A == B ?

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 96648    Accepted Submission(s): 15372 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  

做本题是要判断是否存在前导0存在以及小数点后是否最后几位是0；

#include <stdio.h> #include <string.h> #include <iostream> using namespace std; char a[1000005],b[1000005]; int Judge(char *a, int &x){ int len = strlen(a),i = 0,flag = 0; while(a[i + 1] != '.' && a[i] == '0' && i < len) i ++; x = i; for(i = 0; i < len; i ++) if(a[i] == '.') flag = 1; i = len - 1; while(i >= 0 && a[i] == '0' && flag && a[i] != '.'){ a[i] = '\0'; i --; } if(a[i] == '.') a[i] = '\0'; } int main(){ while(cin >>a >>b){ int x,y,flag = 0; Judge(a,x); Judge(b,y); if(strlen(a) - x != strlen(b) - y){ printf("NO\n"); continue; } else{ while(x < strlen(a)){ if(a[x] != b[y]){ flag = 1; break; } x ++; y ++; } if(flag) printf("NO\n"); else printf("YES\n"); } } return 0; }