PAT 1060. Are They Equal

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一直没做出来，

参照点击打开链接的题解，重做了一遍才对。

先去除没用的先导0, 然后分 [0]+.[#]* 和 [0]*[#]+.[#]*这两种情况讨论。

注意到，在第一种情况里，最后的

if (find_nonzero == false) // 不能省, 否则0.0会被记为*10^-1{exponent = 0;}

是不能略去的，否则不能测试点6 —— 考察0.0的输出是否为0.0 * 10 ^ 0

代码：

#include <string>#include <iostream>#include <sstream>using namespace std;int n;string s1, s2, res1, res2;string normlize(string& s1){string ret;int bits = 0, exponent = 0;size_t i;for (i = 0; i < s1.size() && s1[i] == '0'; ++ i) {}if (i < s1.size() && s1[i] == '.') // 0*.#*{bool find_nonzero = false;for (; i < s1.size(); ++ i){if (s1[i] == '.'){exponent = 1; } else if (s1[i] != '0'){find_nonzero = true;}if (find_nonzero == true){if (bits < n){ret.push_back(s1[i]);++ bits;}} else{-- exponent;}}if (find_nonzero == false) // 不能省, 否则0.0会被记为*10^-1{exponent = 0;}} else  if (i < s1.size())     // 0*#*.#*{bool find_dot = false;for (; i < s1.size(); ++ i){if (find_dot == false && s1[i] != '.'){++ exponent;if (bits < n){ret.push_back( s1[i] );++ bits;}} else if (s1[i] == '.'){find_dot = true;continue;} else if (bits < n){ret.push_back( s1[i] );++ bits;}}}while (bits < n){ret.push_back('0');++ bits;}ret = string("0.") + ret;ret += "*10^";stringstream ss;ss << exponent;string tmp;ss >> tmp;return ret + tmp;}int main(){cin >> n >> s1 >> s2;res1 = normlize(s1);res2 = normlize(s2);if (res1 == res2){cout << "YES " << res2 << endl;} else{cout << "NO " << res1 << " " << res2 << endl;}system("pause");return 0;}

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