hdu 5171 GTY's birthday gift【矩阵快速幂】【思维】【感受矩阵和数论的神奇】

GTY's birthday gift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1176    Accepted Submission(s): 449

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

这真是一个神奇的题，更神奇的是矩阵和数论给我带来的惊喜。矩阵能够求斐波那契数，不仅仅是斐波那契数神奇，矩阵也是如此的神奇。

我们从题目中能够推测：

假设a是序列中最大数，b是次大数，那么每一次多出来的是：a+b,2a+b,3a+2b,5a+3b..................

比原序列多出来的值是：a+b,3a+2b,6a+4b....................

其中我们能发现a前系数为1 2 3 5 8...............

b前系数为:1 1 2 3 5 8......................

很明显的斐波那契数列。如果接触过POJ3070的小伙伴可能马上就有思路了。矩阵快速幂求斐波那契数：

这里初始化矩阵为2阶：

1 1

1 0

自乘：

2 1

1 1

再自乘：

3 2

2 1

再自乘：

5 3

3 2

我们发现第一层上边的两个数一直都是斐波那契数。

这个时候我们根据矩阵乘法的特性，加上各种推论能够构造一个矩阵：

1 1 1

0 1 1

0 1 0

自乘之后

1 3 2

0 2 1

0 1 1

无限幂次都能保证第一行的三个元素分别代表: 1（第一个数永远都是1），比原序列多出的a的个数，比原序列多出的b的个数。

然后我们就能写出对应的代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
#define ll long long int
int const MOD = 10000007;
struct matrix
{
    ll m[3][3];
};
int a[100010];
matrix multiply(matrix x, matrix y)  //矩阵乘法
{
    matrix tmp;
    memset(tmp.m, 0, sizeof(tmp.m));
    for(int i = 0; i < 3; i++)
    {
        for(int j = 0; j < 3; j++)
        {
            if(x.m[i][j] == 0)
                continue;
            for(int k = 0; k < 3; k++)
            {
                if(y.m[j][k] == 0)
                    continue;
                tmp.m[i][k] += x.m[i][j] * y.m[j][k] % MOD;
                tmp.m[i][k] %= MOD;
            }
        }
    }
    return tmp;
}

matrix quickmod(matrix a, int n) //矩阵快速幂
{
    matrix res;
    memset(res.m, 0, sizeof(res.m));
    for(int i = 0; i < 3; i++)
        res.m[i][i] = 1;
    while(n)
    {
        if(n & 1)
            res = multiply(res, a);
        n >>= 1;
        a = multiply(a, a);
    }
    return res;
}
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        ll sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n);
        matrix tmp;
        tmp.m[0][0] = 1; tmp.m[0][1] = 1; tmp.m[0][2] = 1;
        tmp.m[1][0] = 0; tmp.m[1][1] = 1; tmp.m[1][2] = 1;
        tmp.m[2][0] = 0; tmp.m[2][1] = 1; tmp.m[2][2] = 0;
            tmp = quickmod(tmp, k);
            printf("%lld %lld %lld\n",tmp.m[0][0],tmp.m[0][1],tmp.m[0][2]);
        ll output= (tmp.m[0][0] * sum + tmp.m[0][1] * a[n - 1] + tmp.m[0][2] * a[n - 2]) % MOD;
        printf("%lld\n",output);
    }
}