距离的总和

	//假设有n个数，每相邻两个数有一个区间，则有n-1个相邻区间。设每个区间的距离（质数个数）为di，i=0,1,n-2.  



	//在计算所有数据两两之间的距离总和时，会发现第一个距离d0被包含了n-1次（第一个数分别和之后所有n-1个数构成区间）；  



	//第二个距离d1则被包含了2*（n-2）次；  



	//第三个距离d2被包含了3*（n-3）次；  



	//...  



	//第n-1个距离d(n-2)被包含了(n-1)*(n-(n-1))=n-1次  



	//对这些距离求和，即为总距离之和ans=1*(n-1)*d0+...+i*(n-i)*d(i-1)+...+(n-1)*(n-(n-1))*d(n-2)  



	#include <stdio.h>  



	#include <vector> 



	#include <iostream> 



	#include <math.h> 



	using namesapce std; 



	int getprimenum(int a, int b) 



	{ 



	    int count=0;



	    for(int i=a+1; i<b; i++) 



	    { 



	        int j=2; 



	        for(; j<=sqrt(double(i)); j++) 



	        {



	            if(i%j == 0) 



	                break; 



	        } 



	        if(j>sqrt(double(i))) 



	            count++; 



	    } 



	    return count;



	} 



	



	int main(int argc, char** argv)



	{



	    int n;



	    cin>>n;



	    vector<int> data(n);



	    for(int i=0; i<n; i++)



	        cin>>data[i];



	    //统计每两个相邻偶数区间的质数个数（n个质数，有n-1个区间）



	    vector<int> count(n-1);



	    for(int i=1; i<n; i++)



	        count[i-1] = getprimenum(data[i-1], data[i]);



	    //计算所有数据两两之间的距离总和



	    int ans=0;



	    for(int i=1; i<n; i++)



	        ans += i*(n-i)*count[i-1];



	    cout<<ans<<endl;



	    return 0;



	} 

	#include<iostream>
#include<vector>
using namespace std;

int primeCount(vector<int> vec, int num1, int num2)
{
	int count = 0;
	for (int i = vec[num1]+1; i < vec[num2]; i++)
	{
		int j;
		bool flag = true;
		for (j = 2; j <= sqrt(i); j++)
		{
			if (i % j == 0)
			{
				flag = false;
				break;
			}
		}
		if (flag)
			count++;
	}
    return count;
}

int sumOfTwonumber(vector<int> vec, int num1, int num2)
{
	int sum = 0;
	for (int i = num1; i <=num2; i++)
		sum += vec[i];
	return sum;
}
int main()
{
	int n;
	cin >> n;
	vector<int> vec;
	vector<int> primeNumber;
	for (int i = 0; i < n; i++)
	  {
		int number;
		cin >> number;
		vec.push_back(number);
	   }
	int count = 0;
	int p1 = 0;
	int p2 = 1;
	while (p1 < n&&p2 < n)
	  { 
		count = primeCount(vec, p1, p2);
		primeNumber.push_back(count);
		p1++;
		p2++;
	  }
    int result=0;
	for (int i = 0; i < n-1; i++)
	  {
		for (int j = i ; j < n-1; j++)
		{
			int temp = sumOfTwonumber(primeNumber, i, j);
			result += temp;
		}
	  }
	cout << result << endl;
	return 0;
}

#include<iostream>
using namespace std;
//判断是质数
bool IsZhishu(const int &a)
{
 if(a<2)
  return 0;
 for(int i=2;i<a;i++)
  if(a%i==0)
   return 0;
 return 1;
}
//计算距离
int Distance(const int &a,const int &b)
{
 if((a<=2)||(b<=2))
  {
   cout<<"必须大于2"<<endl;
   return -1;
   }
 int count=0;
 if((a%2!=0)||(b%2!=0))
  {
    return count;   
  }
 for(int i=a+1;i<b;i++)
  if(IsZhishu(i))
   count++;
 return count;
}
//main test
int main()
{
 cout<<"样例输入："<<endl;
 int num;
 cin>>num;
 int *a=new int[num];
 for(int i=0;i<num;i++)
  cin>>a[i];
 int sum=0;
 for(int i=0;i<num;i++)
  for(int j=i+1;j<num;j++)
   sum+=Distance(a[i],a[j]);
 cout<<"样例输出："<<endl;
 cout<<sum<<endl;
 delete[] a;
 system("pause");
 return 0;
}

def distant(a,b): #求取两数的距离，距离定义为两数之间素数的个数，a和b为偶数且b>a
    if a>b:
        a,b=b,a
    return distantTozero(b)-distantTozero(a)

