题目描述：
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 0.4Sample Output:
5.00
题目思路：
刚看到题目首先想到dfs枚举子集求和，但是这些数字片段并不是所有的子集。然后想到各个数字的个数是否有一定的规律，
如图，我们发现1和4的个数，2和3的个数是一样的，而他们的位置对称，所以这个思路应该没错，给定一个数，这个数左边也可以放数字，右边也可以放数字，那么该数字出现的次数就应该是 左边的情况 * 右边的情况。
题目代码：
#include <cstdio>
using namespace std;double ans = 0;
double a[100000];
int n;int main(){scanf("%d",&n);for(int i = 0; i < n; i++){scanf("%lf",&a[i]);}for(int i = 0; i < n; i++){ans += a[i] * (i+1) * (n-i);}printf("%.2f\n",ans);return 0;
}