#include <iostream>

 #include <cstdlib>

 using namespace std;

 //二分查找

 bool binarySearch(long a[], long x, int n){

     int left = ,right = n-;

     int middle;

     while (left <= right){

         middle = (left+right)/;

         if (x == a[middle]) return ;

         if (x > a[middle]) left = middle + ;

         else right = middle - ;

     }

     return ;

 }

 int cmp(const void *a, const void *b)  {

     return *(int *)a - *(int *)b;

 }  

 int main(){

     int numprofessor;

     scanf("%ld",&numprofessor);

     long yearProfessor[numprofessor];

     for(int i=;i<numprofessor;i++){

         scanf("%ld",&yearProfessor[i]);

     }

     qsort(yearProfessor,numprofessor,sizeof(long),cmp);

     long numstudent;

     scanf("%ld",&numstudent);

     long result=;

     long yearstudent[numstudent];

     for(int i=;i<numstudent;i++){

         scanf("%ld",&yearstudent[i]);

         if( binarySearch(yearProfessor, yearstudent[i], numprofessor)){

             result++;

         }

     }

     cout<<result;

 } 

 #include <iostream>

 using namespace std;

 int i,j,fi,fj,n;

 int count=,count1=,count2=;

 int finext;           // 表示fi下一个f(i+1)的值 

 int fib1(int j){      /*递归程序1：计算fj时用到的fi和finext的次数  */

       if(j==i){

             count1++;   //记录fi用到的次数

           return ;

       }

       if(j==i+){

             count2++;   //记录finext用到的次数

             return ;

       }

       return  fib1(j-) + fib1(j-);

 }

 long fib3 (int n){

     //* 迭代解法：这里算法复杂度显然为O(n) ，

    // 这个解法目前来看是最好的解法，算法既不复杂，而且在时间复杂度和空间复杂度上都有保证

     if(n==i){

         return fi;

     }

     if(n==i+){

         return finext;

     }

     long long x = fi, y = finext;

     for (int a= i; a<n-; a++){

          y = x + y;

          x = y - x;

     }

     return y;

 }

 int main(){

     cin>>i>>fi>>j>>fj>>n;

     if(j<i){

         int temp=j,temp1=fj;

         j=i;

         fj=fi;

         i=temp;

         fi=temp1;

     }

     int a=fib1(j);

     finext=(fj-count1*fi)/count2;

     cout<<fib3(n)<<endl;

 }

 #include <iostream>

 using namespace std;

 const long long NLAR = ;

 long long i,fi,j,fj,n;

 int main(){

     cin >> i >> fi >> j >> fj >> n;

     if(j<i){     //保证j>i

         int temp=j,temp1=fj;

         j=i;

         fj=fi;

         i=temp;

         fi=temp1;

     }

     // cal f[i+1]

     long long left=-NLAR, right=NLAR, mid;

     long long ti=fi, tj=mid, t;

     while (left+<right){

         mid = (left+right)/;

         ti=fi; tj=mid;

         for (int k=i+; k<=j; ++k){

             t=ti+tj;

             ti=tj;

              tj=t;

             if(t>NLAR* ||t<-NLAR*) break;

         }

         if(tj>=fj) right = mid;

             else left=mid;

     }

     ti=fi; tj=right;

     if(n>i){

         for(int k=i+;k<=n;++k){

             t=ti+tj;

             ti=tj;

             tj=t;

         }

         cout << tj << endl;

     }

     else {

         for(int k=i-; k>=n; ++k){

             t=tj-ti;

             tj=ti;

             ti=t;

         }

         cout << ti << endl;

     }

     return ;

 }

 #include <iostream>

 #include <cstdlib>

 #include <iomanip>

 #include <cmath>

 using namespace std;

 #define EPS 1e-9

 int h,H,L;

 double fx(double x){

     return H/2.0*x/sqrt(h*h+x*x)-x;

 }

 int main(){

     cin>>h>>H>>L;

     double left=,right=H;

     long double isover=;

     while(isover>EPS){

         double m1=left+(right-left)/;

         double m2=right-(right-left)/;

         double fm1=fx(m1);

         double fm2=fx(m2);

         if(fm1<fm2) left=m1;

         else  right=m2;

         isover=right-left;

     }

     cout<<fixed<<setprecision()<<fx(right)<<endl;

 } 

 /*

  * poj-1064 Cable master.cpp

  *      二分， 化为整数存储，下界为1(cm)，上界为最长绳的长度

  *

  */

 #include<cstdio>

 using namespace std;

 const double eps = 1e-;    //注意精度

 const int maxn =  + ;

 int n, k, cable[maxn];

 int main(){

     scanf("%d%d", &n, &k);

     double ftmp;

     int tmpMax = ;

     for(int i=; i<n; i++){

         scanf("%lf", &ftmp);

         cable[i] = int((ftmp+eps) * );    //精度

         if(cable[i] > tmpMax) tmpMax = cable[i];

     }

     int up = tmpMax, low = , mid = -, ans = -;

     int tmpNum = ;

     while(low <= up){

         mid = (up + low) / ;

         tmpNum = ;

         for(int i=; i<n; i++)

             tmpNum += cable[i] / mid;

         if(tmpNum >= k){

             if(mid > ans) ans = mid;

             low = mid + ;

         }

         else

             up = mid - ;

     }

     if(ans < )

         printf("0.00\n");

     else

         printf("%.2lf\n", ans * 0.01);

     return ;

 }

poj1064