1的数目-编程之美132

给定一个十进制数N,写下从1开始,到N的所有整数,然后数一下其中出现的所有'1'的个数
比如:
N=2,写下1,2,这样只出现了1个'1'
N=12,我们写下1,2,3,4,5,6,7,8,9,10,11,12,这样1的个数5
写一个函数f(N),返回1到N之间'1'出现的个数,比如f(12)=5;

以下是作者给的算法

ull Sum1s(ull n)
{
    ull cnt =0;
    ull factor=1;
    ull low = 0;
    ull high = 0;
    ull cur =0;
    while(n/factor != 0)
    {
        low = n -(n/factor)*factor;
        cur = (n/factor)%10;
        high = n/(factor*10);
        switch(cur)
        {
        case 0:
            {
                cnt += high * factor;
            }
            break;
        case 1:
            {
                cnt += high * factor + low +1;
            }
            break;
        default:
            cnt += (high+1)*factor;
        }
        factor*=10;
    }
    return cnt;
}

下面是我的算法,我的算法不单单是能求出1的个数,还可以求出其他数字的个数,比如说7,只需要让num=7即可

typedef unsigned long long ull;
ull SumNs(ull N,ull num =1)
{
    ull sum = 0;
    ull factor = 1;
    ull rest =0;
    while(N!=0)
    {
        ull r = N%10;
        N/=10;
        sum += factor * N;
        if(r>num)
        {
            sum +=factor;
        }
        else if(r==num)
        {
            sum+=rest+1;
        }
        rest += r*factor;
        factor*=10;
    }
    return sum;
}

以下是验证代码

int main()
{
 
    for(ull i =0;i<10000000;++i)
    {
        if(Sum1s(i)!=SumNs(i))
            cout<<Sum1s(i)<<"   "<< SumNs(i)<<endl;
    }
}