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PAT 01-2. Maximum Subsequence Sum (25)
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 Maximum Subsequence is the continuous subsequence which has the largest su原创 2015-02-02 15:39:15 · 371 阅读 · 0 评论 -
PAT 01-1. 最大子列和问题(20)
01-1. 最大子列和问题(20)时间限制10000 ms内存限制65536 kB代码长度限制8000 B判题程序Standard给定K个整数组成的序列{ N1, N2, ..., NK },“连续子列”被定义为{ Ni, Ni+1, ..., Nj },其中 1原创 2015-02-02 15:31:27 · 698 阅读 · 0 评论 -
PAT 02-1. Reversing Linked List (25)
PAT 02-1. Reversing Linked List (25)Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, t原创 2015-08-16 10:41:42 · 636 阅读 · 1 评论 -
PAT 02-2. 一元多项式求导 (25)
02-2. 一元多项式求导 (25)时间限制400 ms内存限制65536 kB代码长度限制8000 B判题程序Standard设计函数求一元多项式的导数。输入格式:以指数递降方式输入多项式非零项系数和指数(绝对值均为不超过1000的整数)。数字间以空格分隔。原创 2015-02-02 15:46:09 · 735 阅读 · 0 评论 -
PAT 02-4. Pop Sequence (25) Python C
02-线性结构4. Pop Sequence (25)题目要求Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, …, N and pop randomly. You are supposed to tell if a given sequence of numbers is原创 2015-08-17 14:53:15 · 434 阅读 · 0 评论 -
03-树1. 二分法求多项式单根(20)
03-树1. 二分法求多项式单根(20)题目要求二分法求函数根的原理为:如果连续函数f(x)在区间[a, b]的两个端点取值异号,即f(a)f(b)<0,则它在这个区间内至少存在1个根r,即f(r)=0。二分法的步骤为:检查区间长度,如果小于给定阈值,则停止,输出区间中点(a+b)/2;否则 如果f(a)f(b)<0,则计算中点的值f((a+b)/2); 如果f((a+b)/2)正好为0,则(a原创 2015-08-18 12:04:23 · 886 阅读 · 0 评论 -
03-树2. List Leaves (25) Python C
03-树2. List Leaves (25)题目要求Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.Input Specification:Each input file contains one test case. For each case, t原创 2015-08-19 14:22:23 · 1052 阅读 · 0 评论