NYOJ：题目524 A-B Problem

题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=860

My思路：

先用两个字符串储存这两个实数，然后再用另外两个字符串储存去掉符号和前后多余的0后的新"实数"，最后只需要比较两个化简后的新字符就ok了。

My代码实现：

 1 #include<iostream>
 2 using namespace std;
 3 string simplify(string s) {  //去字符串s的正负号和首尾多余的0
 4   string a;
 5   int k = s.size()-1, l = 0, t = 0;
 6   for(int i = 0; i < s.size(); i++)  //判断是否是小数,如果是小数就需要去除尾部多余的0
 7     if(s[i] == ‘.‘) {
 8       t = 1;
 9       break;
10     }
11   if(s[0] == ‘+‘ || s[0] == ‘-‘) l++;  //去除符号
12   while(s[l] == ‘0‘) l++;  //去除前面多余的0
13   //cout << l << endl;
14   if(t) {  //去除后面多余的0
15     while(s[k] == ‘0‘) k--;
16   }
17   if(s[k] == ‘.‘) k--;
18   //cout << k << endl;
19   while(l <= k) a += s[l++];
20   return a;
21 }
22 int main() {
23   //string x = "-002540000.0000025000000";
24   //cout << simplify(x);
25   string A, B, a, b;
26   char fa, fb;
27   while(cin >> A >> B) {
28     a = simplify(A);  //去除符号和首尾多余的0
29     b = simplify(B);
30     //cout << a << "\n" << b << endl;
31     if(A[0] != ‘+‘ && A[0] != ‘-‘) fa = ‘+‘;  //正负号拿出来单独判断
32     else fa = A[0];
33     if(B[0] != ‘+‘ && B[0] != ‘-‘) fb = ‘+‘;
34     else fb = B[0];
35     if(((fa == ‘+‘ && fb == ‘-‘) || (fa == ‘-‘ && fb == ‘+‘)) && (a != "" && b != "")) cout << "NO\n";  //这里千万要考虑+0=-0的情况
36     else {
37       int i, p = 1;
38       for(i = 0; i < a.size() && i < b.size(); i++)
39         if(a[i] != b[i]) {
40           cout << "NO\n";
41           p = 0;
42           break;
43         }
44       if(i == b.size() && i == a.size()) cout << "YES\n";
45       else if(p) cout << "NO\n";
46     }
47   }
48 }

代码实现(点击展开)

      开始写于：2016.7.31  ----志银