PAT 1104 Sum of Number Segments python解法

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

题意：给出一个序列，求出它的所有段总和（段：连续的子序列）。比如，给定序列{0.1,0.2,0.3,0.4}，求出10个段：(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) 和 (0.4).然后求和： 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0，注意要保留2位小数，最终结果是5.00

解题思路：通过找规律，可以发现，0.1出现了4次，0.2出现了6次，0.3出现了6次，0.4出现了4次，结果就是0.1*4+0.2*6+0.3*6+0.4*4=5.00。如果扩展到n个数的序列，可以发现，第i个数出现了i*（n+1-i）次（如：0.1出现了1*4次，0.2出现了2*3次，0.3出现了3*2次，0.4出现了4*1次），因为索引从0开始，所以代码中是（n-i）*（i+1）次，接下来只需要求和就行了。

n = int(input())
l = list(map(float,input().split()))
s = 0
for i in range(len(l)):
        s += l[i]*(n-i)*(i+1)
print('%.2f'%s)