1104. Sum of Number Segments

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本文介绍了一种快速计算给定正数序列中所有连续子序列（段）数值之和的方法，并通过具体示例展示了算法的实现过程。该算法利用了每个数在不同子序列中出现次数的规律。

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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

找到每个数出现次数的规律，第i个数出现的次数为i*(n+1-i)，再乘上值求和即可。

#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
	double a;
	int n;
	cin>>n;
	double s=0;
	for(int i=1;i<=n;i++)
	{
		cin>>a;
		s=s+a*(n-i+1)*i;
	}
	printf("%.2lf",s);
	return 0;
}