A == B ?
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 60827 Accepted Submission(s): 9451
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
解题思路：
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。被坑了一次
2.关键在于小数点，去除前导0和多余的后置0。
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2054
1 #include<stdio.h> 2 #include<string.h> 3 char n[100000], m[100000]; 4 int main() 5 { 6 int i, j, len_n, len_m, flag, len_last, len_late,doc, q, p, count, start; 7 char ch; 8 while(1) 9 { 10 memset(n, 0, sizeof(n)); 11 memset(m, 0, sizeof(m)); 12 flag = i = j = doc = count = start = 0; 13 while(1) 14 { 15 if(scanf("%c", &ch) == EOF) return 0; 16 if(ch == ' ' && start == 1) break; 17 else if(ch != ' ' && ch != ' ' && flag == 0) 18 { 19 n[i++] = ch; 20 } 21 else if(ch == ' ' || ch == ' ') 22 { 23 flag = 1; 24 start = 1; 25 } 26 else m[j++] = ch; 27 28 } 29 len_n = strlen(n); 30 len_m = strlen(m); 31 if((n[0] == '-' && m[0] != '-') || (n[0] != '-' && m[0] == '-')) 32 { 33 printf("NO "); 34 continue; 35 } 36 for(i = n[0] != '-'? 0:1; i<len_n && n[i] - '0'== 0; i++); 37 for(j = m[0] != '-'? 0:1; j<len_m && m[j] - '0'== 0; j++); 38 len_n -= i; 39 len_m -= j; 40 len_last = len_n < len_m? len_n:len_m; 41 len_late = len_n + len_m - len_last; 42 { 43 for(; count < len_last; i++, j++, count++) 44 if(n[i] - '0' != m[j] - '0') break; 45 else if(n[i] == '.') doc = 1; 46 if(count >= len_late) 47 { 48 49 printf("YES "); 50 } 51 else 52 { 53 if(len_n > len_m) 54 { 55 if(doc != 1 && n[i] != '.') printf("NO "); 56 else 57 { 58 for(i = n[i] == '.'? i+1: i; i<strlen(n); i++) 59 if(n[i] - '0' != 0 ) break; 60 if(i>=len_n) printf("YES "); 61 else printf("NO "); 62 } 63 64 } 65 else if(len_n < len_m) 66 { 67 if(doc != 1 && m[j] != '.') printf("NO "); 68 else 69 { 70 for(j = m[j] == '.'? j+1: j; j<strlen(m); j++) 71 if(m[j] - '0' != 0 ) break; 72 if(j>=len_m) printf("YES "); 73 else printf("NO "); 74 } 75 } 76 else if(len_n == len_m) printf("NO "); 77 } 78 } 79 80 } 81 return 0; 82 }