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

直接计算每个数出现了几次

公式：i*(n-i+1)

#include <iostream>

#include <string.h>

#include <algorithm>

#include <math.h>

#include <stdio.h>

#include <stdlib.h>

using namespace std;

int n;

double x;

int main()

{

    double ans;

    scanf("%d",&n);

    for(int i=1;i<=n;i++)

    {

        scanf("%lf",&x);

        ans+=x*i*(n-i+1);

    }

    printf("%.2f\n",ans);

    return 0;

}