def distantTozero(x): #求一个自然数到0的距离，距离定义为0到a之间素数的个数
    count=1
    for i in range(2,x):
        for j in range(2,i+1):
            if i%j==0:
                break
            elif j==(i-1):
                count=count+1
                break
    return count

numb=int(input("请输入偶数的个数：\n"))
list_oushu=[]
print("请输入偶数序列：\n")
while True:

    a=0
    a=input()
    if a=="":
        print(a)
        break
    try:
        b=int(a)
        list_oushu.append(b)
    except:
        print("请输入数字！\n")
sum=0
for x in range(len(list_oushu)):
    for y in range(len(list_oushu)):
        a=distant(list_oushu[x],list_oushu[y])
        sum=sum+a
print(sum/2)

input("press any key to exit......")
exit()

#include <iostream>
using namespace std;
//判断是否为质数
bool JudgePrimeNumber(int num)
{
    for (int i = 2; i <= sqrt(num); i++)
        if (num % i == 0)
            return false;
    return true;
}

//计算两个数之间的距离
int DistanseSum(int start, int end)
{
    int count = 0;
    for (int i = start; i <= end; i++)
    {
        if (JudgePrimeNumber(i))
            count++;
    }
    return count;
}

int main()
{
    int N = 0;
    cin >> N;
    int* numbersArray = new int[N];

    //虽然没有在线判定，还是要注意一下输入数据的正确性。
    for (int i = 0; i < N; i++)
    {
        cin >> numbersArray[i];
        if (numbersArray[i] < 2)
            return -1;
        if (i > 0 && numbersArray[i] <= numbersArray[i - 1])
            return -1;
    }

    int sum = 0;
    //遍历任意两点距离
    for (int i = 0; i < N - 1; i++)
        for (int j = i + 1; j < N; j++)
        {
            sum += DistanseSum(numbersArray[i], numbersArray[j]);
        }
    
    delete[] numbersArray;
    //cout << sum << endl; //system("pause");
    return sum;
}

 //system("pause");
 return 0;
}

	/*我这里iostream一直有红色下划线提示，难道是编译环境不认识吗？ 



	大家帮忙看看我的思路对不对。 



	首先对说有偶数排序，然后依次求每个数和所有 后面数的距离，求和



	*/ 



	#include <iostream>



	#include <vector>



	using namespace std;



	 



	int main(){



	    int n;



	    cin >> n;



	    int p;



	    vector<int> o;



	    for(int i = 0; i < n; i++){



	        cin >> p;



	        o.push_back(p);



	    }



	    sort(o.begin(),o.end());



	    int sum = 0;



	    int g = 0;

for (int i = 0; i < o.size()-1; i++){
		g = 1;
		if (g <= o.size() - 1){
			for (int j = o[i]; j < o[i + g]; j++){
				g++;
				for (int q = 2; q <= sqrt(j); q++){
					if (j % q != 0)
						sum++;
				}
			}
		}
	}
	cout << sum << endl;

	return 0;



	}

/*
思路
先将数据升序排序
然后计算每个数据到0的距离存于数组d中
先计算最小的距离d[0]与其他距离的距离差，再计算次小的距离d[1]与其他距离的距离差...以此类推，最后 总距离需要减去（n-1)*d[0]、（n-2)*d[1]、....、1*d[n-2]  (即(n-1-i)*d[i])
总距离需要加上1*d[1]、2*d[2]、....、(n-1)*d[n-1]  (即i*d[i])
*/


#include<iostream>
#include<algorithm>
#include<vector>
#include<string.h>
#include<math.h>
using namespace std;
void Qsort(int a[], int low, int high)//快排
{
	if (low >= high)
		return;
	int first = low;
	int last = high;
	int key = a[first];
	while (first < last)
	{
		while (first < last&&key <= a[last])
			last--;
		a[first] = a[last];
		while (first < last&&key >= a[first])
			first++;
		a[last] = a[first];
	}
	a[first] = key;
	Qsort(a, low, first - 1);
	Qsort(a, first + 1, high);
}
bool func1(int a)//判断素数
{
	if (a == 2 || a == 3)
		return true;
	for (int i = 2; i*i <= a; i++)
		if (a%i == 0)
			return false;
	return true;
}
int func2(int n)//返回小于等于n的素数个数
{
	int count = 0;
	count += 2;
	for (int i = 4; i <= n; i++)
	{
		if (func1(i))
			count++;
	}
	return count;
}
int main(){
	long long n;
	long long res;
	long long i, j,k;
	int x,y,z;
	while (cin >>n)
	{
		res = 0;
		int *a = new int[n];
		int *d = new int[n];
		int *DP1 = new int[n];//前i个a元素中 包含a[i]的连续最长递增子序列
		int *DP2= new int[n];//后N-i个a元素中 包含a[i]的连续最长递增子序列
		for (i = 0; i < n; i++)
			cin >> a[i];
		Qsort(a, 0, n - 1);
		for (i = 0; i < n; i++)
			d[i] = func2(a[i]);
		for (i = 1; i < n; i++)
		{
			res += i*d[i];
		}
		for (i = 0; i < n-1; i++)
		{
			res -= (n-1-i)*d[i];
		}
		cout <<res<< endl;
	}
	return 0;
}

