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

註意點：第一次提交的時候sum = sum+i*（n-i+1）*a； 有兩個測試點不能通過； 修改為sum = sum+1.0*(n-i+1)*i,或者sum=sum+a*i*(n-i+1)就是正確的，並且1.0必須前程才是正確的，。原因不是很明白，以後為了保險起見小數和整數相乘的時候，均把小數寫在前

 1 #include<iostream>
 2 using namespace std;
 3 int main(){
 4   int n, i;
 5   double a, sum=0
 
;
 6   cin>>n;
 7   for(i=1; i<=n; i++){
 8     cin>>a;
 9     sum = sum + a*i*(n-i+1);
10   }
11   printf("%.2f", sum);
12   return 0;
13 }

1104 Sum of Number Segments (20)