HDU 2054 A == B A但是看不太明白,Time Limit:1000/1000 MS (Java/Others) Memory Limit:32768/32768 K (Java/Others)Total Submission(s):38421 Accepted Submission(s):5912#include #include #include char str1[100000],str2[100000];void deal(char *str){

HDU 2054 A == B A但是看不太明白,Time Limit:1000/1000 MS (Java/Others) Memory Limit:32768/32768 K (Java/Others)Total Submission(s):38421 Accepted Submission(s):5912#include #include #include char str1[100000],str2[100000];void deal(char *str){
HDU 2054 A == B A但是看不太明白,
Time Limit:1000/1000 MS (Java/Others) Memory Limit:32768/32768 K (Java/Others)
Total Submission(s):38421 Accepted Submission(s):5912
#include
#include
#include
char str1[100000],str2[100000];
void deal(char *str)
{
int i,j,len;
char *p = str;
while( *p == '0')
p++;
strcpy(str,p);
if(strchr(str,'.'))
{
len = strlen(str);
p = len + str - 1;
while( *p == '0')
*(p--) = 0;
if (*p == '.')
*p = 0;
}
}
int main( )
{
while(scanf("%s%s",str1,str2)!= EOF)
{
deal(str1);
deal(str2);
if (strcmp(str1,str2) == 0)
puts("YES");
else
puts("NO");
}
return 0;
}
Give you two numbers A and B,if A is equal to B,you should print "YES",or print "NO".
Inputeach test case contains two numbers A and B.
Outputfor each case,if A is equal to B,you should print "YES",or print "NO".
Sample Input1 2
2 2
3 3
4 3
Sample OutputNO
YES
YES
NO

HDU 2054 A == B A但是看不太明白,Time Limit:1000/1000 MS (Java/Others) Memory Limit:32768/32768 K (Java/Others)Total Submission(s):38421 Accepted Submission(s):5912#include #include #include char str1[100000],str2[100000];void deal(char *str){
原理是去掉末尾多余的0和小数点（如果能去掉的话）还有多余的前导0
最后比较2字串