The collection of bzoj4547 kit

Description

There is a re-set of size n, kit each operation can be joined a number of a+b (A, a, b all belong to s), after K operations it can obtain the maximum of S and

values. (The data guarantees that the value is non-negative)Input

The first line has two integers n,k represents the initial number of elements and operands, and the second row contains n integers representing the elements that can be re-set initially.

for 100% of data, there are n<=10^5,k<=10^9,|ai|<=10^5Outputoutputs an integer that represents the maximum value. Answer to 10000007 modulo.

Add up to two maximum values in a re-set each time you should take a

If the maximum value is non-negative, the sum of the new number of K operations can be obtained by Fibonacci sequence.

If the maximum value and the second large value of a positive and negative, each time the maximum and sub-large value is added to the K-operation or the second-largest non-negative with the Fibonacci sequence to find the remaining operation.

N

Σfi = fn+2-1

I=1

#include <cstdio>#include<algorithm>intn,k;Const intp=10000007;intv[100005];typedefintmat[2];mat M1,m2;inlineintinput () {intx=0, C=getchar (), f=1;  while(c> $|| c< -){        if(c=='-') f=-1; C=GetChar (); }     while(c> -&&c< -) x=x*Ten+c- -, c=GetChar (); returnx*F;} InlinevoidMul (Mat A,mat b) {intC= (1ll* (a[1]-a[0]) *b[0]+1ll*a[0]*b[1])%Q; intD= (1ll*a[0]*b[0]+1ll*a[1]*b[1])%P; a[0]=c;a[1]=D;} InlinevoidFibintx) {x+=1; m1[0]=0; m1[1]=1; m2[0]=m2[1]=1;  while(x) {if(x&1) Mul (M1,M2);        Mul (M2,M2); X>>=1; }}intans=0;intMain () {n=input (); k=input ();  for(intI=0; i<n;i++) v[i]=input (); Std::nth_element (V,v+n-2, v+N);  for(intI=0; i<n;i++) {Ans+=V[i]; if(ans>=p) ans-=Q; if(ans<p) ans+=P; }    intv1=v[n-1],v2=v[n-2]; if(v1<=0) {Ans= (ans+ (v1+v2) *1ll*k)%Q; if(ans<0) ans+=Q; printf ("%d", Ans); }Else if(v2<0){         while(v2<0&&k) {v2+=v1; Ans+=v2; if(ans>=p) ans-=Q; if(ans<p) ans+=P; --K; }        if(V2&GT;V1) v1^=v2,v2^=v1,v1^=v2; if(k) {fib (k); intAns= ((m1[1]-1) *1ll*v2+ (m1[1]+m1[0]-2) *1ll*v1)%Q; Ans= (Ans+ans)%P; }        if(ans<0) ans+=Q; printf ("%d", Ans); }Else{fib (k); intAns= ((m1[1]-1) *1ll*v2+ (m1[1]+m1[0]-2) *1ll*v1)%P; Ans= (Ans+ans)%P; if(ans<0) ans+=Q; printf ("%d", Ans); }    return 0;}

The collection of bzoj4547 kit