描述
Implement a function that adds two numbers together and returns their sum in binary. The conversion can be done before, or after the addition.
The binary number returned should be a string.
Examples:(Input1, Input2 --> Output (explanation)))
1, 1 --> "10" (1 + 1 = 2 in decimal or 10 in binary)
5, 9 --> "1110" (5 + 9 = 14 in decimal or 1110 in binary)
测试用例
#include <random>
#include <string>
#include <tuple>
#include <cstdint>
#include <sstream>
Describe(FixedTests)
{
It(should_pass_fixed_tests)
{
Assert::That(add_binary(1, 1), Equals("10"), ExtraMessage("Invalid result for a = 1, b = 1"));
Assert::That(add_binary(0, 1), Equals("1"), ExtraMessage("Invalid result for a = 0, b = 1"));
Assert::That(add_binary(1, 0), Equals("1"), ExtraMessage("Invalid result for a = 1, b = 0"));
Assert::That(add_binary(2, 2), Equals("100"), ExtraMessage("Invalid result for a = 2, b = 2"));
Assert::That(add_binary(51, 12), Equals("111111"), ExtraMessage("Invalid result for a = 51, b = 12"));
Assert::That(add_binary(5, 9), Equals("1110"), ExtraMessage("Invalid result for a = 5, b = 9"));
Assert::That(add_binary(10, 10), Equals("10100"), ExtraMessage("Invalid result for a = 10, b = 10"));
Assert::That(add_binary(100, 100), Equals("11001000"), ExtraMessage("Invalid result for a = 100, b = 100"));
Assert::That(add_binary(4096, 1), Equals("1000000000001"), ExtraMessage("Invalid result for a = 4096, b = 1"));
Assert::That(add_binary(0, 2174483647), Equals("10000001100110111111110010111111"), ExtraMessage("Invalid result for a = 0, b = 2174483647"));
}
It(should_handle_edge_case)
{
Assert::That(add_binary(0, 0), Equals("0"), ExtraMessage("Invalid result for a = 0, b = 0"));
}
};
Describe(RandomTests)
{
private:
std::mt19937 engine{ std::random_device{}() };
std::uniform_int_distribution<size_t> dist_l{ 0, 64 };
std::uniform_int_distribution<char> dist_b{ 0, 1 };
std::tuple<uint64_t, uint64_t, std::string> generate_problem()
{
const size_t len = dist_l(engine);
if(len == 0) return { 0, 0, "0" };
std::string res = "1";
uint64_t sum = 1;
for(size_t i = 1; i < len; ++i)
{
const bool d = dist_b(engine);
res += static_cast<char>(d + '0');
sum = sum << 1 | d;
}
const auto a = std::uniform_int_distribution<uint64_t>{ 0, sum }(engine);
return { a, sum - a, res };
}
public:
It(should_pass_random_tests)
{
for(int i = 0; i < 100; ++i)
{
const auto [a, b, res] = generate_problem();
const auto msg = [a = a, b = b]
{
std::ostringstream res;
res << "Incorrect result for a = " << a << ", b = " << b;
return res.str();
};
Assert::That(add_binary(a, b), Equals(res), msg);
}
}
};
解决方案：
format
#include <cstdint>
#include <string>
#include <fmt/core.h>
std::string add_binary(std::uint64_t a, std::uint64_t b) {
return fmt::format("{:b}", a + b);
}
fmt 是一个三方库
字符串拼接
#include <cstdint>
#include <string>
using namespace std;
string add_binary(uint64_t a, uint64_t b) {
a += b;
string output;
do {
output = to_string(a % 2) + output;
a /= 2;
} while(a > 0);
return output;
}
连续拼接字符串会产生大量对象，如果利用append 最后 reverce 会快一点。
bitset
#include <cstdint>
#include <string>
#include <bitset>
using namespace std;
std::string add_binary(uint64_t a, uint64_t b) {
string bi = std::bitset<64>((a + b)).to_string();
return bi.erase(0, min(bi.find_first_not_of('0'), bi.size()-1));
}
bitset库
C++的 bitset 在 bitset 头文件中，它是一种类似数组的结构，它的每一个元素只能是０或１，每个元素仅用１bit空间。
用字符串构造时，字符串只能包含 ‘0’ 或 ‘1’ ，否则会抛出异常。
构造时，需在<>中表明bitset 的大小(即size)。
在进行有参构造时，若参数的二进制表示比bitset的size小，则在前面用０补充；若比bitsize大，参数为整数时取后面部分，参数为字符串时取前面部分
构造函数：
bitset<4> bitset1; //无参构造，长度为４，默认每一位为０
bitset<8> bitset2(12); //长度为８，二进制保存，前面用０补充
string s = "100101";
bitset<10> bitset3(s); //长度为10，前面用０补充
char s2[] = "10101";
bitset<13> bitset4(s2); //长度为13，前面用０补充
方法：
bitset<8> foo ("10011011");
cout << foo.count() << endl; //5 （count函数用来求bitset中1的位数，foo中共有５个１
cout << foo.size() << endl; //8 （size函数用来求bitset的大小，一共有８位
cout << foo.test(0) << endl; //true （test函数用来查下标处的元素是０还是１，并返回false或true，此处foo[0]为１，返回true
cout << foo.test(2) << endl; //false （同理，foo[2]为０，返回false
cout << foo.any() << endl; //true （any函数检查bitset中是否有１
cout << foo.none() << endl; //false （none函数检查bitset中是否没有１
cout << foo.all() << endl; //false （all函数检查bitset中是全部为１
cout << foo.flip(2) << endl; //10011111 （flip函数传参数时，用于将参数位取反，本行代码将foo下标２处"反转"，即０变１，１变０
cout << foo.flip() << endl; //01100000 （flip函数不指定参数时，将bitset每一位全部取反
cout << foo.set() << endl; //11111111 （set函数不指定参数时，将bitset的每一位全部置为１
cout << foo.set(3,0) << endl; //11110111 （set函数指定两位参数时，将第一参数位的元素置为第二参数的值，本行对foo的操作相当于foo[3]=0
cout << foo.set(3) << endl; //11111111 （set函数只有一个参数时，将参数下标处置为１
cout << foo.reset(4) << endl; //11101111 （reset函数传一个参数时将参数下标处置为０
cout << foo.reset() << endl; //00000000 （reset函数不传参数时将bitset的每一位全部置为０
string s = foo.to_string(); //将bitset转换成string类型
unsigned long a = foo.to_ulong(); //将bitset转换成unsigned long类型
unsigned long long b = foo.to_ullong(); //将bitset转换成unsigned long long类型
