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.

4
0.1 0.2 0.3 0.4 

5.00

思路

题目要求求所有子集的和。

1.第i个数在所有子集中出现的次数为(N - i + 1) * i;
2.循环求和就行

代码

#include<iostream>
#include<vector>
#include<iomanip>
using namespace std;
int main()
{
    int N;
    while(cin >> N)
    {
        vector<double> nums(N + 1);
        double sum = 0;
        for(int i = 1;i <= N;i++)
        {
           cin >> nums[i];
           sum += nums[i] * (N - i + 1) * i;
        }
        cout << fixed << setprecision(2) << sum << endl;
    }
}