HDU5171 GTY's birthday gift(矩阵快速幂)

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 23 6 2

 

Sample Output

35

 

Source

BestCoder Round #29

 

Recommend

参考大神博客 ：http://blog.csdn.net/crescent__moon/article/details/43617703

//需要知道的一点是斐波拉契数列  s(n)=a(n+2)-1;#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;#define mod 10000007#define N 100005typedef __int64 ll;struct mat{    ll a[2][2];    mat(){memset(a,0,sizeof(a));}    mat operator *(mat b)    {    mat c;        for(int i=0;i<2;i++)  for(int j=0;j<2;j++){c.a[i][j]=0;for(int k=0;k<2;k++)c.a[i][j]=(c.a[i][j]+a[i][k]*b.a[k][j]%mod)%mod;}return c;    }};ll ans;int a[N];mat s,e;void inint(){ s.a[0][0]=s.a[0][1]=s.a[1][0]=1; s.a[1][1]=0; int i,j; e.a[0][1]=e.a[1][0]=0; for(i=0;i<2;i++)e.a[i][i]=1;}mat powmul(int k){    int i,j;    while(k){if(k&1){e=e*s;}k>>=1;s=s*s;}return e;}int main(){int i,j;ll k,n;while(~scanf("%I64d%I64d",&n,&k)){ans=0;for(i=0;i<n;i++){scanf("%d",&a[i]);ans=(ans+a[i])%mod;}inint();sort(a,a+n);s=powmul(k+2);ans=(ans+(s.a[0][0]-2)*a[n-1]%mod)%mod; //好好看看上面的话，列出数列ans=(ans+(s.a[0][1]-1)*a[n-2]%mod)%mod;printf("%I64d\n",ans);}return 0;}