版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/Ike940067893/article/details/84996397
题面
令
表示
的约数之和求
题目分析
先给结论
可以通过 传送门 类似的证明方法证明
拖更…
AC code
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 1e6 + 1, mod = 1e9 + 7;
int Prime[MAXN/10], Cnt, mu[MAXN];
bool IsnotPrime[MAXN];
inline void init(int n)
{
mu[1] = 1;
for(int i = 2; i <= n; ++i)
{
if(!IsnotPrime[i])
Prime[++Cnt] = i, mu[i] = -1;
for(int j = 1, v; j <= Cnt && i * Prime[j] <= n; ++j)
{
v = i * Prime[j];
IsnotPrime[v] = 1;
if(i % Prime[j] == 0) { mu[v] = 0; break; }
mu[v] = -mu[i];
}
}
for(int i = 1; i <= n; ++i)
mu[i] = (mu[i-1] + i*mu[i]%mod) % mod;
}
map<int, int>s;
inline int f(int i, int j) //i+(i+1)+...+j
{
return ((LL)(i+j) * (j-i+1)/2) % mod;
}
inline int sum(int n) //mu(1)1+mu(2)2+...+mu(n)n
{
if(n < MAXN) return mu[n];
if(s.count(n)) return s[n];
int ret = 1;
for(int i = 2, j; i <= n; i=j+1)
{
j = n/(n/i);
ret = (ret - (LL)f(i,j) * sum(n/i) % mod) % mod;
}
return s[n]=ret;
}
inline int calc(int n)
{
int ret = 0, k;
for(int i = 1, j; i <= n; i=j+1)
{
j = n/(n/i); k = n/i;
ret = (ret + (LL)(j-i+1) * (((LL)k*(k+1)/2)%mod) % mod) % mod;
}
return ret;
}
inline int solve(int n)
{
int ret = 0, last = 0, tmp, tmp2;
for(int i = 1, j; i <= n; i=j+1)
{
tmp = sum(j = n/(n/i)), tmp2 = calc(n/i), tmp2 = (LL)tmp2 * tmp2 % mod;
ret = (ret + (LL)(tmp - last) * tmp2 % mod) % mod;
last = tmp;
}
return ret;
}
int main ()
{
int n; init(MAXN-1);
scanf("%d", &n);
printf("%d\n", (solve(n)+mod)%mod);
}