GTY‘s birthday gift

                                                                      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                         
                          Total Submission(s): 209    Accepted
Submission(s): 71

Problem Description

FFZ‘s birthday is coming. GTY wants to give a gift to ZZF. He asked his
gay friends what he should give to ZZF. One of them said, ‘Nothing is
more interesting than a number multiset.‘ So GTY decided to make a
multiset for ZZF. Multiset can contain elements
with same values. Because GTY wants to finish the gift as soon as
possible, he will use JURUO magic. It allows him to choose two numbers a
and b(a,b∈S),
and add a+b to

the multiset. GTY can use the magic for k times, and he wants the sum
of the multiset is maximum, because the larger the sum is, the happier
FFZ will be. You need to help him calculate the maximum sum of the
multiset.

Input

Multi test cases (about 3) . The first line contains two integers n and k (2≤n≤100000,1≤k≤1000000000).
The second line contains n elements ai (1≤ai≤100000)separated
by spaces , indicating the multiset S .

Output

For each case , print the maximum sum of the multiset (mod 10000007).

Sample Input

3 2

3 6 2

Sample Output

35

???100111110??????Sn−1Fn−1Fn−2???=???SnFnFn−1???

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <cstdlib>
 7 #include <string>
 8 #include <vector>
 9 #include <set>
10 #include <map>
11 #include <stack>
12 #include <queue>
13 using namespace std;
14 const int INF = 0x7fffffff;
15 const int MS = 100005;
16 const double EXP = 1e-8;
17 const int mod = 10000007;
18 typedef long long LL;
19 int a[MS];
20 int n, k;
21 struct Matrix
22 {
23     LL matrix[3][3];
24     Matrix()
25     {
26         memset(matrix, 0, sizeof(matrix));
27     }
28     Matrix operator *(const Matrix &b)const
29     {
30         Matrix x;
31         for (int i = 0; i < 3; i++)
32             for (int j = 0; j < 3; j++)
33                 for (int k = 0; k < 3; k++)
34                     x.matrix[i][j] = ((x.matrix[i][j] + matrix[i][k] * b.matrix[k][j]) % mod) % mod;
35         return x;
36     }
37 };
38 void solve()
39 {
40     sort(a, a + n);
41     LL sum = 0;
42     LL s, t;
43     for (int i = 0; i < n; i++)
44         sum += a[i];
45     s = static_cast<LL>(a[n - 2]);
46     t = static_cast<LL>(a[n - 1]);
47     Matrix m1, m2, ans;
48     m1.matrix[0][1] = m1.matrix[0][0] = m1.matrix[0][2] = 1;
49     m1.matrix[1][1] = m1.matrix[1][2] = 1;
50     m1.matrix[2][1] = 1;
51     ans.matrix[0][0] = ans.matrix[1][1] = ans.matrix[2][2] = 1;
52     while (k)
53     {
54         if (k & 1)
55             ans = ans*m1;
56         k >>= 1;
57         m1 = m1*m1;
58     }
59
60     m2.matrix[0][0] = sum;
61     m2.matrix[1][0] = t;
62     m2.matrix[2][0] = s;
63     ans = ans*m2;
64     cout << ans.matrix[0][0] % mod << endl;
65     return;
66 }
67 int main()
68 {
69     while (cin >> n >> k)
70     {
71         for (int i = 0; i < n; i++)
72             cin >> a[i];
73         solve();
74     }
75     return 0;
76 }

1002 GTY's birthday gift