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, 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
<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] = '\0';
}
return a;
}
int main()
{
char a[15000],b[15000];
while(~scanf("%s%s",a,b))
{
if(strcmp(removepoint(a),removepoint(b))==0) printf("YES\n");
else printf("NO\n");
}
return 0;
}
