尝试读取受保护的MS Access 97文件的密码。
Tries to read the password of protected MS-Access 97 files. (2021-10-10, Pascal, 0KB, 下载0次)
未完成的Win32 BF编译器。，
Unfinished Win32 BF compiler., (2017-02-26, Pascal, 0KB, 下载0次)
ToolBar Components 97 for all delphi
ToolBar Components 97 for all delphi (2020-04-08, Pascal, 137KB, 下载4次)
The source code is used to program PIC microcontroller to drive an lcd display interface.
The source code is used to program PIC microcontroller to drive an lcd display interface. (2014-02-01, Pascal, 3050KB, 下载1次)
Given a positive integer n, the integers a1, a2, ..., An. Consider the sequence of segments a1, a2, ..., AN (a subsequence of consecutive terms) consisting of perfect numbers. Find the largest of the lengths of the segment. Establish a procedure allowing to recognize the perfect number.
Given a positive integer n, the integers a1, a2, ..., An. Consider the sequence of segments a1, a2, ..., AN (a subsequence of consecutive terms) consisting of perfect numbers. Find the largest of the lengths of the segment. Establish a procedure allowing to recognize the perfect number. (2013-08-16, Pascal, 1KB, 下载2次)
exclusive photos of pascal programmer
exclusive photos of pascal programmer (2010-04-15, Pascal, 2710KB, 下载3次)
Pascal基本求值,求长方形的周长和面积.
Pascal basic evaluation, find the perimeter and area of rectangles. (2009-11-09, Pascal, 2KB, 下载4次)
已知一对兔子，每个月可以生一对小兔，而小兔过一个月后也可以生一对小兔。即兔子的对数是：第一个月1对，第二个月2对，第三个月3对，第四个月5对，……，假设兔子的剩余期是12个月，并且不死 ，问一年后，这对兔子有多少对活着的后代？
Are known to a pair of rabbits, a pair of rabbit health every month, while the rabbit can be a month after a pair of rabbit health. Namely, the number of pairs of rabbits are: 1 on the first month, the first month two pairs, the first three months, three pairs, the first four months of 5 pairs ... ..., assuming that the rabbit with a maturity of 12 months, and die, Q. A year later, this is the number of pairs of rabbits alive for future generations? (2009-11-09, Pascal, 1KB, 下载4次)
验证哥德巴赫才:任何一个充分大的偶数N(≥4),可以用两个素数之和表示.
如 4=2+2 6=3+3 8=3+5 98=17+79
Verify Goldbach before: Any one sufficiently large even number N (≥ 4), can be expressed the sum of two primes. Such as 4 = 2+2 6 = 3+3 8 = 3+5 98 = 17+79 (2009-11-09, Pascal, 1KB, 下载24次)
求满足 M > N > 0 的两个正整数之和的最大公约数.
Order to meet the M> N> 0, two positive integers and the greatest common divisor. (2009-11-09, Pascal, 1KB, 下载30次)
Pascal写的用格里公式求派(圆周率).
Gerry seeking to use the formula to send (pi). (2009-11-09, Pascal, 1KB, 下载3次)
N只猴子要选大王,所有猴子按1,2,3...M报数，凡报到M的退出圈外,如此循环,直到圈内只剩一只,就是大王了.
分析:建立一个环形链表,每个猴子就是链表中的一个结点,那么猴子出圈就是对环形链表中的结点的删除,只到最后剩下的一个结点为止.
N monkeys to elect king, all the monkeys reported by a number 1,2,3 ... M, where M to report the withdrawal of outsiders, and so on, until only one circle is the king of. Analysis: the establishment of a ring linked list, each monkey is a list of a node, then the monkey out of the circle is a right circular list node removal, only to the last remaining until a node. (2009-11-08, Pascal, 1KB, 下载5次)
很多不等式在展开以后形成如下的对称形式
sigma(s1^a1*s2^a2*...*sn^an)>=sigma(s1^b1*s2^b2*...*sn^bn)
（当然 作为齐次不等式 a1+a2+....an=b1+b2+...bn 变量s1,s2,...sn非负）
其中sigma表示对称和（也就是说 一共n!项） 例如
sigma(x^3)=x^3y^0z^0+x^3z^0y^0+y^3x^0z^0+y^3z^0x^0+z^3x^0y^0+z^3y^0x^0=2*(x^3+y^3+z^3)
sigma(x^3y^2z^1)=x^3y^2z^1+x^3z^2y^1+y^3x^2z^1+y^3z^2x^1+z^3x^2y^1+z^3y^2x^1
（三元sigma 一共是6项）
有时候 我们把sigma(s1^a1*s2^a2*...*sn*an)写作 [a1,a2,...an]
例如 著名的均值不等式可以写成 [n,0,0...0]>=[1,1,1...1]
又比如x^2+y^2+z^2>=xy+yz+zx 写成[2,0]>=[1,1]
本程序能比较两个完全对称不等式的大小关系。
After a lot of inequality start to form in the following symmetric form of sigma (s1 ^ a1* s2 ^ a2*...* sn ^ an)> = sigma (s1 ^ b1* s2 ^ b2*...* sn ^ bn) (of course, homogeneous inequality as a1+ a2+.... an = b1+ b2+ ... bn variables s1, s2, ... sn non-negative) that the symmetric and one sigma (ie a total of n! items) such as sigma (x ^ 3) = x ^ 3y ^ 0z ^ 0+ x ^ 3z ^ 0y ^ 0+ y ^ 3x ^ 0z ^ 0+ y ^ 3z ^ 0x ^ 0+ z ^ 3x ^ 0y ^ 0+ z ^ 3y ^ 0x ^ 0 = 2* (x ^ 3+ y ^ 3+ z ^ 3) sigma (x ^ 3y ^ 2z ^ 1) = x ^ 3y ^ 2z ^ 1+ x ^ 3z ^ 2y ^ 1+ y ^ 3x ^ 2z ^ 1+ y ^ 3z ^ 2x ^ 1+ z ^ 3x ^ 2y ^ 1+ z ^ 3y ^ 2x ^ 1 (three sigma is a total of six) at times we have sigma (s1 ^ a1* s2 ^ a2*.. .* sn* an) Writing [a1, a2, ... an] such as the well-known inequality can be written in the mean [n, 0,0 ... 0]> = [1,1,1 ... 1] and For example, x ^ 2+ y ^ 2+ z ^ 2> = xy+ yz+ zx written [2,0]> = [1,1] This procedure can compare two completely symmetrical relationship between the si (2009-05-22, Pascal, 1KB, 下载2次)