This time I need you to calculate the f(n) . (3<=n<=1000000)

f(n)= Gcd(3)+Gcd(4)+…+Gcd(i)+…+Gcd(n).
Gcd(n)=gcd(C[n][1],C[n][2],……,C[n][n-1])
C[n][k] means the number of way to choose k things from n some things.
gcd(a,b) means the greatest common divisor of a and b.
Input
There are several test case. For each test case:One integer n(3<=n<=1000000). The end of the in put file is EOF.
Output
For each test case:
The output consists of one line with one integer f(n).
Sample Input
3
26983
Sample Output
3
37556486
題目就是這么短小精悍，這題我實在不知道怎么寫，然后也不會數論的推理，我就打了表，發現跟質數的關系很大，就順著推了一下， 就過了。

#include <iostream> #include<cstdio> #define ll long long #define maxn 1000000 using namespace std; int zs[maxn], t = 0, n; ll f[maxn + 5]; bool v[maxn + 5];int main() {for (int i = 2; i <= maxn; i++){if (!v[i]) f[i] = zs[++t] = i;for (int j = 1,u = zs[j] * i; j <= t && u<= maxn; j++){v[u] = 1;if (!(i % zs[j])){f[u] = f[i];break;}f[u] = 1;u = zs[j+1] * i; }}for (int i = 4; i <= maxn; i++)f[i] += f[i - 1];while (scanf("%d", &n)!=EOF)printf("%lld\n", f[n]);return 0; }

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