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