[hihoCoder] 二进制小数 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET


[hihoCoder] 二进制小数 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET
给定一个十进制小数X,判断X的二进制表示是否是有限确定的。
例如0.5的二进制表示是0.1,0.75的二进制表示是0.11,0.3没有确定有限的二进制表示。

输入

第一行包含一个整数 T (1 ≤ T ≤ 10),表示测试数据的组数。
以下T行每行包含一个十进制小数 X (0 < X < 1)。 X一定是以"0."开头,小数部分不超过100位。

输出

对于每组输入,输出 X 的二进制表示或者NO(如果 X 没有确定有限的二进制表示)。

样例输入

3
0.5
0.75
0.3
样例输出
0.1
0.11
NO

思路: 十进制小数化为二进制小数的一般思路就是不断乘2取整. 可以化成二进制小数的十进制小数有个特点, 即其最后一位必然是5, 因为其是由1/2, 1/(2^2), 1/(2^4).... , 1/(2^n) 构成, 因此我们只需要在运算之前判断最后一位是否是5即可. 因为最高有一百位, 所以正常的double和float都远远不够, 所以我们要用字符串来模拟.

bool kkk = true; char mutiply(string& str) { int len = str.size(), flag = 0, size = 0, k = 0; for(int i = len-1; i >=0; i--) { int x = (str[i]-'0')*2 + flag; str[i] = x%10 +'0'; if(k == 0 && str[i] == '0') size++; if(str[i] != '0') k = 1; flag = x /10; } if(size == 0) { kkk = false; return ' '; } str.resize(len - size); return flag + '0'; } int main() { int T; cin>> T; while(T--) { kkk = true; string tem = "0.", str; cin >> str; str = str.substr(2); while(1) { tem += mutiply(str); if(kkk == false) { cout << "NO" << endl; break;} else if(str.size() == 0) {cout << tem << endl; break;} } } return 0; }
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