ACM HDU 2054 A == B ？

A == B ?

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40699    Accepted Submission(s): 6340

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

 

Source

校庆杯Warm Up

 

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linle

 

#include<stdio.h>
#include<string.h>
char n[100000], m[100000];
int main()
{
    int i, j, len_n, len_m, flag, len_last, len_late,doc, q, p, count, start;
    char ch;
    while(1)
    {
        memset(n, 0, sizeof(n));
        memset(m, 0, sizeof(m));    
        flag = i = j = doc = count = start = 0;
        while(1)
        {
            if(scanf("%c", &ch) == EOF) return 0;
            if(ch == '\n' && start == 1) break;
            else if(ch != ' ' && ch != '\n' && flag == 0)
            {
                n[i++] = ch;    
            }
            else if(ch == ' ' || ch == '\n')
            {
                flag = 1;
                start = 1;
            }
            else m[j++] = ch;
            
        }
        len_n = strlen(n);
        len_m = strlen(m);
        if((n[0] == '-' && m[0] != '-') || (n[0] != '-' && m[0] == '-')) 
        {
            printf("NO\n");
            continue;
        } 
        for(i = n[0] != '-'? 0:1; i<len_n && n[i] - '0'== 0; i++);
        for(j = m[0] != '-'? 0:1; j<len_m && m[j] - '0'== 0; j++);
        len_n -= i;
        len_m -= j;
        len_last = len_n < len_m? len_n:len_m;
        len_late = len_n + len_m - len_last;
        {
            for(; count < len_last; i++, j++, count++)
            if(n[i] - '0' != m[j] - '0') break;
            else if(n[i] == '.') doc = 1;
            if(count >= len_late)
            {
                
                printf("YES\n");
            }
            else
            {
             if(len_n > len_m) 
             {
                 if(doc != 1 && n[i] != '.') printf("NO\n");
                 else
                 {
                     for(i = n[i] == '.'? i+1: i; i<strlen(n); i++)
                     if(n[i] - '0' != 0 ) break;
                     if(i>=len_n) printf("YES\n");
                     else printf("NO\n");
                 }

             }
             else if(len_n < len_m)
             {
                if(doc != 1 && m[j] != '.') printf("NO\n");
                 else
                 {
                     for(j = m[j] == '.'? j+1: j; j<strlen(m); j++)
                     if(m[j] - '0' != 0 ) break;
                     if(j>=len_m) printf("YES\n");
                     else printf("NO\n");
                 }
             }
             else if(len_n == len_m) printf("NO\n");
            }
        }
        
    }
    return 0;    
}

 

解题报告：

1. 一看题目就觉得恶心，题目条件越简单，ＡＣ的条件就越苛刻，数组未开足，  Runtime Error 了一次。

2. 由于我的无知，未考虑小数点的情况，WA = 1 ，其实是有想到小数点的情况，但侥幸的心理对于我这样的菜菜鸟来说是拉不出来的

3.开始考虑小数点，WA + 1；

4. 开始考虑小数点后面的0，WA+1；

5. 考虑数字前面的0, WA + 1；

6. 输入出现了问题，处理换行符

总的来说，能再坚持一点点就坚持下去，给你　！ＡＣ　就是说明你程序出现了问题，当然有时会遇到让你趾高气扬的时候，就是题目和AC的代码有点难以恰和的时候，但更多的情况是自己的代码真的出问题，所以你得谦虚！