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c++
- 计算任意一个多位数的位反序数； 由两个平方三位数获得三个平方二位数； 求车速； 阿姆斯特朗数； 求素数表中1~1000之间的所有素数。-Calculation of arbitrary bit more than one anti-median ordinal number by two three-digit access to three square square double-digit for speed
1002
- Sicily上的1002题，主要是为一些初学者设计的-Problem Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-p
anti
- 输入一个整数n,将n!分解为质因子的幂的乘积，例如：输入4，则4!=2^3*3-Enter an integer n, will be decomposed into prime factors n! power of the product, such as: type 4, then 4! = 2 ^ 3* 3
Anti-prime-Sequences
- 中山大学Sicily Online Judge里面的1002题目的源代码，大家可以参考学习一下-Sun Yat-sen Sicily Online Judge inside the 1002 title of source code, we can refer to learn about. . ....
anti
- 输入一个整数n,将n!分解为质因子的幂的乘积，例如：输入4，则4!=2^3*3-Enter an integer n, will be decomposed into prime factors n! power of the product, such as: type 4, then 4! = 2 ^ 3* 3