昨天的bc被坑惨了= =
本来能涨rating的大好机会又浪费了。。。大号已弃号
A：第一反应是高精度，结果模板找不到了= =，然后现学现卖拍了个java的BigInteger+快速幂，调了好半天不说还TLE。貌似这题就在卡java
实际上尼玛等号两边取log不就完了么。。。
卒
B：A题调了半天，开始做B的时候已经没多少时间了。。。
找出了斐波那契数列+前缀和的规律，结果把用矩阵快速幂求斐波那契前n项和的那个梗又忘了
最后out of submit time
卒
事实证明还是要多做成套的题，这样才能发现很多平时难以察觉到的问题。
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补上AC Code：
A:高你妹妹的高精度
#include <iostream>
#include <cstdio>
#include <cmath>
#define eps 1e-8
using namespace std; int fcmp(double a,double b)
{
double tm=fabs(a-b);
if (tm<eps) return ; //a==b
else
return a<b?-:;
} int a,b,c,d; int main()
{
while(~scanf("%d%d%d%d",&a,&b,&c,&d))
{
double aa=log(a),cc=log(c);
aa=aa*b; cc=cc*d;
switch (fcmp(aa,cc))
{
case :printf("=\n");
break;
case -:printf("<\n");
break;
case :printf(">\n");
break;
}
} return ;
}
B:事实证明即使只有三组数据也可能出现cin TLE而scanf AC的情况= =
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
#define ULL long long
#define MOD 10000007 ULL ans,mx1,mx2;
ULL n,k,x;
typedef vector<ULL> vec;
typedef vector<vec> mat;
int a[]; mat mul(mat &A,mat &B) //return A*B
{
mat C(A.size(),vec(B[].size()));
for (int i=;i<(int)A.size();i++)
{
for (int k=;k<(int)B.size();k++)
{
for (int j=;j<(int)B[].size();j++)
{
C[i][j]=(C[i][j]+A[i][k]*B[k][j])%MOD;
}
}
}
return C;
} mat m_pow(mat A,int m) //return A^m
{
mat B(A.size(),vec(A.size()));
for (int i=;i<(int)A.size();i++)
B[i][i]=;
while (m>)
{
if (m&) B=mul(B,A);
A=mul(A,A);
m>>=;
}
return B;
} int main()
{ //while (cin>>n>>k)
while (~scanf("%d%d",&n,&k))
{
mat AA(,vec());
mat A(,vec());
mat B(,vec());
mat C(,vec());
mat D(,vec());
mat T(,vec());
T[][]=; T[][]=; T[][]=;
A[][]=; A[][]=; A[][]=;
A[][]=; A[][]=; A[][]=;
A[][]=; A[][]=; A[][]=;
AA=A;
A=m_pow(A,k-);
C=mul(A,T);
B=mul(A,AA);
//B=m_pow(B,k-1);
D=mul(B,T);
ULL FK=C[][],FKK=D[][]-;
if (k==) {FK=; FKK=;}
if (k==) {FK=; FKK=;}
//cout<<FK<<" -- "<<FKK<<endl; /*
LL fk=1,fkk=2,FK=2,FKK=4;
for (int i=3;i<=k;i++)
{
//fk:f[k] fkk:f[k+1]
LL tmp=fk+fkk;
fk=fkk; fkk=tmp;
fk=fk%MOD;
fkk=fkk%MOD;
FK+=fk;
FKK=FK+fkk;
FK=FK%MOD;
FKK=FKK%MOD;
}
FKK-=1; if (FKK<0) FKK+=MOD;
*/ ans=;
mx1=; mx2=;
for (int i=;i<=n;i++)
{
cin>>x;
ans+=x;
if (x>mx1)
{
mx2=mx1;
mx1=x;
}
else if ((x<=mx1)&&(x>mx2))
{
mx2=x;
}
}
/*
for (int i=1;i<=n;i++)
{
cin>>a[i];
ans+=a[i];
}
sort(a+1,a+n+1);
mx1=a[n]; mx2=a[n-1];
*/ //cout<<mx1<<" "<<mx2<<" = "<<ans<<endl;
/*
for (int i=1;i<=k;i++)
{
ans=ans+mx1+mx2;
ans=ans%MOD;
LL tmp=mx1;
mx1=mx1+mx2;
mx2=tmp;
}
*/
//cout<<ans<<" - "<<FK<<" "<<FKK<<endl;
mx1=mx1*FKK; mx1=mx1%MOD;
mx2=mx2*FK; mx2=mx2%MOD;
ans=ans+mx1+mx2;
ans=ans%MOD;
//cout<<ans<<endl;
printf("%I64d\n",ans);
}
return ;
}