The Balance POJ - 2142

首先,可以知道题目要求解一个\(ax+by=c\)的方程,且\(x+y\)最小。
感性证明:
当\(a>b\)时,\(y\)取最小正整数解,\(b\)减的多,\(a\)增的少,此时\(x+y\)取最小值。(类似比热容与温度之间)
反之亦然。

#include<iostream>
#include<cmath>
#include<cstdlib>
using namespace std;
int a, b, c;
int exgcd(int a, int b, int &x, int &y) {
    if(b == 0) {
        x = 1, y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int tmp = x;
    x = y;
    y = tmp - (a / b) * y;
    return g;
}
void solve(int &x, int &y, int a, int b, int c) {
    int g = exgcd(a, b, x, y);
    x *= (c / g);
    int t = b / g;
    x = (x % t + t) % t;
    y = (a * x - c) / b;
    y = abs(y);
}
int main() {
    while(cin >> a >> b >> c) {
        if(!a && !b && !c) break;
        int x1, y1, x2, y2;
        solve(x1, y1, a, b, c);
        solve(x2, y2, b, a, c);
        if(x1 + y1 < x2 + y2) cout << x1 << " " << y1 << "\n";
        else cout << y2 << " " << x2 << "\n";
    }
    return 0;
} 

原文地址:https://www.cnblogs.com/yangxuejian/p/12000277.html

时间: 2025-01-16 17:24:05

The Balance POJ - 2142的相关文章

The Balance POJ 2142 扩展欧几里得

Description Ms. Iyo Kiffa-Australis has a balance and only two kinds of weights to measure a dose of medicine. For example, to measure 200mg of aspirin using 300mg weights and 700mg weights, she can put one 700mg weight on the side of the medicine an

POJ 2142 The Balance(exgcd)

嗯... 题目链接:http://poj.org/problem?id=2142 AC代码: 1 #include<cstdio> 2 #include<iostream> 3 4 using namespace std; 5 6 inline int _abs(int x){ 7 if(x < 0) return -x; 8 return x; 9 } 10 11 inline void exgcd(int a, int b, int &g, int &x,

POJ 2142 The Balance

The Balance Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 6325   Accepted: 2787 Description Ms. Iyo Kiffa-Australis has a balance and only two kinds of weights to measure a dose of medicine. For example, to measure 200mg of aspirin usi

POJ 2142 - The Balance [ 扩展欧几里得 ]

题意: 给定 a b n找到满足ax+by=n 的x,y 令|x|+|y|最小(等时令a|x|+b|y|最小) 分析: 算法一定是扩展欧几里得. 最小的时候一定是 x 是最小正值 或者 y 是最小正值 (简单的证明应该是分x,y 符号一正一负,和x,y符号都为正来考虑) 扩欧解的方程为 ax+by = gcd(a, b) 先简化问题,等价为扩欧求的是 a'x+b'y = 1 则原方程等价为 a'x+b'y = n' (a, b, n 全部除以gcd(a, b) ) 先解x为最小正值的时候 x =

POJ 2142 The Balance【扩展欧几里德】

题意:有两种类型的砝码,每种的砝码质量a和b给你,现在要求称出质量为c的物品,要求a的数量x和b的数量y最小,以及x+y的值最小. 用扩展欧几里德求ax+by=c,求出ax+by=1的一组通解,求出当x取最小合法正整数解时y的取值,当y小于0时,说明应该放在a的另一边,变为正值.同理当y取最小时,可得到另一组解,比较两组解,取最小即可. #include<stdio.h> int ex_gcd(int a,int b,int &x,int &y){ if(!b){ x=1,y=

poj 2142

Ms. Iyo Kiffa-Australis has a balance and only two kinds of weights to measure a dose of medicine. For example, to measure 200mg of aspirin using 300mg weights and 700mg weights, she can put one 700mg weight on the side of the medicine and three 300m

Balance POJ - 1837 地推

Gigel has a strange "balance" and he wants to poise it. Actually, the device is different from any other ordinary balance. It orders two arms of negligible weight and each arm's length is 15. Some hooks are attached to these arms and Gigel wants

POJ 2142:The Balance_扩展欧几里得(多组解)

先做出两个函数的图像,然后求|x|+|y|的最小值.|x|+|y|=|x0+b/d *t |+|y0-a/d *t| 这个关于t的函数的最小值应该在t零点附近(在斜率大的那条折线的零点附近,可以观察出来).以下三种情况中,函数最小值都应该出现在B点附近./* 对于不定整数方程xa+yb=c,若 c mod Gcd(a, b)=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x * a+y * b = Gcd(a, b)的一组解x0,y0后 ,/*x * a+y

POJ 2142 TheBalance 模线性方程求解

题目大意: 就是将两种砝码左右摆放,能够在物品放置在天平上时保持平衡 很容易得到 ax + by = t的模线性方程 按题目要求,希望首先满足 |x| + |y| 最小 , 如果有多种情况,再满足所有砝码质量最小,也就是a|x| + b|y|最小 x = x0 + b/g * k y = y0 - a/g * k 这里可以通过画一个2维坐标图进行观察 x , y 对于k的直线,我假定 b > a ,初始如果 a>b就交换两者数据,记得最后答案交换回来 因为a,b为砝码重量都大于0 所以x是递增