1104. Sum of Number Segments (20)

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本文介绍了一种针对正数序列的高效算法，该算法能够计算序列所有可能子段的数值总和。通过实例演示了如何利用每个元素在不同长度子段中出现的规律来简化计算过程。

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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

规律是第i个值出现次数为(i+1)*(n-i)，根据这个规律遍历一遍数组累加即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <vector>
using namespace std;


int main()
{
	int n;
	scanf("%d",&n);
	vector<double>num(n);
	double sum=0;
	for(int i=0;i<n;i++)
	{
		cin>>num[i];
	}
	
	for(int i=0;i<n;i++)
	{
		sum+=(num[i]*(n-i)*(i+1));
	}
	
	printf("%.2f",sum);
}