Educational Codeforces Round 41 (Rated for Div. 2) G. Partitions

G. Partitions

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of n elements indexed from 1 to n. The weight of i-th element is wi. The weight of some subset of a given set is denoted as . The weight of some partition R of a given set into k subsets is  (recall that a partition of a given set is a set of its subsets such that every element of the given set belongs to exactly one subset in partition).

Calculate the sum of weights of all partitions of a given set into exactly k non-empty subsets, and print it modulo 109?+?7. Two partitions are considered different iff there exist two elements x and y such that they belong to the same set in one of the partitions, and to different sets in another partition.

Input

The first line contains two integers n and k (1?≤?k?≤?n?≤?2·105) — the number of elements and the number of subsets in each partition, respectively.

The second line contains n integers wi (1?≤?wi?≤?109)— weights of elements of the set.

Output

Print one integer — the sum of weights of all partitions of a given set into k non-empty subsets, taken modulo 109?+?7.

Examples

input

Copy

4 22 3 2 3

output

Copy

160

input

Copy

5 21 2 3 4 5

output

Copy

645

Note

Possible partitions in the first sample:

1. {{1,?2,?3},?{4}}, W(R)?=?3·(w1?+?w2?+?w3)?+?1·w4?=?24;
2. {{1,?2,?4},?{3}}, W(R)?=?26;
3. {{1,?3,?4},?{2}}, W(R)?=?24;
4. {{1,?2},?{3,?4}}, W(R)?=?2·(w1?+?w2)?+?2·(w3?+?w4)?=?20;
5. {{1,?3},?{2,?4}}, W(R)?=?20;
6. {{1,?4},?{2,?3}}, W(R)?=?20;
7. {{1},?{2,?3,?4}}, W(R)?=?26;

Possible partitions in the second sample:

1. {{1,?2,?3,?4},?{5}}, W(R)?=?45;
2. {{1,?2,?3,?5},?{4}}, W(R)?=?48;
3. {{1,?2,?4,?5},?{3}}, W(R)?=?51;
4. {{1,?3,?4,?5},?{2}}, W(R)?=?54;
5. {{2,?3,?4,?5},?{1}}, W(R)?=?57;
6. {{1,?2,?3},?{4,?5}}, W(R)?=?36;
7. {{1,?2,?4},?{3,?5}}, W(R)?=?37;
8. {{1,?2,?5},?{3,?4}}, W(R)?=?38;
9. {{1,?3,?4},?{2,?5}}, W(R)?=?38;
10. {{1,?3,?5},?{2,?4}}, W(R)?=?39;
11. {{1,?4,?5},?{2,?3}}, W(R)?=?40;
12. {{2,?3,?4},?{1,?5}}, W(R)?=?39;
13. {{2,?3,?5},?{1,?4}}, W(R)?=?40;
14. {{2,?4,?5},?{1,?3}}, W(R)?=?41;
15. {{3,?4,?5},?{1,?2}}, W(R)?=?42.

思路一：考虑每个点对整体的贡献。也就是SUM (size * S(n - size, k - 1) )* wi， S(n, k)为第二类斯特林数。但这样需求出所有的S(i, k - 1)，暂时不会nlogn求法。复杂度O(n*k)。

思路二：定义g(n, k, i, j)表示n个数分成k个非空集合且i与j在一个集合中的方案数。单独对一个点ai考虑，对于每一个合法的划分，它的贡献要有size次，size为ai所在集合的大小。那么对于SUM(g(n, k, i, j)),其中j从1遍历到n，在这些所有的方案中，我们之前考虑的特定的那个划分也正好出现了size次，两个数刚好等价了。所以答案就是g(n, k, i, i) + SUM(g(n, k, i, j) (j != i)) = S(n, k) + (n - 1) * S(n - 1, k)。对于单点斯特林数，可以通过容斥加快速幂nlogn求出。

S(n, k) = SUM((-1) ^ i * C(k, i) * (k - i) ^ n) / k!。复杂度nlogn。

 1 #include <iostream>
 2 #include <fstream>
 3 #include <sstream>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <string>
 8 #include <cstring>
 9 #include <algorithm>
10 #include <queue>
11 #include <stack>
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <list>
16 #include <iomanip>
17 #include <cctype>
18 #include <cassert>
19 #include <bitset>
20 #include <ctime>
21
22 using namespace std;
23
24 #define pau system("pause")
25 #define ll long long
26 #define pii pair<int, int>
27 #define pb push_back
28 #define mp make_pair
29 #define clr(a, x) memset(a, x, sizeof(a))
30
31 const double pi = acos(-1.0);
32 const int INF = 0x3f3f3f3f;
33 const int MOD = 1e9 + 7;
34 const double EPS = 1e-9;
35
36 /*
37 #include <ext/pb_ds/assoc_container.hpp>
38 #include <ext/pb_ds/tree_policy.hpp>
39
40 using namespace __gnu_pbds;
41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
42 */
43
44 int n, k;
45 ll W, N[400015];
46 ll mpow(ll x, ll y) {
47     if (!y) return 1;
48     ll res = mpow(x, y >> 1);
49     if (y & 1) return res * res % MOD * x % MOD;
50     return res * res % MOD;
51 }
52 void pre() {
53     N[0] = 1;
54     for (int i = 1; i <= 400000; ++i) {
55         N[i] = N[i - 1] * i % MOD;
56     }
57 }
58 ll inv(ll x) {
59     return mpow(x, MOD - 2);
60 }
61 ll C(int n, int k) {
62     return N[n] * inv(N[n - k]) % MOD * inv(N[k]) % MOD;
63 }
64 ll stirling(ll x, ll y) {
65     ll res = 0;
66     for (int i = 0; i < y; ++i) {
67         if (i & 1) res -= C(y, i) * mpow(y - i, x) % MOD;
68         else res += C(y, i) * mpow(y - i, x) % MOD;
69     }
70     res = (res % MOD + MOD) % MOD;
71     return res * inv(N[y]) % MOD;
72 }
73 int main() {
74     pre();
75     scanf("%d%d", &n, &k);
76     for (int i = 1, w; i <= n; ++i) {
77         scanf("%d", &w);
78         W += w;
79     }
80     printf("%lld\n", W % MOD * (stirling(n, k) + (n - 1) * stirling(n - 1, k) % MOD) % MOD);
81     return 0;
82 }

原文地址：https://www.cnblogs.com/BIGTOM/p/9149528.html