hdu 5171 GTY's birthday gift (BestCoder Round #29)

GTY's birthday gift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536K (Java/Others)Total Submission(s): 209 Accepted Submission(s): 71Problem DescriptionFFZ'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 elementswith 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 tothe 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.InputMulti 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)separatedby spaces , indicating the multiset S .OutputFor each case , print the maximum sum of the multiset (mod 10000007).Sample Input

3 2
3 6 2

Sample Output

35

题解：

显然每次会从可重集中选择最大的两个进行操作,设这两数为a,b(a>=b),操作之后的数一定是操作后集合中最大的,下一次选取的数一定是a+b和a,这就形成了一个类似于斐波那契数列的东西,矩阵乘法快速幂求前n项和即可,转移矩阵如下

1 0 1     sum

0 1 1  *  a+b 

0 1 0      a

设矩阵A为

1 0 1

0 1 1

0 1 0

B为

sum

a+b

a

ans=A^k*B，接着矩阵快速幂

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int mod=10000007;typedef struct matrix{long long ma[5][5];}matrix;matrix multi(matrix x,matrix y){matrix ans;memset(ans.ma,0,sizeof(ans.ma));for(int i=1;i<=3;i++){for(int j=1;j<=3;j++){if(x.ma[i][j])for(int k=1;k<=3;k++){ans.ma[i][k]=(ans.ma[i][k]+(x.ma[i][j]*y.ma[j][k])%mod)%mod;}}}return ans;}int main(){long long sum;long long n,k;while(~scanf("%I64d%I64d",&n,&k)){sum=0;long long t, maxf=0,maxs=0;for(int i=0;i<n;i++){scanf("%I64d",&t);sum=(sum+t)%mod;if(t>maxf){maxs=maxf;maxf=t;}else if(t>maxs)maxs=t;}matrix a,b,ans;memset(ans.ma,0,sizeof(ans.ma));for(int i=1;i<=3;i++)for(int j=1;j<=3;j++)if(i==j)ans.ma[i][j]=1;memset(a.ma,0,sizeof(a.ma));a.ma[1][1]=a.ma[1][2]=a.ma[2][2]=a.ma[2][3]=a.ma[3][2]=1;while(k){if(k&1)ans=multi(a,ans);a=multi(a,a);k=k>>1;}memset(b.ma,0,sizeof(b.ma));b.ma[1][1]=sum;b.ma[2][1]=maxf+maxs;b.ma[3][1]=maxf;ans=multi(ans,b);printf("%I64d\n",ans.ma[1][1]);}return 0;}

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