例子入手：求abmodc算法1.直接设计这个算法：int ans = 1;for(int i = 1;i0) { if(b % 2 = = 1) ans = (ans * a) % c; b = b/2; a = (a * a) % c; } return ans;}本算法的时间复杂度为O（logb），能在几乎所有的程序设计... 阅读全文
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Problem DescriptionA number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.Given A, B, and n, you are to calculate the value of f(n).InputThe input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 using 阅读全文
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Problem DescriptionContest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.This year, they decide to l 阅读全文
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Problem DescriptionGiven a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.InputThe first line of the input contains an integer T(1#includeusing namespace std;void solve 阅读全文
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Problem DescriptionA simple mathematical formula for e iswhere n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.OutputOutput the approximations of e generated by the above formula for the values of n from 0 to 9. The beginni 阅读全文
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DescriptionProblems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.This problem requires that you write a program to compute the exact value of Rnwhere R is 阅读全文
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Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.InputThe first line of the input contains an integer T(1 0) { a = cin.nextBigInteger(); b = cin.nextBigInteger(); System.out.println("Case "... 阅读全文