题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2054
1 2 2 2 3 3 4 3
NO YES YES NO
思路:
此题关键是找小数点,找到小数点把最后面无效的零去掉再比较就OK;
代码如下:
#include <cstdio>
#include <cstring>
char a[100017], b[100017];
void re(char s[])
{
int len = strlen(s);
int p = 0;
for(int i = 0; i < len; i++)
{
if(s[i] == '.')
{
p = 1;
break;
}
}
if(p)
{
for(int i = len-1; i >= 0; i--)
{
if(s[i] == '0')
s[i] = '\0';
else
break;
len--;
}
if(s[len-1] == '.')
s[len-1] = '\0';
}
}
int main()
{
while(~scanf("%s%s",a,b))
{
re(a);
// printf("%s\n",a);
re(b);
// printf("%s\n",b);
if(strcmp(a,b))
printf("NO\n");
else
printf("YES\n");
}
return 0;
}
原文地址:http://blog.csdn.net/u012860063/article/details/38975841