C++中想输入两个数据，第二个始终参与无法运算

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
using std::cin;         using std::endl;
using std::cout;        using std::vector;
using std::sort;        using std::iterator;

int main(void)
{   cout << "请输入一个数集：" << endl;

 
    float x;
    vector<float>  intsets;
    while (cin>> x)
   
    {
        intsets.push_back(x);
    }

 
    // 获取整数集中整数的个数
    typedef vector<float>::size_type vec_sz;
    vec_sz size = intsets.size();

 
    // 判断整数集是否为空
    if (size <= 3)
    {
        cout << "你输入的数集小于3，数集需"
                "大于3，请重新输入！！" << endl;
        return 1;
    }
    // 整数集从小到大排序
    sort(intsets.begin(), intsets.end());

 
    // 输出排序后的结果
    cout << "排序后的数集是：" << endl;

 
    vector<float>::iterator iter;
    for (iter = intsets.begin(); iter != intsets.end(); iter++)
    {
       cout << *iter << ' ';
    }
    cout << endl;

 
    // 中位数和相应的索引
    vec_sz mid, mid1, mid3;
    double median, median1, median3;

 
    mid = size / 2;
    // 求得中值
    median = size % 2 == 0 ? (intsets[mid] + intsets[mid-1]) / 2.0: intsets[mid];

 
    //mid1 = size % 2 == 0 ? (mid-1)/2 : mid/2;
    mid1 = mid / 2;
    mid3 = size%2 == 0  ? (mid + mid1) : (mid+mid1+1);

 
    median1 = mid % 2 == 0 ? (intsets[mid1] + intsets[mid1-1]) / 2.0: intsets[mid1];
    median3 = mid % 2 == 0 ? (intsets[mid3] + intsets[mid3-1]) / 2.0: intsets[mid3];

 
    cout << "中分位数为：" << endl;
    cout << "median1 = " << median1 << endl;
    cout << "median2 = " << median << endl;
    cout << "median3 = " << median3 << endl;
    float a=median3+1.5*(median3-median1);
    cout << "a = " << a << endl;
    float b=median3-1.5*(median3-median1);
    cout << "b = " << b  << endl;
   cout << "请输入一个数n：" << endl;
   int n;
   int s=0;
   for(int i=0;i<n;i++)
   cout<<"s= "<<intsets[i];
   {if(intsets[n-1]>a)
   s++;}

 cout<<endl<<"这个数组中大于指定值的元素的个数是"<<s<<endl;

 int z=0;
   for(int i=0;i<n;i++)
   cout<<"z= "<<intsets[i];
   {if(intsets[n-1]<b)
   z++;}

 cout<<endl<<"这个数组中小于指定值的元素的个数是"<<z<<endl;

    return 0;
}