甲乙二人交替在黑板上写正整数，新写的数不能是黑板上已有数的整数组合，写1者就输了。甲写5，乙写6，甲再写…, 问甲/乙有必胜策略吗？用计算机求解这个问题。A 称为a1, a2, …, an的整数组合，当且仅当存在一组非负整数I1, I2, …, In, 使下式成立:A= I1a1 + I2a2 + … + Inan.使用C++ 计算最优策略
B duo alternately write on the blackboard a positive integer, new written can not be on the blackboard for several combinations of integer, write 1 or lose. A write 5, B write A write ..., Q A/B has a winning strategy? Using a computer to solve this problem. A called a1, a2, ..., an integer combination of AN, if and only if there exists a set of non-negative integers I1, I2, ..., In, so that the following equation is established: A = I1a1 PPLS I2a2 PPLS ... PPLS Inan of use C++ Calculation optimal strategy (2012-11-11, C++ Builder, 2KB, 下载4次)
DLT645 检验和、+/-33计算小软件，附带源码。
DLT645 inspection and+/-33 computing software, with source code. (2012-07-07, C++ Builder, 489KB, 下载22次)
最佳矩阵连乘 给定n个矩阵｛A1,A2,…An｝，其中Ai与A i+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2…An。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q矩阵，B是一个q×r矩阵，则其乘积C=AB是一个p×r矩阵,需要pqr次数乘。
由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序。例如，设3个矩阵{A1,A2,A3}的维数分别为10×100，100×5，和5×50。若按加括号方式((A1A2)A3)计算，3个矩阵连乘积需要的数乘次数为10×100×5+10×5×50=7500。若按加括号方式(A1(A2A3))计算，3个矩阵连乘积总共需要10×5×50+10×100×50=75000次数乘。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。
矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵｛A1,A2,…An｝(其中矩阵Ai的维数为pi-1×p，i=1,2，…，n)，确定计算矩阵连乘积A1,A2,…An的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
best matrix continually multiply given n matrix (A1, A2, ... An), Ai and A is a mere i, i = 1, 2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p q matrix B is a q-r-matrix, its product C = AB is a p r matrix, the number required by pqr. Because matrix multiplication meet the law of combination, it's even calculated matrix product can be calculated in many different priorities. For example, the matrix-based 3 (A1, A2, A3) dimension of 10 100, 100 5 5 and 50. If bracketed by the way ((A1A2) A3), even three product matrix multiplication in the number of 10 100 10 5 50 = 7,500. If bracketed by the way (A1 (A2A3)), three matrix product even need a total of 10 5 50 (2005-11-28, C++ Builder, 6KB, 下载18次)