uva 498 - Polly the Polynomial

【uva 498 - Polly the Polynomial】的更多相关文章

uva 498 - Polly the Polynomial

UVa 498: Polly the Polynomial | MathBlog #include <cstdio> #include <cstdlib> using namespace std; ]; int main() { int n; while((temp = getchar()) != EOF) { ; while(temp != '\n') { ') { ungetc(temp, stdin); scanf("%d", &c[count +…

uva 10951 - Polynomial GCD(欧几里得)

题目链接:uva 10951 - Polynomial GCD 题目大意:给出n和两个多项式,求两个多项式在全部操作均模n的情况下最大公约数是多少. 解题思路:欧几里得算法,就是为多项式这个数据类型重载取模运算符,须要注意的是在多项式除多项的过程中,为了保证各项系数为整数,须要将整个多项式的系数总体扩大至一定倍数,碰到先除后模的时候要用逆元. #include <cstdio> #include <cstring> const int maxn = 105; int M; void…

UVA 10951 - Polynomial GCD(数论）

UVA 10951 - Polynomial GCD 题目链接题意:给定两个多项式,求多项式的gcd,要求首项次数为1,多项式中的运算都%n,而且n为素数. 思路:和gcd基本一样,仅仅只是传入的是两个多项式,因为有%n这个条件.所以计算过程能够用乘法逆去计算除法模,然后最后输出的时候每项除掉首项的次数就是答案了. 代码: #include <stdio.h> #include <string.h> #include <vector> using namespace…

【数论】UVa 10586 - Polynomial Remains

Problem F: Polynomial Remains Given the polynomial a(x) = an xn + ... + a1 x + a0, compute the remainder r(x) when a(x) is divided by xk+1. The input consists of a number of cases. The first line of each case specifies the two integers n and k (0 ≤ n…

UVA 12009 - Avaricious Maryanna(数论)

UVA 12009 - Avaricious Maryanna 题目链接题意:给定一个n.求出n个数位组成的数字x,x^2的前面|x|位为x 思路:自己先暴力打了前几组数据,发现除了1中有0和1以外,其它数据都是由前一项往上再加入一位得到的,因此设新数字为(a∗10k+x)2=(a∗10k)2+x2+2∗a∗10k∗x 因此(a∗10k+x)=((a∗10k)2+x2+2∗a∗10k∗x)/10k%10 化简后得到x2/10k%10+2∗a∗x%10 因此仅仅要能求出x2/10k%10.然后再…

Polynomial Library in OpenCascade

Polynomial Library in OpenCascade eryar@163.com 摘要Abstract:分析幂基曲线即多项式曲线在OpenCascade中的计算方法,以及利用OpenSceneGraph来显示OpenCascade的计算结果,加深对多项式曲线的理解. 关键字Key Words:OpenCascade.PLib.OpenSceneGraph.Polynomial Library 一. 概述 Overview CAGD(Computer Aided Geometry D…

1Z0-053 争议题目解析498

1Z0-053 争议题目解析498 考试科目:1Z0-053 题库版本:V13.02 题库中原题为: 498.The database Is configured in ARCHIVELOG mode. The database needs to be up 24 X 7. You want to perform user managed backup for the data files of the HR_DATA tablespace. To accomplish the task, yo…