solution Of 1104. Sum of Number Segments (20)

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本文介绍了一种计算给定正数序列所有子序列元素总和的算法，通过计算每个元素在其所在子序列中的出现次数来得出结果，并提供了一个具体的实现案例。

1104. Sum of Number Segments (20)

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

结题思路 :
题意要求我们计算一个确定数列的所有子序列的元素总和。
要求1：子序列非空；
要点2：最后总和保留两位小数。

程序步骤：
第一步、对每个输入的序列元素，计算其在左右子序列中出现的个数；
第二步、累加全部数据以后，输出最终结果。

(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).

统计数据：
index value times
1 0.1 4
2 0.2 6
3 0.3 6
4 0.4 4

可以简单得到

times=index（num-index+1）

具体程序(AC)如下：

#include <iostream>
using namespace std;
int main()
{
    int num;
    cin>>num;
    double sum=0;
    int pre=1;
    double value;
    for(int i=0;i<num;++i)
    {
        cin>>value;
        sum+=value*pre*(num-i);
        ++pre;
    }
    printf("%.2lf\n",sum);
}