1104. Sum of Number Segments (20) consecutive subsequence 每个数出现的次数

最新推荐文章于 2021-09-29 16:08:04 发布

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最新推荐文章于 2021-09-29 16:08:04 发布

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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序列的元素之和。通过输入一个正数序列，算法将输出所有子序列元素总和，并给出具体实现代码示例。

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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

#include <iostream>
using namespace std;
/*
 sum += t * i * (n - i + 1);
 */

int main(int argc, const char * argv[]) {
    int n;
    double sum = 0.0, t;
    cin >> n;

    for (int i = 1; i <= n; i++) {
        cin >> t;
        sum += t * i * (n - i + 1);
    }
    printf("%.2lf\n", sum);
    
    return 0;
}