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D. Steps to One（概率DP，莫比乌斯反演）
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D. Steps to One
設f[i]f[i]f[i]為gcd?\gcdgcd為iii，還需要多少個數(shù)，那么有f[i]=1+∑j=1mf[gcd?(i,j)]mf[i] = 1 + \frac{\sum\limits_{j = 1} ^{m} f[\gcd(i, j)]}{m}f[i]=1+mj=1∑m?f[gcd(i,j)]?，
f[1]=0f[1] = 0f[1]=0，考慮化簡∑j=1mf[gcd?(i,j)]\sum\limits_{j = 1} ^{m} f[\gcd(i, j)]j=1∑m?f[gcd(i,j)]，
∑d∣if[d]∑j=1m[gcd?(i,j)=d]給定i,d，求∑j=1m[gcd?(i,j)=d]∑j=1md[gcd?(id,j)=1]∑k∣idμ(k)mkd則原式為∑d∣if[d]∑k∣idμ(k)mkd\sum_{d \mid i} f[d] \sum_{j = 1} ^{m}[\gcd(i, j) = d]\\ 給定i, d，求\sum_{j = 1} ^{m}[\gcd(i, j) = d]\\ \sum_{j = 1} ^{\frac{m}ozvdkddzhkzd}[\gcd(\frac{i}ozvdkddzhkzd, j) = 1]\\ \sum_{k \mid \frac{i}ozvdkddzhkzd} \mu(k) \frac{m}{kd}\\ 則原式為 \sum_{d \mid i} f[d] \sum_{k \mid \frac{i}ozvdkddzhkzd} \mu(k) \frac{m}{kd}\\ d∣i∑?f[d]j=1∑m?[gcd(i,j)=d]給定i,d，求j=1∑m?[gcd(i,j)=d]j=1∑dm??[gcd(di?,j)=1]k∣di?∑?μ(k)kdm?則原式為d∣i∑?f[d]k∣di?∑?μ(k)kdm?
#include <bits/stdc++.h>using namespace std
;const int N
= 1e5 + 10, mod
= 1e9 + 7;int f
[N
], prime
[N
], mu
[N
], m
, cnt
;vector
<int> fac
[N
];bool st
[N
];int quick_pow(int a
, int n
) {int ans
= 1;while (n
) {if (n
& 1) {ans
= 1ll * ans
* a
% mod
;}a
= 1ll * a
* a
% mod
;n
>>= 1;}return ans
;
}inline int inv(int a
) {return quick_pow(a
, mod
- 2);
}void init() {mu
[1] = 1;for (int i
= 2; i
< N
; i
++) {if (!st
[i
]) {prime
[++cnt
] = i
;mu
[i
] = mod
- 1;}for (int j
= 1; j
<= cnt
&& 1ll * i
* prime
[j
] < N
; j
++) {st
[i
* prime
[j
]] = 1;if (i
% prime
[j
] == 0) {break;}mu
[i
* prime
[j
]] = (mod
- mu
[i
]) % mod
;}}for (int i
= 1; i
< N
; i
++) {for (int j
= i
; j
< N
; j
+= i
) {fac
[j
].push_back(i
);}}
}int main() {init();scanf("%d", &m
);int inv_m
= inv(m
);for (int i
= 1; i
<= m
; i
++) {for (auto &d
: fac
[i
]) {if (d
== i
) {continue;}int cur
= 0;for (auto &k
: fac
[i
/ d
]) {cur
= (cur
+ 1ll * mu
[k
] * (m
/ (k
* d
)) % mod
) % mod
;}f
[i
] = (f
[i
] + 1ll * f
[d
] * cur
% mod
) % mod
;}f
[i
] = 1ll * (m
+ f
[i
]) * inv(m
- m
/ i
) % mod
;}int ans
= 0;for (int i
= 1; i
<= m
; i
++) {ans
= (ans
+ f
[i
]) % mod
;}cout
<< (1 + 1ll * ans
* inv_m
) % mod
<< "\n";return 0;
}
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