题解:www.cygmasot.com/index.php/2015/08/17/hdu_3970
给定n
求连续整数[0,n), 中任意选一些数使得选出的数和为n的倍数的方法数
。。。并不会如何递推。。
思路:
然后这是公式：点击打开链接
a(n) = 1/n * sum_{d divides n and d is odd} 2^(n/d) * phi(d).
d最大是n，也就是1e9,要计算1e9的phi，所以容斥来算phi.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <vector>
#include <string>
#include <time.h>
#include <math.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <set>
#include <map>
const int inf = 1e9;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x < 0) { putchar('-'); x = -x; }
if (x > 9) pt(x / 10);
putchar(x % 10 + '0');
}
using namespace std;
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
typedef pair<int, int> pii;
int Pow(int x, int y){//快速幂
int ans = 1;
while (y){
if (y & 1)ans = (ll)ans*x%mod;
y >>= 1;
x = (ll)x*x%mod;
}
return ans;
}
int prime[N], primenum;//素数表
void PRIME(int Max_Prime){
primenum = 0;
prime[primenum++] = 2;
for (int i = 3; i <= Max_Prime; i += 2)
for (int j = 0; j<primenum; j++)
if (i%prime[j] == 0)break;
else if (prime[j]>sqrt((double)i) || j == primenum - 1)
{
prime[primenum++] = i;
break;
}
}
void add(int &x, int y){ x += y; if (x >= mod)x -= mod; }//加法
void go(int x, vector<int>&Pri, vector<int>&Num){//分解质因数
Pri.clear(); Num.clear();
while (!(x & 1))x >>= 1;
for (int i = 1; (ll)prime[i] * prime[i] <= x; i++){
if (x%prime[i])continue;
Pri.push_back(prime[i]);
Num.push_back(0);
while (x%prime[i] == 0)x /= prime[i], Num[Num.size() - 1]++;
}
if (x != 1 && (x&1))Pri.push_back(x), Num.push_back(1);
}
vector<int>_pri, _num;
void cal(int id, int mul, int siz, int sor, int &now){//容斥算欧拉函数
if (id == _pri.size()){
if (mul == 1)return;
if (siz & 1)now += sor / mul;
else now -= sor / mul;
return;
}
cal(id + 1, mul, siz, sor, now);
cal(id + 1, mul*_pri[id], siz + 1, sor, now);
}
int phi(int x){
if (x == 1)return 1;
go(x, _pri, _num);
int now = 0;
cal(0, 1, 0, x, now);
return x - now;
}
int ans, n;
vector<int>pri, num;
void dfs(int id, int d){
if (id == pri.size())
{
add(ans, (ll)Pow(2, n / d) * phi(d) % mod);
return;
}
for (int i = 0; i <= num[id]; i++){
dfs(id + 1, d);
d *= pri[id];
}
}
int main(){
PRIME(1e5);
int T; rd(T);
while (T--){
rd(n);
go(n, pri, num);
ans = 0;
dfs(0, 1);
pt((ll)ans*Pow(n, mod - 2) % mod); puts("");
}
return 0;
}
