HDU 2054 A == B ?

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2054

A == B ?

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 66431 Accepted Submission(s): 10353



Problem Description
Give you two numbers A and B, if A is equal to B, you should print "YES", or print "NO".


Input
each test case contains two numbers A and B.



Output
for each case, if A is equal to B, you should print "YES", or print "NO".


Sample Input

1 2
2 2
3 3
4 3

Sample Output

NO
YES
YES
NO

就是考虑高精度小数喽，记得处理负号和小数点，考察你的处理细节的能力。

比较简单，我直接上J***A的代码

import java.util.*;
import java.math.*;
import java.io.*;
public class Main {
	public static void main(String args[]){
		Scanner cin=new Scanner(new BufferedInputStream(System.in));
		BigDecimal a,b;
		int n;
		while(cin.hasNext()){
			a=cin.nextBigDecimal();b=cin.nextBigDecimal();
			String sa=a.stripTrailingZeros().toPlainString();
			String sb=b.stripTrailingZeros().toPlainString();
			if(sa.compareTo(sb)==0)  System.out.println("YES");
			else System.out.println("NO");
		}
	}
}

C++版，最后还是发现C++的string好用。

1.考虑正负号

2.去除前导零

3.去除后道零

4.一位位比对

算法结束

#include<iostream>
using namespace std;
int main()
{
    string a,b;
    while(cin>>a>>b)
    {
        bool fa=1,fb=1;
        if(a[0]!='-') fa=0;//判断a是否非负
        if(b[0]!='-') fb=0;
        if(fa!=fb)
        {
            cout<<"NO"<<endl;
            continue;
        }
        if(a[0]=='+'||a[0]=='-') a=a.substr(1);//如果有符号则去掉符号
        if(b[0]=='+'||b[0]=='-') b=b.substr(1);
        int la=0,lb=0,ra=a.size(),rb=b.size(),pa=1<<30,pb=1<<30;
        for(int i=0;i<ra;i++)//找出a中'.'的位置
            if(a[i]=='.')
            {
                pa=i;
                break;
            }
        while(a[la]=='0'&&la+1<pa&&la+1<ra) la++;
        while((a[--ra]=='0'||a[ra]=='.')&&ra>=pa) ;

        for(int i=0;i<rb;i++)
            if(b[i]=='.')
            {
                pb=i;
                break;
            }
        while(b[lb]=='0'&&lb+1<pb&&lb+1<rb) lb++;
        while((b[--rb]=='0'||b[rb]=='.')&&rb>=pb) ;

        a=a.substr(la,ra-la+1);//截取有效部分
        b=b.substr(lb,rb-lb+1);

        if(a!=b) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;;
    }
    return 0;
}