cf1097D. Makoto and a Blackboard(期望dp)
阿新 • • 發佈:2019-01-05
題意
Sol
首先考慮當\(n = p^x\),其中\(p\)是質數,顯然它的因子只有\(1, p, p^2, \dots p^x\)(最多logn個)
那麼可以直接dp, 設\(f[i][j]\)表示經過了\(i\)輪,當前數是\(p^j\)的概率,轉移的時候列舉這一輪的\(p^j\)轉移一下
然後我們可以把每個質因子分開算,最後乘起來就好了
至於這樣為什麼是對的。(開始瞎扯),考慮最後的每個約數,它一定是一堆質因子的乘積,而每個質因子的概率我們是知道的,所以這樣乘起來算是沒問題的。
其實本質上還是因為約數和/個數都是積性函式
時間複雜度:\(O(\sqrt{n} + k log^2 n)\)
#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 (())) ( */