Title description
Input a decimal number N, convert it into R-based number and output.
Input
input data contains multiple test cases, each test case contains two integers N (32-bit integer) and R (2<=R<=16, R<>10).
Output
outputs the converted number for each test instance, and each output occupies one line. If R is greater than 10, the corresponding number rule refers to hexadecimal (for example, 10 is represented by A, etc.).
Sample Input
7 2
23 12
-4 3
Sample Output
111
1B
-11
Ideas
Convert the decimal number N into an R number and output. Here I wrote a hexadecimal conversion function. Of course, there are many ways to solve this problem, and I don’t have to use my method.
C++ code 1 (decimal conversion function template):
#include<bits/stdc++.h>
using namespace std;
char num[100];
char s[36] = {
'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
int n, m, step;
void solve(int a, int b) //十进制数a转化为b进制数
{
step = 0;
if(a < 0)
{
cout << "-";
a = -a;
}
while(a)
{
num[step++] = s[a % b];
a /= b;
}
for(step -= 1; step >= 0; step--)
cout << num[step];
cout << endl;
}
int main ()
{
while(cin >> n >> m) //十进制数n转化为m进制数
solve(n, m);
return 0;
}
Of course you can also write like this:
#include<bits/stdc++.h>
using namespace std;
char num[10005];
int main()
{
ios::sync_with_stdio(false);
long long int n, r;
while(cin >> n >> r)
{
memset(num, 0, sizeof(num));
int i = 0, m, flag = 0;
if(n < 0)
{
flag = 1;
n = -n;
}
while(n >= r)
{
m = n % r;
if(m < 10) num[i++] = m + 48;
else if(m == 10) num[i++] = 'A';
else if(m == 11) num[i++] = 'B';
else if(m == 12) num[i++] = 'C';
else if(m == 13) num[i++] = 'D';
else if(m == 14) num[i++] = 'E';
else if(m == 15) num[i++] = 'F';
n /= r;
}
if(0 < n && n < 10) num[i++] = n + 48;
else if(n == 10) num[i++] = 'A';
else if(n == 11) num[i++] = 'B';
else if(n == 12) num[i++] = 'C';
else if(n == 13) num[i++] = 'D';
else if(n == 14) num[i++] = 'E';
else if(n == 15) num[i++] = 'F';
i--;
if(flag) cout << "-";
for(; i >= 0; i--)
cout << num[i];
cout << endl;
}
return 0;
}
It can also be written like this:
#include<bits/stdc++.h>
using namespace std;
int main()
{
char s1[] = "0123456789ABCDEF", s2[100]; //最大为十六进制，所以只需到F就可以了
int t, r;
while(cin >> t >> r)
{
int flag = 0,x;
if(t < 0)
{
t = -t; //判断是否是负数
flag = 1;//标记
}
for(int i = 0; ; i++)
{
if(t == 0) break;
s2[i] = s1[t % r]; //求余数并赋值s2，一直循环，直到t=0
t /= r;
x = i; //记录一下字符串的长度
}
if(flag == 1) cout << '-';
for(int i = x; i >= 0; i--)
cout << s2[i]; //逆序输出
cout << "\n";
}
return 0;
}