https://leetcode.com/problems/fraction-to-recurring-decimal/description/
模拟，想下啥时间会循环
class Solution {
public:
string fractionToDecimal(int num, int de) {
if (num == 0) return "0";
string result;
if ( (num <0 ) ^ (de < 0) ) {
result = result + '-';
}
long long a = abs((long long)num), b = abs((long long)de);
result = result + to_string(a/b);
a = a % b;
if (!a) return result;
result = result + '.';
map< pair<long long, long long>, int > pos;
while (a) {
cout << "a:" << a ;
long long old = a;
a *= 10;
long long r = a % b;
pair<long long, long long> tmp = make_pair(old, r);
cout << " r:" << r << endl;
if (pos.find(tmp) != pos.end()) {
result.insert( pos[tmp], 1, '(' );
result = result + ')';
break;
}
result = result + to_string( a/b );
a %= b;
pos[ tmp ] = result.size() - 1;
}
return result;
}
};https://blog.csdn.net/u012162613/article/details/41998617 这里的代码更简洁
class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
if(numerator==0) return "0";
string result;
if(numerator<0 ^ denominator<0 ) result+='-'; //异或，numerator<0和denominator<0仅有一个为真
//转化为正数，INT_MIN转化为正数会溢出，故用long long。long long int n=abs(INT_MIN)得到的n仍然是负的，所以写成下面的形式。
long long int n=numerator,d=denominator;
n=abs(n);d=abs(d);
result+=to_string(n/d); //整数部分
long long int r=n%d; //余数r
if(r==0) return result;
else result+='.';
//下面处理小数部分，用哈希表
unordered_map<int,int> map;
while(r){
//检查余数r是否在哈希表中，是的话则开始循环了
if(map.find(r)!=map.end()){
result.insert(map[r],1,'('); //http://www.cplusplus.com/reference/string/basic_string/insert/
result+=')';
break;
}
map[r]=result.size(); //这个余数对应于result的哪个位置
//正常运算
r*=10;
result+=to_string(r/d);
r=r%d;
}
return result;
}
};
有一个case需要测试的 1/6