引入
前几天参加湖南多校的比赛，其中有这样一道题，需要使用高精度，同时需要排序，如果用c++实现的话，重载运算符很麻烦，于是直接学习了一发怎样用Java写大数，同时也算是学习Java基本常识了
题目
Description
Today the Intergalactic Council of Pebble Coins (ICPC) conducted an intergalactic auction of the Neutronium Chaos Pebble Coin (NCPC). This coin, which was forged in the Ancient Coin Machine (ACM), is rumored to be the key to ruling the universe.
Due to the extremely competitive nature of the auction, as well as the odd mechanics of the intergalactic currency used (far too advanced for mere mortals to understand), the auction was conducted with the following rules:
- only one participant was allowed to make a bid at a time,
- each participant was only allowed to make one bid, and
- a participant making a bid had to bid at least twice the amount of the highest bid at the time.
The first participant making a bid was allowed to make a bid of any positive amount.
After the auction there were a lot of sore losers – understandably, having just lost their chance at world domination. To make the losers feel a little better and prevent possible rioting, the ICPC has decided to hold a lottery for the participants. The winners of the lottery are determined as follows. The ICPC picks a random number (s). A group of participants is called winning if the sum of their bets from the auction is equal to (s). A participant wins the lottery and receives a prize – a shiny Pebble Coin – if they belong to any winning group of participants.
Given the names of the participants, the bets that they made, and the random number (s) chosen by the ICPC, help them determine which participants won the lottery.
Input
The first line of input contains two integers (n) and (s), where (1 le n le 1\,000) is the number of participants, and (1 le s < 10^{1\,000}) is the random number chosen by the ICPC.
Then follow (n) lines describing the participants. Each line contains a string (t) and an integer (b), where (t) is the name of a participant, and (1 le b < 10^{1\,000}) is the amount of his bet. The name of each participant is unique and consists of between (1) and (20) letters from the English alphabet.
Output
Output an integer (k) denoting the number of participants that won the lottery. Then output (k) lines containing the names of the participants that won the lottery, one per line, in any order.
Examples
Input
5 63
Vader 3
Voldemort 7
BorgQueen 20
Terminator 40
Megatron 101
Output
3
Terminator
BorgQueen
Vader
Input
4 1112
Blorg 10
Glorg 1000
Klorg 1
Zlorg 100
Output
0
题解
首先这个题说了一堆没用的废话，但有几点是必须要关注的，每一次竞标的价格一定是上次价格的两倍及以上，这样我们可以知道如果一个数在s之内，如果不选它，剩下的所有数加起来一定凑不够它，所以它必选，这样我们就可以排一遍序后直接按竞标价格从大到小枚举，小于当前要凑的数就选上，否则不选，问题就在于这个数很大，要用到高精度，于是就开始了学习java之旅
java输入输出
输入
要用到Scanner类，首先声明一个scanner变量，并用new运算符实例化Scanner，实例化Scanner时，需要传入System.in对象，Scanner通过传入的System.in获取用户输入，并对用户输入的字符进行处理，屏蔽了获取用户输入的复杂操作。
示例：
Scanner cin = new Scanner(System.in);
定义了一个名为cin的Scanner
输入不同类型数据的示例：
int n = cin.nextInt();//输入一个整数n
BigInteger sum = cin.nextBigInteger();//输入一个大整数sum
cin.next();//输入一个字符串，以空格为分隔
cin.nextLine();//输入一个字符串，以回车为分隔
输出
可以使用c中的printf，也可以使用新的println, print，区别是printf可以用来控制格式，println打印后换行，print不换行，均为System.out下的方法
示例：
System.out.println("不能加逗号，输出多个变量用加号连接" + "!");
System.out.print("。。。");
System.out.printf("%d", a);
Java定义变量
示例:
int n;//定义一个整数
BigInteger sum;//定义一个大整数
Vector<node> a = new Vector<node>();//定义一个Vector，类型为node，node为自己定义的类
Vector<String> ans = new Vector<String>();//定义一个类型为String的Vector
String[] name = new String[n + 1];//定义一个长度为n+1的字符串数组
Java实现结构体(Java中的类)
首先注意一点，静态类调用不了动态类的方法，如果在main类外面定义一个类，main中要调用这个类中的方法，要把这个类定义为static，就像下面这个
public static class node {
String name;
BigInteger bid;
node (String name, BigInteger bid) {
this.name = name;
this.bid = bid;
}
}
注意调用构造函数的时候要加new
Java中Vector的使用
添加一个元素：
for (int i = 1; i <= n; i++) {
a.add(new node(cin.next(), cin.nextBigInteger()));
}
对其排序，并自定义比较方法：
Collections.sort(a, new Comparator<node>() {
public int compare(node a, node b) {
return b.bid.compareTo(a.bid);
}
});
注意括号位置，这个排序是按照bid从大到小排序
遍历:
for (int i = 0; i < ans.size(); i++) {
System.out.println(ans.get(i));
}
使用get(i)获取在i处的值，如果是类型是类，可以再加.访问成员变量
Java大数操作
a = a.add(b);//加：a和b都是大数类型
a.subtract(b);//减
a.multiply(b);//乘
a.divide(b);//除：整除
a.compareTo(b);//比较大小，a小于b返回-1，等于返回0，大于返回1
a.mod(b);//求余
a.gcd(b);//求最大公约数
a.max(b);//求最大值
a.min(b);//求最小值
BigInteger.valuerOf();//将括号内数的转换为大整数
最后附上本题AC代码：
AC代码
import java.util.*;
import java.math.*;
public class test {
public static class node {
String name;
BigInteger bid;
node (String name, BigInteger bid) {
this.name = name;
this.bid = bid;
}
}
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int n = cin.nextInt();
BigInteger sum = cin.nextBigInteger();
Vector<node> a = new Vector<node>();
for (int i = 1; i <= n; i++) {
a.add(new node(cin.next(), cin.nextBigInteger()));
}
Collections.sort(a, new Comparator<node>() {
public int compare(node a, node b) {
return b.bid.compareTo(a.bid);
}
});
Vector<String> ans = new Vector<String>();
for (int i = 0; i < n; i++) {
if (a.get(i).bid.compareTo(sum) <= 0) {
sum = sum.subtract(a.get(i).bid);
ans.add(a.get(i).name);
}
}
if (sum.compareTo(BigInteger.valueOf(0)) != 0) ans.removeAllElements();
System.out.println(ans.size());
for (int i = 0; i < ans.size(); i++) {
System.out.println(ans.get(i));
}
}
}