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 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

思路

• 包含第 i 个数的片段，其首指针有 i 中选择：1, 2, 3, … i，其尾指针有 n - i + 1 中选择：i, i + 1, … n。所以包含第 i 个数的片段有 i * (n - i + 1) 种。将 i 从 1 到 n 遍历，将包含每个数的片段数相加即可。

#include <cstdio>

const int maxn = 100000;

int main() {
    int n;
    scanf("%d", &n);
    double temp, ans = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%lf", &temp);
        ans += temp * i * (n - i + 1);
    }
    printf("%.2f", ans);
    return 0;
}