编程之美--2.4 1的数目之扩展问题

思路与书中2.4的类似，只是把十进制换成了二进制，只需在iFactor*10的地方，改成iFactor*2即可。

代码如下：

#include <iostream>#include<bitset>using namespace std;const unsigned SYSTEM = 2;unsigned long Sum1s(unsigned long n){unsigned long iCount = 0;unsigned long iFactor = 1;unsigned long iLowerNum = 0;unsigned long iCurrNum = 0;unsigned long iHigherNum = 0;while (n / iFactor != 0){iLowerNum = n - (n / iFactor) * iFactor;iCurrNum = (n / iFactor) % SYSTEM;iHigherNum = n / (iFactor * SYSTEM);switch (iCurrNum){case 0:iCount += iHigherNum * iFactor;break;case 1:iCount += iHigherNum * iFactor + iLowerNum + 1;break;default:iCount += (iHigherNum + 1) * iFactor;break;}iFactor *= SYSTEM;}return iCount;}int main(){unsigned long n = 0x4;cout << "binary number: " << bitset<8>(n) << endl;cout << "number of 1: " << Sum1s(n) << endl;system("pause");return 0;}