1104 Sum of Number Segments （20 分）

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本文介绍了一道编程题目，要求计算给定正数数列中所有连续子序列（段）的元素之和。通过输入数列大小及元素，利用规律计算并输出所有段的总和，精确到小数点后两位。

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1104 Sum of Number Segments （20 分）

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) and (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

水题，找下规律就可以

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
using namespace std;
int main(){
	int n;
	cin >> n;
	double a[100002];
	for(int i = 0;i < n;i++){
		cin >> a[i];
	}
	double sum = 0;
	for(int i = 0;i < n;i++){
		sum += a[i]*(n-i)*(i+1);
	}
	printf("%.2lf",sum);
	return 0;
}