bc#29 做题笔记

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。

-------------------------------------------------

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 0;   //a==b
    else
        return a<b?-1:1;
}

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  0:printf("=
");
                    break;
            case -1:printf("<
");
                    break;
            case  1:printf(">
");
                    break;
        }
    }

    return 0;
}

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[100010];

mat mul(mat &A,mat &B)      //return A*B
{
    mat C(A.size(),vec(B[0].size()));
    for (int i=0;i<(int)A.size();i++)
    {
        for (int k=0;k<(int)B.size();k++)
        {
            for (int j=0;j<(int)B[0].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=0;i<(int)A.size();i++)
        B[i][i]=1;
    while (m>0)
    {
        if (m&1)    B=mul(B,A);
        A=mul(A,A);
        m>>=1;
    }
    return B;
}

int main()
{

    //while (cin>>n>>k)
    while (~scanf("%d%d",&n,&k))
    {
        mat AA(3,vec(3));
        mat A(3,vec(3));
        mat B(3,vec(3));
        mat C(3,vec(1));
        mat D(3,vec(1));
        mat T(3,vec(1));
        T[0][0]=2;  T[1][0]=1;  T[2][0]=1;
        A[0][0]=1;  A[0][1]=1;  A[0][2]=1;
        A[1][0]=0;  A[1][1]=1;  A[1][2]=1;
        A[2][0]=0;  A[2][1]=1;  A[2][2]=0;
        AA=A;
        A=m_pow(A,k-2);
        C=mul(A,T);
        B=mul(A,AA);
        //B=m_pow(B,k-1);
        D=mul(B,T);
        ULL FK=C[0][0],FKK=D[0][0]-1;
        if (k==2)   {FK=2;  FKK=3;}
        if (k==1)   {FK=1;  FKK=1;}
        //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=0;
        mx1=0;  mx2=0;
        for (int i=1;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
",ans);
    }
    return 0;
}