1104. Sum of Number Segments (20)

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本文介绍了一种针对正数序列求所有连续子序列之和的算法实现，并提供了完整的C++代码示例。该算法利用了序列中元素位置与求和的关系，高效地解决了问题。

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1104. Sum of Number Segments (20)
时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CAO, Peng

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

注意:

数据类型一定要用double，用long double跑不过，我也不知道为什么，哪位大佬知道原因告知一下，十分感谢

#define _CRT_SECURE_NO_WARNINGS
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <map>

using namespace std;

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    int n;
    double sum = 0, input;
    cin >> n;
    for (int i = 0; i < n; ++i)
    {
        cin >> input;
        sum += input * (i + 1) * (n - i);
    }

    cout << setiosflags(ios::fixed) << setprecision(2) << sum;

    return 0;
}