1104. Sum of Number Segments (20)【数学题】——PAT (Advanced Level) Practise

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本文介绍了一种快速计算给定正数数列所有连续子序列（段）元素之和的方法，并提供了一个C++实现示例。对于输入的数列，算法能够高效地计算出所有可能子序列的总和。

题目信息

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
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 10^5. 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

解题思路

直接看代码吧，，不想多说，

AC代码

#include <cstdio>

int main()
{
    int n;
    long long st = 0;
    double s = 0, v;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i){
        scanf("%lf", &v);
        st += n - i - i;
        s += v * st;
    }
    printf("%.2f\n", s);
    return 0;
}