ICPC2008哈尔滨-E-Gauss Elimination
题目描述
“Do you know how far our voyage is?” The captain asks. Li Zhixiang feels ashamed because he can not answer. Then the captain says with a smile, “5050 miles. Do you still remember the story of 5050?” This time the young man really blushes. The old captain continues saying:” You definitely know the story of 5050. When the German mathematician, “the prince of mathematicians”, Gauss was 10 years old …” Young man remembers this story and goes on to tell, “ When Gauss was 10 years old, he could add a list of integers from 1 to 100 in a few seconds, which shocked the teachers.” The old captain adds, “Gauss has many other stories like this. When he entered the university at the age of 17, he was able to construct heptadecagon by compass and straightedge. His university teachers were also impressed by his ability. Not only could college graduate students fail to do it, but also they felt hard to understand Gauss’s constructing process.”
At this time, vice-captain greets the old captain. The old captain says to Li Zhixiang: “Come over to my office tonight, let’s continue the conversation.” It is still calm and tranquil in the evening. The freighter travels smoothly on the sea in the silver moonlight. The captain tells the young man the following words.
Among the mathematicians through the ages, there are three greatest mathematicians: Archimedes, Newton and Gauss. Most of Gauss’s mathematical achievements are difficult to understand. Nevertheless, there are some comparatively easy. For instance, when it comes to solving multivariate system of linear equations, there is a solution called “Gauss Elimination”. In the navigation business, many problems can be solved by “Gauss elimination”. If you are interested in it, I will show you a simple question. Try it.”
输入
输出
样例输入
2
1000000000000000000000000 1000000000000000000000000 1000000000000000000000000
-1000000000000000000000000 1000000000000000000000000 0
1
0 4
样例输出
1/2
1/2 No solution.
大数分数高斯消元
import java.math.BigInteger;
import java.util.Scanner; class Number{
BigInteger a,b;
Number() {
a=BigInteger.valueOf(1);
b=BigInteger.valueOf(1);
} Number(BigInteger x,BigInteger y) {
a=x;
b=y;
} Number sub(Number x){
Number c=new Number();
c.b=b.multiply(x.b);
c.a=a.multiply(x.b).subtract(x.a.multiply(b));
BigInteger d=c.a.gcd(c.b);
if (d.compareTo(BigInteger.valueOf(0))!=0){
c.a=c.a.divide(d); c.b=c.b.divide(d);
}
return c;
} Number mul(Number x){
Number c=new Number();
c.b=b.multiply(x.b);
c.a=a.multiply(x.a);
BigInteger d=c.a.gcd(c.b);
if (d.compareTo(BigInteger.valueOf(0))!=0){
c.a=c.a.divide(d); c.b=c.b.divide(d);
}
return c;
} Number div(Number x) {
Number c=new Number();
c.b=b.multiply(x.a);
c.a=a.multiply(x.b);
BigInteger d=c.a.gcd(c.b);
if (d.compareTo(BigInteger.valueOf(0))!=0){
c.a=c.a.divide(d); c.b=c.b.divide(d);
}
return c;
} int com(Number x) {
BigInteger p=a.multiply(x.b);
BigInteger q=x.a.multiply(b);
if (p.compareTo(BigInteger.valueOf(0))<0) p=p.multiply(BigInteger.valueOf(-1));
if (q.compareTo(BigInteger.valueOf(0))<0) q=q.multiply(BigInteger.valueOf(-1)); return p.compareTo(q);
}
}
public class Main { public static boolean Guss(int n,Number a[][],Number b[]){
int k=1,col=1;
while (k<=n && col<=n) {
int max_r=k;
for (int i=k+1;i<=n;i++)
if (a[i][col].com(a[max_r][col])>0)
max_r=i;
if (a[max_r][col].com(new Number(BigInteger.valueOf(0),BigInteger.valueOf(1)))==0) return false;
if (k!=max_r) {
for (int j=col;j<=n;j++) {
Number tmp=a[k][j];
a[k][j]=a[max_r][j];
a[max_r][j]=tmp;
}
Number tmp=b[k]; b[k]=b[max_r]; b[max_r]=tmp;
} b[k]=b[k].div(a[k][col]);
for (int j=col+1;j<=n;j++) a[k][j]=a[k][j].div(a[k][col]);
a[k][col].a=BigInteger.valueOf(1);
a[k][col].b=BigInteger.valueOf(1); for (int i=1;i<=n;i++) {
if (i!=k) {
b[i]=b[i].sub(b[k].mul(a[i][col]));
for (int j=col+1;j<=n;j++) a[i][j]=a[i][j].sub(a[k][j].mul(a[i][col]));
a[i][col].a=BigInteger.valueOf(0);
}
}
k++; col++;
}
return true;
} public static void main(String[] args) {
Number a[][] = new Number[105][105];
Number b[] = new Number[105]; for (int i=1;i<=100;i++) {
for (int j=1;j<=100;j++) a[i][j]=new Number();
b[i]=new Number();
}
int n;
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
n=in.nextInt();
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
a[i][j].a = in.nextBigInteger();
a[i][j].b = BigInteger.valueOf(1);
}
b[i].a=in.nextBigInteger();
b[i].b=BigInteger.valueOf(1);
} if (Guss(n,a,b)==true) {
for (int i=1;i<=n;i++) {
BigInteger d=b[i].a.gcd(b[i].b);
if (d.compareTo(BigInteger.valueOf(0))!=0){
b[i].a=b[i].a.divide(d); b[i].b=b[i].b.divide(d);
}
// System.out.println(1+" "+b[i].b+" "+b[i].b.compareTo(BigInteger.valueOf(0)));
if (b[i].b.compareTo(BigInteger.valueOf(0))<0){
// System.out.println("*");
b[i].b=b[i].b.multiply(BigInteger.valueOf(-1));
b[i].a=b[i].a.multiply(BigInteger.valueOf(-1));
}
// System.out.println(2+" "+b[i].b+" "+b[i].b.compareTo(BigInteger.valueOf(0)));
if (b[i].a.compareTo(BigInteger.valueOf(0))==0) b[i].b=BigInteger.valueOf(1);
if (b[i].b.compareTo(BigInteger.valueOf(1))==0) System.out.println(b[i].a);
else System.out.println(b[i].a+"/"+b[i].b);
}
} else System.out.println("No solution."); System.out.println();
}
}
}
ICPC2008哈尔滨-E-Gauss Elimination的更多相关文章
- Gauss elimination Template
Gauss elimination : #include <iostream> #include <cstdlib> #include <cstring> #inc ...
