Given a sequence of positive numbers, a segment is defined to be a consecutive subsequence. For example, given the sequence {0.1, 0.2, 0.3, 0.4}, we have 10 segments: (0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) (0.4).

Now given a sequence, you are supposed to find the sum of all the numbers in all the segments. For the previous example, the sum of all the 10 segments is 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N, the size of the sequence which is no more than 10⁵. The next line contains N positive numbers in the sequence, each no more than 1.0, separated by a space.

Output Specification:

For each test case, print in one line the sum of all the numbers in all the segments, accurate up to 2 decimal places.

Sample Input:

4 0.1 0.2 0.3 0.4

Sample Output:

5.00

num 输入队列有多少个数

依次输入num个in，都是不大于1.0的小数，

要求输出这个队列非空子串的和，保留两位小数，有没有换行符结束都可以通过。

注意：如果 sum += (num - i + 1)*i*in;就有两个错误，详见后面。double型in要在前面，可能是数据大溢出了吧

串的各部分术语http://xujiayu317.blog.163.com/blog/static/25475209201601895222262/

评测结果

测试点

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
  double sum=0, in;
  int num;
  cin >> num;
  for (int i = num; i > 0; i--)
  {
    cin >> in;
    sum += in*(num - i + 1)*i;
  }
  cout << setiosflags(ios::fixed) << setprecision(2)
    << sum ;
  system("pause");
  return 0;
}

评测结果

测试点

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
  double sum=0, in;
  int num;
  cin >> num;
  for (int i = num; i > 0; i--)
  {
    cin >> in;
    sum += (num - i+1)*i*in;
  }
  cout << setiosflags(ios::fixed) << setprecision(2)
    << sum ;
  system("pause");
  return 0;
}