资源列表
mumfunction
- ACM程序竞赛中会有不少题目会用到母函数的技巧，而母函数的模板却很少人会打。本源码就是一个母函数模板-ACM contest procedures used there will be many topics generating function skills, and generating function of the template are very few people would have to fight. The source is a generating function
sd
- 大家看看 0-1背包的分支界限程序-We look at the 0-1 Knapsack procedures branch boundaries
tj
- 统计输入的一串字符中汉字字母数字的个数。-Statistics entered a string of alphanumeric characters in the number of Chinese characters.
shuzhifenxi
- 数值分析的作业，里面有详细的方法，主要有迭代法，多项式的插值震荡，误差的传播与算法稳定-Numerical Analysis of the operation, there are detailed methods, mainly iterative method, polynomial interpolation concussion, the spread of error and algorithm stability
shuzhi3
- 龙贝格求积公式，求解定积分 牛顿迭代公式，求方程的近似解 牛顿-科特斯求积公式，求定积分-Romberg quadrature formula for solving the definite integral of Newton s iterative formula, and the approximate solution equation Newton- cortez quadrature formula, and the definite integral
shuzhi4
- 雅克比迭代，求解方程近似解 秦九昭算法 幂法 高斯塞德尔 -Jacobian iteration, solving equations approximate solution algorithm Qin Zhao nine high-power law斯塞德尔
C1
- Newton Methom 迭代4次，精度达到了7.977e-12，程序执行时间<0.01s在知道解的大致值的情况下，Newton Methom是相当好的解法-Newton Methom iteration 4 times the accuracy of 7.977e-12, program execution time <0.01s in the know the approximate value solution circumstances, Newton Methom i
shiyan12
- 编制具有如下原型的函数prime，用来判断整数n是否为素数：bool prime(int n) 而后编制主函数，任意输入一个大于4的偶数d，找出满足d=d1+d2的所有数对，其中要求d1与d2均为素数（通过调用prime来判断素数）。如偶数18可以分解为11+7以及13+5；而偶数80可以分解为：43+37、61+19、67+13、73+7。 提示：i与d-i的和恰为偶数d，而且只有当i与d-i均为奇数时才有可能成为所求的“数对”。 -Prepared with the follow
544pp
- 用VB程序编写的数值计算解雅克比迭代公式。-VB programmers using numerical solution of the Jacobian iterative formula.
baiqianbaiji
- 经典题目——百钱百鸡问题，虽然很简单但是也是ACM培训初期必须掌握的呀-Classical topics- money 100 100 chickens, although very simple but the ACM is also the early training must master it
qipanfugai
- 小弟今天上传几个程序想和大家交流一下,我十分想和大家学习点东西.棋盘覆盖-From today, a few procedures boy would like to exchange, I very much would like to learn something. Chessboard coverage
acm
- 关于ACM的资料，如果大家希望参加ACM的竞赛的话，就可以下载下来学习-Information on the ACM, if you wish to participate in the ACM contest, it can be downloaded from learning