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hdu2054 A == B ?【大数】

发布时间:2020-12-14 03:56:37 所属栏目:大数据 来源:网络整理
导读:A == B ? Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51920 Accepted Submission(s): 8014 Problem Description Give you two numbers A and B,if A is equal to B,you should print "YES",or

A == B ?

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51920 Accepted Submission(s): 8014


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,or print "NO".

Sample Input
  
  
1 2 2 2 3 3 4 3

Sample Output
  
  
NO YES YES NO
<pre name="code" class="java">import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.MathContext;
import java.util.Scanner;
public class Main{
    public static void main(String args[]){
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext()){
            BigDecimal a = cin.nextBigDecimal();
            BigDecimal b = cin.nextBigDecimal();
            if(a.equals(BigDecimal.valueOf(0.0)))
                a = BigDecimal.ZERO;
            if(b.equals(BigDecimal.valueOf(0.0)))
                b = BigDecimal.ZERO;
            if(a.stripTrailingZeros().toPlainString().equals(b.stripTrailingZeros().toPlainString()))
                System.out.println("YES");
            else {
                System.out.println("NO");
            }
        }
        cin.close();
    }
}
#include<stdio.h>
#include<string.h>
char *removepoint(char a[])//不用考虑前导0...这让我很困惑啊..Dev结果都不对还能AC
{
	int len = strlen(a);
	//strchr的功能 : 
	//返回首次出现c的位置的指针,返回的地址是字符串在内存中
	//随机分配的地址再加上你所搜索的字符在字符串位置,如果s中不存在c则返回NULL 
	if(strchr(a,'.')!=NULL)
	{
		while(a[--len] == '0');
		if(a[len]=='.') len--;
		a[len+1] = '';
	}
	return a;
}
int main()
{
	char a[15000],b[15000];
	while(~scanf("%s%s",a,b))
	{
		if(strcmp(removepoint(a),removepoint(b))==0) printf("YESn");
		else printf("NOn");
	}
	return 0;
}

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