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
- 【记录】ELK之logstash同步mysql数据到Elasticsearch ,配置文件详解
本文出处:https://my.oschina.net/xiaowangqiongyou/blog/1812708#comments 截取部分内容以便学习 input { jdbc { # mysql ...
- openssl部分解读
前言 openssl是个开源的加密库.可以对文件进行加密解密. 小知识 术语: 单词: Encryption 加密 Decryption 解密 ssl 安全socket层 tsl 最 ...
- ubuntu 搜狗输入法内存占用太多,卡顿不够处理办法
1. 输入 free -m 查看是否内存不够导致卡顿 2. 输入 gnome-system-monitor 打开ubuntu 任务管理器 找到搜狗输入法结束进程 3. 完美解决
- JavaSE---System类
1.概述 1.1 System类 代表当前java程序的运行平台: 1.2 System类 提供的类方法: getenv():获取系统所有的环境变量: getenv(String name):获取 ...
- 编译php-5.5.15出错,xml2-config not found
今天在centos上编译php-5.5.15, cd php-5.5.15 ./configure --prefix=/usr/local/php/ --with-config-file-path=/ ...
- apue 第4章 文件和目录
获取文件属性 #include <sys/types.h> #include <sys/stat.h> #include <unistd.h> int stat(c ...
- hdu1059&poj1014 Dividing (dp,多重背包的二分优化)
Problem Description Marsha and Bill own a collection of marbles. They want to split the collection a ...
- bzoj 3881 [Coci2015]Divljak——LCT维护parent树链并
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=3881 对 S 建 SAM ,每个 T 会让 S 的 parent 树的链并答案+1:在 T ...
- nginx 全面优化 负载均衡
修改nginx.conf文件,它保存有nginx不同模块的全部设置.如果是原生安装的话应该在服务器的 /etc/nginx 目录找到 nginx.conf ,使用其它安装包的话也可以自行查找nginx ...
- Django的流程如何理解(餐厅点餐举例)
去饭店(商场)吃饭的步骤: 告诉前台服务员,来一小碗牛肉拉面,菜单上勾上一个牛肉拉面(url) 服务员去拉面窗口,告诉后厨,一碗牛肉拉面),后厨(view)开始准备. 后厨给打杂小弟说,给我一份儿面条 ...