这是一个相对简单的模拟,因为运算规则已经告诉了我们,并且比较简单,不要被吓到…… 思路:多项式除以另外一个多项式,如果能除,那么他的最高次一定被降低了,如果最高次不能被降低,那说明已经无法被除,就是题目要求输出的膜了,降低最高次的方法很简单,只要被除式的最高次 >= 除式的最高次,就将除式的最高次升高到与被除式一样高,然后让被除式减去它,直到不满足上述关系为止. 代码如下: #include<cstdio> #include<algorithm> #include<io…

题目来源:http://poj.org/problem?id=1060 题目大意: 考虑系数为0和1的多项式.两个多项式的加法可以通过把相应次数项的系数相加而实现.但此处我们用模2加法来计算系数之和.一个实例: (x^6 + x^4 + x^2 + x + 1) + (x^7 + x + 1) = x^7 + x^6 + x^4 + x^2 同样,多项式的减法我们在求系数之差时也用模2减法,例如: (x^6 + x^4 + x^2 + x + 1) - (x^7 + x + 1) = x^7 +…

Modular multiplication of polynomials Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3239   Accepted: 1459 Description Consider polynomials whose coefficients are 0 and 1. Addition of two polynomials is achieved by 'adding' the coeffici…

Modular multiplication of polynomials Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4377   Accepted: 1980 Description Consider polynomials whose coefficients are 0 and 1. Addition of two polynomials is achieved by 'adding' the coeffici…

题意:给出f(x),g(x),h(x)的 (最高次幂+1)的值,以及它们的各项系数,求f(x)*g(x)/h(x)的余数. 这里多项式的系数只有1或0,因为题目要求:这里多项式的加减法是将系数相加/减后再模2,这样其实也就可以用异或运算来代替加减法. 思路:看代码吧,水题一个,主要在于把除法转化成减法,一次一次减就行. #include <iostream> #include <stdio.h> #include <string.h> using namespace s…

In data structure Hash, hash function is used to convert a string(or any other type) into an integer smaller than hash size and bigger or equal to zero. The objective of designing a hash function is to "hash" the key as unreasonable as possible.…

This time, you are supposed to find A*B where A and B are two polynomials. Input Specification: Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 … NK aNK, wher…

Stack Machine Executor 题目链接: http://acm.hust.edu.cn/vjudge/problem/26628 Description http://7xjob4.com1.z0.glb.clouddn.com/18113a5cf78f108997460e36f7079fc6 Input The input contains several instances. Each instance consists of two lines, the first lin…

意甲冠军  由于矩阵乘法计算链表达的数量,需要的计算  后的电流等于行的矩阵的矩阵的列数  他们乘足够的人才  非法输出error 输入是严格合法的  即使仅仅有两个相乘也会用括号括起来  并且括号中最多有两个 那么就非常easy了 遇到字母直接入栈  遇到反括号计算后入栈  然后就得到结果了 #include<cstdio> #include<cctype> #include<cstring> using namespace std; const int N = 10…

Dance Problem Description For a dance to be proper in the Altered Culture of Machinema, it must abide by the following rules: 1. A dip can only appear 1 or 2 steps after a jiggle, or before a twirl, as in:* ...jiggle dip...* ...jiggle stomp dip...* .…