A == B ?

超、凢脫俗 2022-08-05 00:48 162阅读 0赞

# A == B ? #

**Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 73036 Accepted Submission(s): 11508**

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

Author

8600 && xhd

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

#include<stdio.h>
    #include<string.h>
    #define M 500000
    int main()
    {
        void deletes(char *a);
        int i,len;char a[M],b[M];
        while(scanf("%s%s",a,b)!=EOF)
        {
            len=strlen(a);
            for(i=0;i<len;i++)
            {
                if(a[i]=='.')
                    deletes(a);
            }
            len=strlen(b);
            for(i=0;i<len;i++)
            {
                if(b[i]=='.')
                    deletes(b);
            }
            if(strcmp(a,b))
                printf("NO\n");
            else
                printf("YES\n");
        }
        return 0;
    }
    void deletes(char *a)
    {
        int len,i;
        len=strlen(a)-1;
        for(i=len;i>=0;i--)
        {
            if(a[i]=='0')     
                len--;                           
            else
                break;
        }
        if(a[len]=='.')   
            len--;
        a[len+1]='\0';    
    }