本题是极其裸的EXGCD AX+BY+C=0 给你a b c 和x与y的区间范围,问你整数解有几组 作为EXGCD入门,题目比较简单 主要需要考虑区间范围的向上.向下取整,及正负符号的问题 问题是这正负号判断考虑让我WA无数次 我好菜阿 补充:关于使用扩展欧几里德算法解决不定方程的办法 对于不定整数方程pa+qb=c,若 c mod Gcd(p, q)=0,则该方程存在整数解,否则不存在整数解. 上面已经列出找一个整数解的方法,在找到p * a+q * b = Gcd(p, q)的一组解p0,q…

思路:看题就知道用扩展的欧几里得算法做!!! 首先我们可以求出ax+by=gcd(a,b)=g的一个组解(x0,y0).而要使ax+by=c有解,必须有c%g==0. 继而可以得到ax+by=c的一个组解x1=c*x0/g , y1=c*y0/g. 这样可以得到ax+by=c的通解为:                   x=x1+b*t;                   y=y1-a*t; 再就是要注意符号问题!!! 代码如下: #include<cstdio> #include<…

[lightoj P1306] Solutions to an Equation You have to find the number of solutions of the following equation: Ax + By + C = 0 Where A, B, C, x, y are integers and x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2. Input Input starts with an integer T (≤ 10000), denoting th…

1306 - Solutions to an Equation    PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB You have to find the number of solutions of the following equation: Ax + By + C = 0 Where A, B, C, x, y are integers and x1 ≤ x ≤ x2 and y1 …

Solutions to an Equation LightOJ - 1306 一个基础的扩展欧几里得算法的应用. 解方程ax+by=c时,基本就是先记录下a和b的符号fla和flb(a为正则fla为1,为负则fla为-1,flb相同),然后对a和b取绝对值.求出ax+by=gcd(a,b)的一组解x=ansx,y=ansy,那么只有当c是gcd(a,b)的倍数时原方程才可能有解.设g=gcd(a,b),通解是x=ansx*(c/g)*fla+k*b/g*fla,y=ansy*(c/g)*flb…

The Solutions of Nonlinear Equation 本文主要介绍几种用于解非线性方程$f(x)=0$的一些方法. (1) Bisection Method. 算法: step 1: 初始化$a,b(b>a)$,使$f(a),f(b)$异号. step 2: while (停止条件不满足) $p=a+\frac{b-a}{2}$: 若 $f(p)f(a)<0$,$b=p$:否则$a=p$. end while step 3: 返回的$p$为方程$f(x)=0$的解. 停止条件…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1306 You have to find the number of solutions of the following equation: Ax + By + C = Where A, B, C, x, y are integers and x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2. Input Input starts with an integer T (≤ ),…

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27938#problem/E 题目大意:Given, n, count the number of solutions to the equation x+2y+2z=n, where x,y,z,n are non negative integers. 解题思路:只对一个枚举由此推算出另外两个的种类,千万不要都枚举!!! #include<iostream> #include<…

Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only '+', '-' operation, the variable x and its coefficient. If there is no solution for the equation, return "No solution". If th…

Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only '+', '-' operation, the variable x and its coefficient. If there is no solution for the equation, return "No solution". If th…