Problem 36</h2>
31 January 2003</div>
The decimal number, 585 = 10010010012</sub> (binary), is palindromic in both bases.
Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
(Please note that the palindromic number, in either base, may not include leading zeros.)</p>
</div>
</div>
Answer:</div></td>
872187</b></td>
</tr>
</tbody>
</table>
#include
using namespace std;
int digits[6];
int bins[20];
int binNum=0;
bool ispaldec(long i)
{
int j=0;
for (; i>0; i/=10)
digits[j++]=i%10;
for (i = 0; i < j/2; i++) {
if(digits[i]!=digits[j-i-1])
return false;
}
return true;
}
/*
* alternaltive way recursive
void ispaldectobin(long i)
{
binNum=0;
if(i<=1)
{
bins[binNum++]=i;
return;
}
ispaldectobin(i/2);
bins[binNum++]=i%2;
}
*/
//while loop
void ispaldectobin(long i)
{
binNum=0;
do {
bins[binNum++]=i%2;
i/=2;
}while(i);
}
bool ispalbin(int *bins)
{
int i=0;
for (i = 0; i < binNum/2; i++) {
if(bins[i]!=bins[binNum-i-1])
return false;
}
return true;
}
int main (int argc, char const* argv[]) {
int i=1,j=0,sum=0;
for (i = 1; i < 1000000; i++) {
ispaldectobin(i);
if(ispalbin(bins)&&ispaldec(i))
{
ispaldectobin(i);
sum+=i;
}
}
cout<
}
分析预测模型在机器学习中的应用之线性回归
回归能做任何事情 Continue reading
android中的线程与进程模型
Published on June 15, 2016
android内存优化之图片加载
Published on June 11, 2016
The decimal number, 585 = 10010010012</sub> (binary), is palindromic in both bases.
Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
(Please note that the palindromic number, in either base, may not include leading zeros.)</p>
</div>
</div>
|
Answer:</div></td>
872187</b></td> | </tr> </tbody> </table> #include using namespace std; int digits[6]; int bins[20]; int binNum=0; bool ispaldec(long i) { int j=0; for (; i>0; i/=10) digits[j++]=i%10; for (i = 0; i < j/2; i++) { if(digits[i]!=digits[j-i-1]) return false; } return true; } /* * alternaltive way recursive void ispaldectobin(long i) { binNum=0; if(i<=1) { bins[binNum++]=i; return; } ispaldectobin(i/2); bins[binNum++]=i%2; } */ //while loop void ispaldectobin(long i) { binNum=0; do { bins[binNum++]=i%2; i/=2; }while(i); } bool ispalbin(int *bins) { int i=0; for (i = 0; i < binNum/2; i++) { if(bins[i]!=bins[binNum-i-1]) return false; } return true; } int main (int argc, char const* argv[]) { int i=1,j=0,sum=0; for (i = 1; i < 1000000; i++) { ispaldectobin(i); if(ispalbin(bins)&&ispaldec(i)) { ispaldectobin(i); sum+=i; } } cout< } 分析预测模型在机器学习中的应用之线性回归回归能做任何事情 Continue reading android中的线程与进程模型Published on June 15, 2016android内存优化之图片加载Published on June 11, 2016 |