Stern-Brocot Tree

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Problem Description 　　

vcr9wdBGoaM8YnI+CqGhoaHP1tTax+vE+rHgs8y8xsvjtdpu0NC1xMr9wdBGtcS49sr9oaM8YnI+CgoKIAo8YnI+CgpJbnB1dAoKoaGhocrkyOuw/LqstuDX6bLiytTTw8D9o6zDv9fpyuTI68r9vt3Kx9K7uPbV/dX7yv1uo6huPD0xMDAwMDAwo6mhowoKIAo8YnI+CgpPdXRwdXQKCqGhoaG21NPaw7/X6bXEsuLK1Mr9vt1uo6zH68rks/a12m7Q0LXEyv3B0Ea1xLj2yv2howoKIAo8YnI+CgpTYW1wbGUgSW5wdXQKCjxwcmUgY2xhc3M9"brush:java;">1 2 4 6

Sample Output

仔细看图可以发现：对于每一行都可以看成是关于1/1对称的两部分，所以只需求出1/1左边的个数就可求出这一行的个数。而左边全部都是真分数，分母为x的真分数的个数就是x的欧拉函数值。n最大为1000000，所以可以递推打表。

#include
   
    
const int N = 1000001;
int e[N];
__int64 a[N], res = 0;
void euler()
{
    for(int i = 2; i < N; i++)
        e[i] = 0;
    e[1] = 1;
    for(int i = 2; i < N; i++)
        if(!e[i])
        {
            for(int j = i; j < N; j += i)
            {
                if(!e[j])
                    e[j] = j;
                e[j] = e[j] / i * (i-1);
            }
        }
}
int main()
{
    int i, n;
    euler();
    for(i = 1; i < N; i++)
    {
        res += e[i];
        a[i] = res * 2 + 1;
    }
    while(~scanf("%d", &n))
        printf("%I64d\n", a[n]);
    return 0;
}

hdu 4556 Stern-Brocot Tree

Stern-Brocot Tree