cf1139D. Steps to One(dp)
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2022-04-28 10:46:57
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 ......
题意
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; }
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