c/c++语言开发共享cf1097D. Makoto and a Blackboard(期望dp)

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 (())) ( */