我决定使用字符串来实现加法,因为我的数字可能很大。我面临的问题是如何将字符串中的十进制数字转换为二进制。如果lol
是加法的结果,我正在尝试使用以下方法将其转换为二进制形式:
unsigned long long function_arg = atoi( lol.c_str() );
随着位数的增加,例如n>14,它失败了,所以甚至long long
太短了。你能建议我做什么?这实际上是我的程序的代码(使用十进制系统):
#include <iostream>
#include <string>
#include <bitset>
#include <conio.h>
using namespace std;
string add (string &s1, string &s2) {
int carry=0,sum,i;
string min=s1,
max=s2,
result = "";
if (s1.length() > s2.length()) {
max = s1;
min = s2;
} else {
max = s2;
min = s1;
}
for (i = min.length()-1; i>=0; i--) {
sum = min[i] + max[i + max.length() - min.length()] + carry - 2*'0';
carry = sum/10;
sum %=10;
result = (char)(sum + '0') + result;
}
i = max.length() - min.length()-1;
while (i>=0) {
sum = max[i] + carry - '0';
carry = sum/10;
sum%=10;
result = (char)(sum + '0') + result;
i--;
}
if (carry!=0) {
result = (char)(carry + '0') + result;
}
return result;
}
string Dex_To_bin(unsigned long long number){
string result;
unsigned long long bitmask = 1ULL << 63;
do
result.push_back(static_cast<bool>(number & bitmask));
while (bitmask >>= 1);
return result;
}
int main () {
string a,b;
cin >> a >> b;
add(a,b);
cout << add(a,b) << endl;
cout << endl;
cout << endl;
string lol=add(a,b);
unsigned long long funtion_arg = atoi( lol.c_str() );
cout << Dex_To_bin(funtion_arg) << bitset<64>(funtion_arg) << endl;
cin.get();
cin.ignore();
getch();
return 0;
}
功能add
很好用,问题出在Dec_to_Bin
:(