1:
ULONGLONG Count1InAInteger(ULONGLONG n)
{
ULONGLONG iNum = 0;
while (n != 0)
{
iNum += (n % 10 == 1) ? 1 : 0;
n /= 10;
}
return iNum;
}
ULONGLONG f(ULONGLONG n)
{
ULONGLONG iCount = 0;
for (ULONGLONG i = 1; i <= n; i++)
{
iCount += Count1InAInteger(i);
}
return iCount;
}
2:找规律:只需要分析N就可以得到f(N),避开了从1到N的遍历
LONGLONG Sum1s(ULONGLONG n)
{
ULONGLONG iCount = 0;
ULONGLONG iFactor = 1;
ULONGLONG iLowerNum = 0;
ULONGLONG iCurrNum = 0;
ULONGLONG iHigherNum = 0;
while (n / iFactor != 0)
{
iLowerNum = n - (n / iFactor) * iFactor;
iCurrNum = (n / iFactor) % 10;
iHigherNum = n / (iFactor * 10);
switch (iCurrNum)
{
case 0:
iCount += iHigherNum * iFactor;
break;
case 1:
iCount += iHigherNum * iFactor + iLowerNum + 1;
break;
default:
iCount += (iHigherNum + 1) * iFactor;
break;
}
iFactor *= 10;
}
return iCount;
}
原创:https://blog.csdn.net/ndzjx/article/details/84404471