GTY's birthday gift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1708    Accepted Submission(s): 662 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 .   1 Output For each case , print the maximum sum of the multiset ( mod 10000007       ).   Sample Input 3 2 3 6 2   Sample Output 35  

给n个数，k次操作，每次操作取两个数加入集合，问最后的集合所有数的和（%MOD）最大， 解法：要使得结果最大，最优的办法就是每次取最大的两个数，加入集合，递推几个，便可以发现结果只与最大两个数的个数有关。 加入的数：第一次：max1+max2　　　 sum多加入的数max1+max2 　　　　　第二次：2*max1+max2 　　sum多加入的数3*max1+2*max2 　　　　　第三次：3*max1+2*max2 　sum多加入的数6*max1+4*max2 　　　　　第四次：5*max1+3*max2 　sum多加入的数11*max1+7*max2 　　　　　第五次：8*max1+5*max2 　sum多加入的数19*max1+12*max2

所以： f1=1,f2=2 多加入的数max1为有递推式 fn+1=fn+fn−1+2 /max2的n+1项-1； 多加入的数max2为有递推式 fn=fn−1+fn−2+1 直接对max2构造矩类似于斐波拉契数列的阵快速幂的矩阵，优化运算 #include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #include<cmath> using namespace std; #define LL long long #define MOD 10000007 struct Mat{ int n,m; LL mat[9][9]; }; Mat operator *(Mat a,Mat b){ Mat c; memset(c.mat,0,sizeof(c.mat)); c.n = a.n,c.m = b.m; for(int i=1;i<=a.n;i++){ for(int j=1;j<=b.m;j++){ for(int k=1;k<=a.m;k++){ c.mat[i][j] += (a.mat[i][k]*b.mat[k][j])%MOD; c.mat[i][j] %= MOD; } } } return c; } Mat operator +(Mat a,Mat b){ Mat c; memset(c.mat,0,sizeof(c.mat)); c.n = a.n,c.m = a.m; for(int i=1;i<=a.n;i++){ for(int j=1;j<=a.m;j++){ c.mat[i][j] = (a.mat[i][j]+b.mat[i][j])%MOD; } } return c; } Mat operator ^(Mat a,int k){ Mat c; memset(c.mat,0,sizeof(c.mat)); c.n = a.n,c.m = a.n; for(int i=1;i<=a.n;i++)c.mat[i][i] = 1; while(k){ if(k&1){ c = c*a; } a = a*a; k>>=1; } return c; } void out(Mat a){ for(int i=1;i<=a.n;i++){ for(int j=1;j<=a.m;j++){ printf(j==a.m? "%I64d\n":"%I64d ",a.mat[i][j]); } } } int main(){ int n,k; while(scanf("%d %d",&n,&k)!=EOF){ int x; LL ans=0; int max1=0,max2=0; while(n--){ scanf("%d",&x); ans+=x; if(x> max2)max2=x; if(max1<max2) swap(max1,max2); } if(k==1){ ans+=(max1+max2); ans%=MOD; } else if(k==2){ ans+=(3*max1+2*max2); ans%=MOD; } else{ Mat d; d.n=3,d.m=3; d.mat[1][1]=1,d.mat[1][2]=1,d.mat[1][3]=0; d.mat[2][1]=1,d.mat[2][2]=0,d.mat[2][3]=0; d.mat[3][1]=1,d.mat[3][2]=0,d.mat[3][3]=1; Mat temp=d^(k-2); //cout<<2*temp.mat[1][1]+temp.mat[2][1]+temp.mat[3][1]<<endl; ans+=(2*temp.mat[1][1]+temp.mat[2][1]+temp.mat[3][1])*max2%MOD; temp=temp*d; ans%=MOD; //cout<<2*temp.mat[1][1]+temp.mat[2][1]+temp.mat[3][1]-1<<endl; ans+=(2*temp.mat[1][1]+temp.mat[2][1]+temp.mat[3][1]-1)*max1%MOD; ans%=MOD; } printf("%I64d\n",ans); } return 0; }