- 高斯消元法(Gauss Elimination)【超详解&模板】
高斯消元法,是线性代数中的一个算法,可用来求解线性方程组,并可以求出矩阵的秩,以及求出可逆方阵的逆矩阵.高斯消元法的原理是:若用初等行变换将增广矩阵 化为 ,则AX = B与CX = D是同解方程组. ...
- HDU2449 Gauss Elimination 高斯消元 高精度 (C++ AC代码)
原文链接https://www.cnblogs.com/zhouzhendong/p/HDU2449.html 题目传送门 - HDU2449 题意 高精度高斯消元. 输入 $n$ 个 $n$ 元方程 ...
- ICPC2008哈尔滨-A-Array Without Local Maximums
题目描述 Ivan unexpectedly saw a present from one of his previous birthdays. It is array of n numbers fr ...
- LU分解(1)
1/6 LU 分解 LU 分解可以写成A = LU,这里的L代表下三角矩阵,U代表上三角矩阵.对应的matlab代码如下: function[L, U] =zlu(A) % ZLU ...
- 线性代数-矩阵-【5】矩阵化简 C和C++实现
点击这里可以跳转至 [1]矩阵汇总:http://www.cnblogs.com/HongYi-Liang/p/7287369.html [2]矩阵生成:http://www.cnblogs.com/ ...
- 线性代数-矩阵-【1】矩阵汇总 C和C++的实现
矩阵的知识点之多足以写成一本线性代数. 在C++中,我们把矩阵封装成类.. 程序清单: Matrix.h//未完待续 #ifndef _MATRIX_H #define _MATRIX_H #incl ...
- 高斯消元 & 线性基【学习笔记】
高斯消元 & 线性基 本来说不写了,但还是写点吧 [update 2017-02-18]现在发现真的有好多需要思考的地方,网上很多代码感觉都是错误的,虽然题目通过了 [update 2017- ...
- bingoyes' tiny dream
Gauss Elimination bool Gauss(){ int now=1,nxt; double t; R(i,1,n){ //enumerate the column for(nxt=no ...
随机推荐
- Codeforces 1195E. OpenStreetMap (单调队列)
题意:给出一个n*m的矩形.询问矩形上所有的a*b的小矩形的最小值之和. 解法:我们先对每一行用单调栈维护c[i][j]代表从原数组的mp[i][j]到mp[i][j+b-1]的最小值(具体维护方法是 ...
- EventBus总结(原)
1.EventBus的作用 EventBus is a publish/subscribe event bus for Android and Java. EventBus可以被用来在各种自定义的监听 ...
- eclipse maven install后查看报错信息
- 【vlfeat】O(n)排序算法——计数排序
今天想在网上找一个实现好的er算法来着,没啥具体的资料,无奈只能看vlfeat的mser源码,看能不能修修补补实现个er. 于是,看到某一段感觉很神奇,于是放下写代码,跑来写博客,也就是这段 /* - ...
- python读取配置文件(ini、yaml、xml)
python读取配置文件(ini.yaml.xml)
- python用户名密码限定次数登录
""" 1. 用户输入帐号密码进行登陆 2. 用户信息保存在文件内 3. 用户密码输入错误三次后锁定用户"""" test.txt ...
- openwrt 编译支持sqlite3
编译版本加载lib库 ------------------------------Libraries----------------------------------- Filesystem -- ...
- Mysql查看编码方式
查看数据库的字符集 show variables like 'character\_set\_%'; 输出: +--------------------------+--------+ | Varia ...
- vue基础七
事件处理器 1.监听事件 可以用 v-on 指令监听 DOM 事件来触发一些 JavaScript 代码. <div id="example-1"> <butto ...
- Spring 容器中bean的加载过程
bean 的加载过程大致可以分为以下几个步骤: 1.获取配置的资源文件 2.对获取到的xml资源文件进行解析 3.获取包装资源 4.解析处理包装之后的资源 5.加载 提取bean 并进行注册(添加到b ...