cf1139D. Steps to One(dp)

题意

题目链接

从\([1, m]\)中随机选数，问使得所有数gcd=1的期望步数

sol

一个很显然的思路是设\(f[i]\)表示当前数为\(i\)，期望的操作轮数，转移的时候直接枚举gcd

\(f[i] = 1 + \frac{ \sum_{j=1}^n f[gcd(i, j)]}{n}\)

然后移一下项就可以算出\(f[i]\)了。

发现gcd相同的有很多，可以预处理一下。

复杂度\(o(跑的过)\)

还有一种反演做法表示推不出来qwq

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define ull unsigned long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
const double eps = 1e-9;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
template <typename a, typename b> inline ll fp(a a, b p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
template <typename a, typename b> inline a gcd(a x, b y) {return !y ? x : gcd(y, x % y);}
int inv(int x) {
    return fp(x, mod - 2);
}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, f[maxn], invn;
vector<int> d[maxn], cnt[maxn];
void sieve() {
    for(int i = 1; i <= n; i++) 
        for(int k = i; k <= n; k += i) d[k].push_back(i);
    for(int i = 1; i <= n; i++) {
        cnt[i].resize(d[i].size() + 1);
        for(int j = d[i].size() - 1; ~j; j--) {
            cnt[i][j] = n / d[i][j];
            for(int k = j + 1; k < d[i].size(); k++)
                if(!(d[i][k] % d[i][j])) cnt[i][j] -= cnt[i][k];
        }
        //for(int j = 0; j < d[i].size(); j++)
        //  printf("%d %d %d\n", i, d[i][j], cnt[i][j]);
    }
            
}
signed main() {
    n = read(); invn = inv(n);
    sieve();
    int ans = 0;
    for(int i = 2; i <= n; i++) {
        int lf = n, tmp = 0;
        /*
        for(int j = 1, t = 1; j <= n; j++) {
            if((t = gcd(i, j)) == i) lf--;
            else add2(tmp, f[t]);
        }
        */
        for(int j = 0; j < d[i].size(); j++) {
            if(d[i][j] == i) lf -= cnt[i][j];
            else add2(tmp, mul(cnt[i][j], f[d[i][j]]));
        }
        f[i] = add(n, tmp); 
        mul2(f[i], inv(lf));
    }
    for(int i = 1; i <= n; i++) add2(ans, f[i] + 1);
    cout << mul(ans, invn);
    return 0;
}