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。

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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 ;
 }