SDNUOJ 1301.判断相等

给定两个数A和B（长度不超过100），判断A和B是否相等。

有几个特殊样例需要过：

-0 = 0

-0001.10 = -1.1

01.0020 = 1.0020000000

做法：

符号单独看。以小数点为分界线，处理左边和右边，去除前导或者后导零。

#include
typedef long long ll;
using namespace std;
 
string a, b;
bool solve1()
{
	int pos = 1000, pos2 = 1000;
	std::vector<int> A, B;
	for (int i = 0; i < a.size(); i ++ ) {
		if(a[i] == '.'){pos = i; break;}
		A.push_back(a[i] - '0');
	}
	reverse(A.begin(), A.end());
	while(A.size() > 1 && A.back() == 0) A.pop_back();
 
	for (int i = 0; i < b.size(); i ++ ) {
		if(b[i] == '.'){pos2 = i; break;}
		B.push_back(b[i] - '0');
	}
	reverse(B.begin(), B.end());
	while(B.size() > 1 && B.back() == 0) B.pop_back();
	if(A.size() != B.size()) return false;
	for (int i = 0; i < A.size(); i ++ ) {
		if(A[i] != B[i]) return false;
	}
	A.clear();
	B.clear();
	for (int i = pos + 1; i < a.size(); i ++ ) {
		A.push_back(a[i] - '0');
	}
	for (int i = pos2 + 1; i < b.size(); i ++ ) {
		B.push_back(b[i] - '0');
	}
	while(A.size() > 0 && A.back() == 0) A.pop_back();
	while(B.size() > 0 && B.back() == 0) B.pop_back();
	for (int i = 0; i < A.size(); i ++ ) {
		if(A[i] != B[i]) return false;
	}
	return true;
}
 
bool solve2()
{
	int pos = 1000, pos2 = 1000;
	std::vector<int> A, B;
	for (int i = 1; i < a.size(); i ++ ) {
		if(a[i] == '.'){pos = i; break;}
		A.push_back(a[i] - '0');
	}
	reverse(A.begin(), A.end());
	while(A.size() > 1 && A.back() == 0) A.pop_back();
 
	for (int i = 1; i < b.size(); i ++ ) {
		if(b[i] == '.'){pos2 = i; break;}
		B.push_back(b[i] - '0');
	}
	reverse(B.begin(), B.end());
	while(B.size() > 1 && B.back() == 0) B.pop_back();
	if(A.size() != B.size()) return false;
	for (int i = 0; i < A.size(); i ++ ) {
		if(A[i] != B[i]) return false;
	}
	A.clear();
	B.clear();
	for (int i = pos + 1; i < a.size(); i ++ ) {
		A.push_back(a[i] - '0');
	}
	for (int i = pos2 + 1; i < b.size(); i ++ ) {
		B.push_back(b[i] - '0');
	}
	while(A.size() > 0 && A.back() == 0) A.pop_back();
	while(B.size() > 0 && B.back() == 0) B.pop_back();
	for (int i = 0; i < A.size(); i ++ ) {
		if(A[i] != B[i]) return false;
	}
	return true;
}
 
 
int main(){
	//std::ios::sync_with_stdio(false);
    //std::cin.tie(nullptr);
 	while(cin >> a >> b)
 	{
 		int cnt = 0;
 		if(a[0] == '-') cnt ++;
 		if(b[0] == '-') cnt ++;
 		if(cnt == 1) {
 			bool f = 0;
 			for (int i = 1; i < a.size(); i ++ ) {
 				if(a[i] != '0' && a[i] != '.') f = 1;
 			}
 			for (int i = 1; i < b.size(); i ++ ) {
 				if(b[i] != '0' && b[i] != '.') f = 1;
 			}
 			cout << (f == 1 ? "no\n" : "yes\n");
 		}
 		else {
 			if(cnt == 2) 
 				if(solve2())cout << "yes\n"; 
 				else cout << "no\n";
 			else if(solve1())cout << "yes\n";
 				else cout << "no\n";
 		}
 	}    
	return 0;
}