int primOfCount(int num1,int num2)
{
	int count = 0;
	for (num1 = num1+1;num1 < num2;num1+=2)
	{
		bool flag = true;  
		for(int i = 2;i <= sqrt(num1);i++)//
		{
			if (num1%i == 0)
			{
				flag = false;
				break;
			}
		}
		if (flag)
		{
			//cout<<"num1 "<<num1<<endl;
			count++;
		}

	}
	return count;
}

int main()
{
	int n,a;
	int cnt = 0;
	cin>>n;
	vector<int>vec;
	for (int i = 0; i < n; i++)
	{
		cin>>a;
		vec.push_back(a);
	}
	for (int i = 0; i < n - 1; i++)
	{
		for (int j = i+1; j < n ; j++)
		{
	       cnt += primOfCount(vec[i],vec[j]);
		}
	}
	cout<<cnt<<endl;
	return 0;
}

//判断出规律，每个数据段的素数计算了多少次
class Solution {
public:
	int distance(int len,int arr[])
	{
		int result = 0;
		if (len < 2)
			return 0;
		for (int i = 0; i < len-1; i++)
		{
			result += (i+1)*(len - i - 1)*primenum(arr[i], arr[i + 1]);
		}
		return result;
	}

	//给出这个数据段的素数个数
	int primenum(int left, int right)
	{
		int cnt = 0;
		for (int i = left + 1; i <= right; i += 2)
		{
			if (isprime(i))
				cnt++;
		}
		return cnt;
	}

	//判断数字是否为素数
	bool isprime(int num)
	{
		bool result = true;
		for (int i = 2; i < num / 2+1;i++)
		if (num%i == 0)
		{
			result = false;
			break;
		}
		return result;
	}

	void test()
	{
		int n;
		cin >> n;
		int *arr = new int[n];
		for (int i = 0; i < n; i++)
			cin >> arr[i];
		cout << distance(n, arr);
	}
};

	
	

		
		

			
			

				
				

					
					

						
						

							
							

								#include<iostream>
							

							

								#include<cstring>
							

							

								usingnamespacestd;
							

							

								 
							

							

								intfun(intm, intn)//两个数之间的质数个数
							

							

								{
							

							

								    intnum = 0;
							

							

								     
							

							

								    for(inti = m; i <= n; i++)
							

							

								    {
							

							

								        inttemp = 0;
							

							

								        for(intj = 2; j <= sqrt(i); j++)
							

							

								        {
							

							

								            if(i%j == 0) temp++;
							

							

								        }
							

							

								 
							

							

								        if(temp == 0) num++;
							

							

								    }
							

							

								    returnnum;
							

							

								}
							

							

								 
							

							

								intmain()
							

							

								{
							

							

								    intfun(intm, intn);
							

							

								    intgeshu;
							

							

								    cin >> geshu;
							

							

								    intt = 0;
							

							

								    intnumber;
							

							

								    intshuzu[100];
							

							

								    for(intt = 0; t < geshu; t++)
							

							

								    {
							

							

								        cin >> number;
							

							

								        shuzu[t] = number;
							

							

								         
							

							

								    }
							

							

								 
							

							

								    //计算两两数之间的质数，并求和。
							

							

								    ints=0;
							

							

								    intsum = 0;
							

							

								    for(inti = 0; i < geshu; i++)
							

							

								    {
							

							

								        for(intj = i + 1; j < geshu; j++)
							

							

								        {
							

							

								             
							

							

								                s = fun(shuzu[i], shuzu[j]);
							

							

								                sum = sum + s;
							

							

								             
							

							

								        }
							

							

								    }
							

							

								     
							

							

								     
							

							

								    cout << sum << endl;
							

							

								}
							

							

								
							

						

					

				

			

		

	

}