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
  
 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 int main(){
 4     long long n, count;
 5     double res=0, tmp;
 6     scanf("%lld", &n);
 7     for (int i = 0; i<n; i++){
 8         count = (n - i)*(i + 1);
 9         scanf("%lf", &tmp);
10         res += tmp*count;
11     }
12     printf("%.2lf", res);
13 
14     system("pause");
15 }
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注意点：找到规律以后还要注意用long long，题目n不超过10e5，两个10e5乘起来就超过int范围了