HDU 5171 GTY's birthday gift (矩阵快速幂)

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本文介绍了一个经典的算法问题——如何通过特定操作使一个数字集合的总和最大化，并提供了一种使用矩阵快速幂的方法来解决该问题。通过对输入数字进行排序并应用矩阵运算，可以有效地计算出在指定次数的操作下集合的最大可能总和。

GTY's birthday gift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 421 Accepted Submission(s): 164

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

Source

BestCoder Round #29

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5171

题目大意：从一个数字集合中取出两个最大的相加得到的结果和原来的两个数再放回集合，求经过k次操作后，集合元素和的最大值

题目分析：因为每次肯定找最大的两个相加，开始有a b假设a比b小
num a b a+b a+2b 2a+3b...
sum a a+b 2a+2b 3a+4b 5a+7b...
类似斐波那契数列sum[i] = sum[i - 1] + num[i - 1] + num[i - 2]
由此我们推出转移矩阵：

ans = (A^k * B) % MOD，采用矩阵快速幂

#include <cstring>
#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
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) != EOF)
    {
        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);
        ll ans = (tmp.m[0][0] * sum + tmp.m[0][1] * a[n - 1] + tmp.m[0][2] * a[n - 2]) % MOD;
        printf("%lld\n", ans);
    }
}