cf1097D. Makoto and a Blackboard(期望dp)

时间：2023-07-14 22:49:53

题意

题目链接

Sol

首先考虑当\(n = p^x\)，其中\(p\)是质数，显然它的因子只有\(1, p, p^2, \dots p^x\)(最多logn个)

那么可以直接dp, 设\(f[i][j]\)表示经过了\(i\)轮，当前数是\(p^j\)的概率，转移的时候枚举这一轮的\(p^j\)转移一下

然后我们可以把每个质因子分开算，最后乘起来就好了

至于这样为什么是对的。(开始瞎扯)，考虑最后的每个约数，它一定是一堆质因子的乘积，而每个质因子的概率我们是知道的，所以这样乘起来算是没问题的。

其实本质上还是因为约数和/个数都是积性函数

时间复杂度:\(O(\sqrt{n} + k log^2 n)\)

#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 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;}

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, K, top, inv[MAXN];

Pair st[MAXN];

int f[10001][1001];

int fp(int a, int p) {

    int base = 1;

    while(p) {

        if(p & 1) base = mul(base, a);

        a = mul(a, a); p >>= 1;

    }

    return base;

}

int solve(int a, int b) {//a^b

    memset(f, 0, sizeof(f));

    f[0][b] = 1;

    for(int i = 1; i <= K; i++)

        for(int j = 0; j <= b; j++)

            for(int k = j; k <= b; k++)

                add2(f[i][j], mul(f[i - 1][k], inv[k + 1]));

    int res = 0;

    for(int i = 0; i <= b; i++) add2(res, mul(f[K][i], fp(a, i)));

    return res;

}

signed main() {

    for(int i = 1; i <= 666; i++) inv[i] = fp(i, mod - 2);

    cin >> N >> K;

    for(int i = 2; i * i <= N; i++) {

        if(!(N % i)) {

            st[++top] = MP(i, 0);

            while(!(N % i)) st[top].se++, N /= i;

        }

    }

    if(N) st[++top] = MP(N, 1);

    int ans = 1;

    for(int i = 1; i <= top; i++)

        ans = mul(ans, solve(st[i].fi, st[i].se));

    cout << ans;

    return 0;

}

/*

2

(()))

(

*/

相关文章