PAT 1104. Sum of Number Segments (20)

最新推荐文章于 2022-03-06 20:59:54 发布

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最新推荐文章于 2022-03-06 20:59:54 发布

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本文介绍了一种针对正数序列所有可能子序列之和的计算方法，并提供了一个C++实现示例。该算法通过遍历输入序列，计算每个元素在不同长度子序列中出现的次数来得出最终结果。

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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

算法分析：

十分简单也比较容易想到的解法。

不过，第三个和第四个case一直无法通过，网上找了下源代码发现本质都一样。仔细看是在进行求和过程中sum += a*(n-i)*(i+1); //如果写的sum += (n-i)*(i+1)*a，则会后面两个case无法通过

#include<iostream>  
using namespace std;  
  
int main(){  
    int n;  
    double a,sum = 0;  
    cin>>n;  
    for(int i = 0;i<n;i++){  
        cin>>a;  
		sum += a*(n-i)*(i+1);  //如果写的sum += (n-i)*(i+1)*a，则会后面两个case无法通过
    }  
    printf("%.2lf\n",sum);  
    return 0;  
}