我有一个十进制字符串,长度小于5000:
std::string decimalString = "555";
有没有一种标准的方法将这个字符串转换为二进制表示形式?就像这样:
std::string binaryString = "1000101011";
更新。 这篇文章 对我很有帮助。
因为10不是2的幂(或者说反过来),所以你需要在十进制中实现算术运算。你需要以下两个操作:
两者可以通过相同的算法计算。或者,你可以使用各种C++大整数库,如GNU MP或Boost.Multiprecision。
#include <iostream>
#include <bitset>
#include <limits>
#include <algorithm>
std::string BinaryAdd(std::string First, std::string Second)
{
int Carry = 0;
std::string Result;
while(Second.size() > First.size())
First.insert(0, "0");
while(First.size() > Second.size())
Second.insert(0, "0");
for (int I = First.size() - 1; I >= 0; --I)
{
int FirstBit = First[I] - 0x30;
int SecondBit = Second[I] - 0x30;
Result += static_cast<char>((FirstBit ^ SecondBit ^ Carry) + 0x30);
Carry = (FirstBit & SecondBit) | (SecondBit & Carry) | (FirstBit & Carry);
}
if (Carry)
Result += 0x31;
std::reverse(Result.begin(), Result.end());
return Result;
}
std::string BinaryMulDec(std::string value, int amount)
{
if (amount == 0)
{
for (auto &s : value)
{
s = 0x30;
}
return value;
}
std::string result = value;
for (int I = 0; I < amount - 1; ++I)
result = BinaryAdd(result, value);
return result;
}
std::int64_t CalcPowers(std::int64_t value)
{
std::int64_t t = 1;
while(t < value)
t *= 10;
return t;
}
std::string ToBinary(const std::string &value)
{
std::vector<std::string> sets;
std::vector<int> multipliers;
int Len = 0;
int Rem = value.size() % 4;
for (auto it = value.end(), jt = value.end(); it != value.begin() - 1; --it)
{
if (Len++ == 4)
{
std::string t = std::string(it, jt);
sets.push_back(std::bitset<16>(std::stoull(t)).to_string());
multipliers.push_back(CalcPowers(std::stoull(t)));
jt = it;
Len = 1;
}
}
if (Rem != 0 && Rem != value.size())
{
sets.push_back(std::bitset<16>(std::stoull(std::string(value.begin(), value.begin() + Rem))).to_string());
}
auto formula = [](std::string a, std::string b, int mul) -> std::string
{
return BinaryAdd(BinaryMulDec(a, mul), b);
};
std::reverse(sets.begin(), sets.end());
std::reverse(multipliers.begin(), multipliers.end());
std::string result = sets[0];
for (std::size_t i = 1; i < sets.size(); ++i)
{
result = formula(result, sets[i], multipliers[i]);
}
return result;
}
void ConcatenateDecimals(std::int64_t* arr, int size)
{
auto formula = [](std::int64_t a, std::int64_t b) -> std::int64_t
{
return (a * CalcPowers(b)) + b;
};
std::int64_t val = arr[0];
for (int i = 1; i < size; ++i)
{
val = formula(val, arr[i]);
}
std::cout<<val;
}
int main()
{
std::string decimal = "64497387062899840145";
//6449738706289984014 = 0101100110000010000100110010111001100010100000001000001000001110
/*
std::int64_t arr[] = {644, 9738, 7062, 8998, 4014};
ConcatenateDecimals(arr, 5);*/
std::cout<<ToBinary(decimal);
return 0;
}
在最坏情况下,执行时间约为0.039秒。由于使用10^9作为计数系统,乘法、加法和除法算法非常快,但是实现可以进行很好的优化。 源链接ai -> aj
2 <= i,j <= 36; 0 <= a <= 10^1000
时间限制:1秒
#include <iostream>
#include <string>
#include <vector>
using namespace std;
#define sz(x) (int((x).size()))
typedef vector<int> vi;
typedef long long llong;
int DigToNumber(char c) {
if( c <= '9' && c >= '0' )
return c-'0';
return c-'A'+10;
}
char NumberToDig(int n) {
if( n < 10 )
return '0'+n;
return n-10+'A';
}
const int base = 1000*1000*1000;
void mulint(vi& a, int b) { //a*= b
for(int i = 0, carry = 0; i < sz(a) || carry; i++) {
if( i == sz(a) )
a.push_back(0);
llong cur = carry + a[i] * 1LL * b;
a[i] = int(cur%base);
carry = int(cur/base);
}
while( sz(a) > 1 && a.back() == 0 )
a.pop_back();
}
int divint(vi& a, int d) { // carry = a%d; a /= d; return carry;
int carry = 0;
for(int i = sz(a)-1; i >= 0; i--) {
llong cur = a[i] + carry * 1LL * base;
a[i] = int(cur/d);
carry = int(cur%d);
}
while( sz(a) > 1 && a.back() == 0 )
a.pop_back();
return carry;
}
void add(vi& a, vi& b) { // a += b
for(int i = 0, c = 0, l = max(sz(a),sz(b)); i < l || c; i++) {
if( i == sz(a) )
a.push_back(0);
a[i] += ((i<sz(b))?b[i]:0) + c;
c = a[i] >= base;
if( c ) a[i] -= base;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int from, to; cin >> from >> to;
string s; cin >> s;
vi res(1,0); vi m(1,1); vi tmp;
for(int i = sz(s)-1; i >= 0; i--) {
tmp.assign(m.begin(), m.end());
mulint(tmp,DigToNumber(s[i]));
add(res,tmp); mulint(m,from);
}
vi ans;
while( sz(res) > 1 || res.back() != 0 )
ans.push_back(divint(res,to));
if( sz(ans) == 0 )
ans.push_back(0);
for(int i = sz(ans)-1; i >= 0; i--)
cout << NumberToDig(ans[i]);
cout << "\n";
return 0;
}
如何将字符串s从“from”转换为“to”:
PS我注意到线程发起者的昵称是Denis。 我认为这link也可能有用。
binaryString
和decimalString
之间有什么明显的对应关系吗?我看不出来。 - Paul R