【每日一题】6月9日Steps to One
Steps to One
https://ac.nowcoder.com/acm/problem/113552
题目意思
给出一个空数列a,从1到m随机选一个数进入a数组,当a数组中全部数的gcd是1的时候退出程序,问平均期望长度。
解题思路
数论,期望dp
[参考1,zzugzx大佬](https://blog.nowcoder.net/n/6fc91cc5671f457d9389dc3bd9303aa1)
[参考2,superj7大佬](https://blog.nowcoder.net/n/fb90d1cb191a4432bb908a53857c8a25)
和gcd有关显然加入数据不会把当前最小的gcd减小,只会不变和变小,所以考虑动态规划解题。
剩下的就是推化式子了……这对一个高数要挂科的小白来说。。太太太困难了。
迷迷糊糊的看出来预处理数组,期望dp,其余的就是参考第二位大佬的式子了。。
#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const int MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int N = 1e5 + 7;
ll dp[N];
vector<ll> d[N], cnt[N];
int main() {
int n = read();
// 预处理
for (int i = 1; i <= n; ++i)
for (int j = i; j <= n; j += i)
d[j].push_back(i), cnt[j].push_back(0);
for (int i = 1; i <= n; ++i)
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] == 0) cnt[i][j] -= cnt[i][k];
}
//期望dp
dp[1] = 0;
for (int i = 2; i <= n; ++i) {
ll sum = 0, x = 0;
for (int j = 0; j < d[i].size(); ++j)
if (d[i][j] == i) (sum += cnt[i][j]) %= MOD;
else (x += dp[d[i][j]] * cnt[i][j] % MOD) %= MOD;
dp[i] = (x + n) % MOD * qpow((n - sum + MOD) % MOD, MOD - 2, MOD) % MOD;
}
ll ans = 0;
for (int i = 1; i <= n; ++i) (ans += dp[i] + 1) %= MOD;
(ans *= qpow(n, MOD - 2, MOD)) %= MOD;
write(ans);
return 0;
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