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

NYOJ:题目524 A-B Problem-LMLPHP

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:

 #include<iostream>
using namespace std;
string simplify(string s) { //去字符串s的正负号和首尾多余的0
string a;
int k = s.size()-, l = , t = ;
for(int i = ; 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
if(s[i] == '.') {
t = ;
break;
}
if(s[] == '+' || s[] == '-') l++; //去除符号
while(s[l] == '') l++; //去除前面多余的0
//cout << l << endl;
if(t) { //去除后面多余的0
while(s[k] == '') k--;
}
if(s[k] == '.') k--;
//cout << k << endl;
while(l <= k) a += s[l++];
return a;
}
int main() {
//string x = "-002540000.0000025000000";
//cout << simplify(x);
string A, B, a, b;
char fa, fb;
while(cin >> A >> B) {
a = simplify(A); //去除符号和首尾多余的0
b = simplify(B);
//cout << a << "\n" << b << endl;
if(A[] != '+' && A[] != '-') fa = '+'; //正负号拿出来单独判断
else fa = A[];
if(B[] != '+' && B[] != '-') fb = '+';
else fb = B[];
if(((fa == '+' && fb == '-') || (fa == '-' && fb == '+')) && (a != "" && b != "")) cout << "NO\n"; //这里千万要考虑+0=-0的情况
else {
int i, p = ;
for(i = ; i < a.size() && i < b.size(); i++)
if(a[i] != b[i]) {
cout << "NO\n";
p = ;
break;
}
if(i == b.size() && i == a.size()) cout << "YES\n";
else if(p) cout << "NO\n";
}
}
}

代码实现(点击展开)

      开始写于:2016.7.31  ----志银

05-11 11:04
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