PAT A1104 Sum of Number Segments (20point(s))

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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) 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 $10^5$. 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

题意

给出一个序列,计算所有连续子集的和。

解答

本题是数学问题,只要计算每一项的出现次数就可以了,给定一个序列{a1, a2, a3,..., an}和ai(1<=i<=n),则包含ai的子集个数为i*(n-i+1),因为子集的第一个位置可以是1到i,有i个位置,最右位置可以是i到n,有n-i+1个位置。

#include<iostream>
#include<vector>
using namespace std;
int main()
{
	int n;
	cin >> n;
	double sum = 0.0, k;
	for (int i = 1; i <= n; i++) {
		cin >> k;
		sum += k * i * (n - i + 1);
	}
	printf("%.2f", sum);
	return 0;
}