hdu2054（坑爹题）

来源：互联网发布：网络尖兵和星空极速编辑：程序博客网时间：2024/06/14 12:19

A == B ?

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 100767 Accepted Submission(s): 16018

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 22 23 34 3

Sample Output

NOYESYESNO

Author

8600 && xhd

Source

校庆杯Warm Up

Recommend

linle | We have carefully selected several similar problems for you: 2072 2055 2057 2056 2058

大清早起来写到这道题，坑爹死了，考虑少了一种情况，字符串处理而已，暴力过··

注意坑点：

1.999999999999999999999999999999999999999999992（这组数据接收了为什么要有字符串来输入）1.3001.3（这组数据解释了为什么要在把字符串的结尾符'\0'改为'0'）1.00001（这组数据解释了为什么要在把'.'改为'0'）考虑给出的这三种情况就对了。对了，小心数组越界，要给字符串一个结尾符'\0'。

#include<stdio.h>#include<math.h>#include<string.h>char a[1000005],b[1000005];int main(){int i;while(scanf("%s%s",a,b)!=EOF){int flag=0;int len1=strlen(a);int len2=strlen(b);    for(i=0;i<len1;i++){    if(a[i]=='.') {    flag=1;break;}}if(flag){for(int i=len1-1;i>=0;i--){if(a[i]=='0') a[i]='\0';else break;len1--;}if(a[len1-1]=='.') a[len1-1]='\0';}flag=0;//bfor(i=0;i<len2;i++){    if(b[i]=='.'){    flag=1;break;} }if(flag){for(int i=len2-1;i>=0;i--){if(b[i]=='0') b[i]='\0';else break;len2--;}if(b[len2-1]=='.') b[len2-1]='\0';}if(strcmp(a,b)==0) printf("YES\n");else printf("NO\n");}}

对两个数组处理，最好写成一个函数，由于我懒就直接复制粘贴了。。。

0 0